Tng hp kin thc c bn ho hc 8Cc khi nim:1. Vt th, cht. - Vt th: L ton b nhng g xung quanh chng ta v trong khng gian. Vt th gm 2 loi: Vt th t nhin v vt th nhn to- Cht: l nguyn liu cu to nn vt th. Cht c khp mi ni, u c vt th l c cht.- Mi cht c nhng tnh cht nht nh. Bao gm tnh cht vt l v tnh cht ho hc.o Tnh cht vt l: Trng thi (R,L,K), mu sc, mi v, tnh tan, tnh dn in, dn nhit, nhit si (t0s), nhit nng chy (t0nc), khi lng ring (d)o Tnh cht ho hc: L kh nng b bin i thnh cht khc: Kh nng chy, n, tc dng vi cht khc2. Hn hp v cht tinh khit. - Hn hp l 2 hay nhiu cht trn li vi nhau. Mi cht trong hn hp c gi l 1 cht thnh phn.- Hn hp gm c 2 loi: hn hp ng nht v hn hp khng ng nht- Tnh cht ca hn hp: Hn hp c tnh cht khng n nh, thay i ph thuc vo khi lng v s lng cht thnh phn.- Cht tinh khit l cht khng c ln cht no khc. Cht tinh khit c tnh cht nht nh, khng thay i.- Khi tch ring cc cht ra khi hn hp ta thu c cc cht tinh khit. tch ring cc cht ra khi hn hp ngi ta c th s dng cc phng php vt l v ho hc: tch, chit, gn, lc, cho bay hi, chng ct, dng cc phn ng ho hc3. Nguyn t. a. nh ngha: L ht v cng nh, trung ho v in, cu to nn cc chtb. Cu to: gm 2 phn Ht nhn: to bi 2 loi ht: Proton v Ntron- Proton: Mang in tch +1, c khi lng 1 vC, k hiu: P- Ntron: Khng mang in, c khi lng 1 vC, k hiu: N V: cu to t cc lp Electron- Electron: Mang in tch -1, c khi lng khng ng k, k hiu: eTrong nguyn t, cc e chuyn ng rt nhanh v sp xp thnh tng lp t trong ra.+ Lp 1: c ti a 2e+ Lp 2,3,4 tm thi c ti a 8eKhi lng nguyn t = s P + s N + s e = s P + s N (v e c khi lng rt nh)4. Nguyn t ho hc. 1L tp hp nhng nguyn t cng loi, c cng s P trong ht nhnNhng nguyn t c cng s P nhng s N khc nhau gi l ng v ca nhau5. Ho tr. L con s biu th kh nng lin kt ca nguyn t hay nhm nguyn tQuy tc ho tr:a bx yABta c: a.x = b.y(vi a, b ln lt l ho tr ca nguyn t A v B)So snh n cht v hp chtn cht hp chtVD St, ng, oxi, nit, than chNc, mui n, ngK/N L nhng cht do 1 nguyn t ho hc cu to nnL nhng cht do 2 hay nhiu nguyn t ho hc cu to nnPhn loiGm2loi: Kimloi vphi kim.Gm 2 loi: hp cht v c v hp cht hu cPhn t(ht i din)- Gm 1 nguyn t: kim loi v phi kim rn- Gmcc nguynt cng loi: Phi kim lng v kh- Gm cc nguyn t khc loi thuc cc nguyn t ho hc khc nhauCTHH - Kim loi v phi kim rn:CTHH KHHH (A)- Phi kim lng v kh:CTHH =KHHH +ch s (Ax)CTHH = KHHH ca cc nguyn t + cc ch s tng ngAxBySo snh nguyn t v phn tnguyn t phn tnh nghaL ht v cng nh, trung hovin, cutonn cc chtL ht v cng nh, i din chochtvmangy tnh cht ca chtS bin i trong phn ng ho hc.Nguyn t c bo ton trong cc phn ng ho hc.Lin kt gia cc nguyn t trong phn t thay i lm cho phn t ny bin i thnh phn t khcKhi l-ngNguyn t khi (NTK) cho bit nng nh khc nhau gia cc nguyn t v Phn t khi (PTK) l khi l-ng ca 1 phn t tnh bng n v Cacbon2l i lng c trng cho mi nguyn tNTK l khi lng ca nguyn t tnh bng n v CacbonPTK =tng khi lng cc nguyn t c trong phn t.p dng quy tc ho tr1. Tnh ho tr ca 1 nguyn t- Gi ho tr ca nguyn t cn tm (l a)- p dng QTHT: a.x = b.y a = b.y/x- Tr li2. Lp CTHH ca hp cht.- Gi cng thc chung cn lp- p dng QTHT: a.x = b.y ''x b by a a - Tr li.*** C th dng quy tc cho lp nhanh 1 CTHH: Trong CTHH, ho tr ca nguyn t ny l ch s ca nguyn t kia.Lu : Khi cc ho tr cha ti gin th cn ti gin trc6. Phn ng ho hc. L qu trnh bin i cht ny thnh cht khc.Cht b bin i gi l cht tham gia, cht c to thnh gi l sn phmc biu din bng s : A + B C + D c l: A tc dng vi B to thnh C v DA + B C c l A kt hp vi B to thnh CA C + D c l A b phn hu thnh C v D3-Ngoi ra c th chia axit thnh axit mnh v axit yu

Axit mnh Axit trung bnh Axit yu Axit rt yu4Hp cht v cOxit (AxOy)Axit (HnB)Baz- M(OH)nMui (MxBy)Oxit axit: CO2, SO2, SO3, NO2, N2O5, SiO2, P2O5Oxit baz: Li2O, Na2O, K2O, CaO, BaO, CuO,Fe2O3Oxit trung tnh: CO, NOOxit lng tnh: ZnO, Al2O3, Cr2O3 Axit khngcoxi (Hidraxit): HCl, HBr, H2S, HFAxit c oxi (Oxaxit): HNO3, H2SO4, H3PO4 .Baztan(Kim):NaOH, KOH, Ca(OH)2, Ba(OH)2Baz khng tan: Mg(OH)2, Cu(OH)2, Fe(OH)3 Mui axit: NaHSO4, NaHCO3, Ca(HCO3)2 Mui trung ho: NaCl, KNO3, CaCO3 Phn loiHCVCHNO3H2SO4HClH3PO4H2SO3CH3COOHH2CO3H2Soxit axit baz muinh nghaL hp cht ca oxi vi 1 nguyn t khcL hp cht m phn t gm1 hay nhiu nguyntHlinkt vi gc axitL hp cht m phn tgm1nguynt kim loi lin kt vi 1 hay nhiu nhm OHL hp cht m phn t gm kim loi lin kt vi gc axit.CTHHGi nguyn t trong oxit l A ho tr n. CTHH l:- A2On nu n l- AOn/2 nu n chnGi gc axit lBc ho tr n. CTHH l: HnBGi kimloi lMc ho tr nCTHH l: M(OH)nGi kim loi l M, gc axit l BCTHH l: MxByTn giTn oxit = Tn nguyn t + oxitLu : Km theo ho tr ca kim loi khi kim loi c nhiu ho tr.Khi phi kim c nhiu ho tr th km tip u ng.- Axit khng c oxi: Axit +tnphi kim+ hidric- Axit c t oxi: Axit + tn phi kim + (r)- Axit c nhiu oxi: Axit + tn phi kim + ic (ric)Tn baz = Tn kim loi + hidroxitLu: Kmtheo ho trcakimloi khi kim loi c nhiu ho tr.Tn mui = tn kim loi + tn gc axitLu : Km theo ho tr ca kim loi khi kim loi c nhiu ho tr.TCHH 1. Tc dng vi nc- Oxit axit tc dng vi nc to thnh dd Axit- Oxit baztcdng vi nc to thnh dd Baz2. Oxax + dd Baz to thnh mui v nc3.Oxbz + dd Axit to thnh mui v nc4. Oxax +Oxbz to thnh mui1. Lm qu tm hng2. Tc dng vi Baz Mui v nc3. Tc dng vi oxit baz mui v nc4. Tc dng vi kim loi mui v Hidro5. Tc dng vi mui mui mi v axit mi1. Tc dng vi axit mui v nc2. dd Kim lm i mu cht ch th- Lmqu tmxanh- Lm dd phenolphtalein khng mu hng3. dd Kim tc dng vi oxax mui v n-c1. Tc dng vi axit mui mi + axit mi2. dd mui + dd Kimmui mi + baz mi3. dd mui + Kim loi Mui mi + kim loi mi4. dd mui + dd mui 2 mui mi5. Mt s mui b nhit phn54. dd Kim + dd mui Mui + Baz5. Baz khng tan b nhit phnoxit + ncLu - Oxit lng tnh c th tc dng vi c dd axit v dd - HNO3, H2SO4 c c cc tnh cht ring- Bazlngtnhc thtc dngvi c dd axit v - Mui axit c th phn ng nh 1 axit6Tnh cht ho hc ca cc hp cht v c+ dd Mui+ axit+ dd baz+ kim loit0+ dd muit0+ axit + Oxax+ Oxit Baz+ Baz+ dd Mui+ KL + Nc + NcOxit axitOxit bazMui + ncaxit KimMui+ dd Axit + dd BazAxitMui + H2OQu tm Mui + h2Mui + AxitMuiBazKim k.tanQu tm xanhPhenolphalein k.mu hngMui + h2Ooxit + h2OMui + axitMui + bazMui + muiMui + kim loi Ccsn phm khc nhauTchh ca oxit Tchh ca AxitTchh ca mui Tchh ca bazLu : Thng ch gp 5 oxit baz tan c trong nc l Li2O, Na2O, K2O, CaO, BaO. y cng l cc oxit baz c th tc dng vi oxit axit.i vi baz, ccctnhchtchungchoc2loi nhngc nhng tnh cht ch ca Kim hoc baz khng tanMt s loi hp cht c cc tnh cht ho hc ring, trong ny khng cp ti, c th xem phn c thm hoc cc bi gii thiu ring trong sgk.Mui + baz7Mi quan h gia cc loi hp cht v cCc phng trnh ho hc minh ho thng gp4Al + 3O2 2Al2O3CuO + H2 0t Cu + H2OFe2O3 + 3CO 0t 2Fe + 3CO2S + O2 SO2CaO + H2O Ca(OH)2Cu(OH)2 0t CuO + H2OCaO + 2HCl CaCl2 + H2OCaO + CO2 CaCO3Na2CO3 + Ca(OH)2 CaCO3 + 2NaOHNaOH + HCl NaCl + H2O2NaOH + CO2 Na2CO3 + H2OBaCl2 + Na2SO4 BaSO4 + 2NaClSO3 + H2O H2SO4P2O5 + 3H2O 2H3PO4P2O5 + 6NaOH 2Na3PO4 + 3H2ON2O5 + Na2O 2NaNO3BaCl2 + H2SO4 BaSO4 + 2HCl2HCl + Fe FeCl2 + H22HCl + Ba(OH)2 BaCl2 + 2H2O6HCl + Fe2O3 2FeCl3 + 3H2OPhn hu+ H2O+ dd Kim+ Oxbaz+ Baz+ Axit+ Kim loi + dd Kim+ Axit+ Oxaxit+ dd Muit0+ H2O+ Axit+ Oxi + H2, CO + OxiMui + h2OOxit axit Oxit bazBazKim k.tan+ OxaxitKim loiPhi kim + Oxbaz+ dd MuiAxitMnh yuLu :- Mt s oxit kim loi nh Al2O3, MgO, BaO, CaO, Na2O, K2O khng b H2, CO kh.- Ccoxitkimloi khi trng thi hotrcaoloxit axit nh: CrO3, Mn2O7,- Cc phn ng ho hc xy ra phi tun theo cc iu kin ca tng phn ng.- Khi oxitaxittcdngvi dd Kim th tu theo t l s mol s toramui axit hay mui trung ho.VD: NaOH + CO2 NaHCO32NaOH + CO2 Na2CO3 + H2O- Khi tc dng vi H2SO4 c, kimloi s thhinhotr caonht, khnggii phng HidroVD:82HCl + CaCO3 CaCl2 + 2H2Oiu ch cc hp cht v c `192021131415161718126789101112354Kim loi + oxiPhi kim + oxiHp cht + oxioxitNhit phn muiNhit phn baz khng tanBazPhi kim + hidroOxit axit + ncAxit mnh + muiKim + dd muiOxit baz + ncin phn dd mui(c mng ngn)Axit1. 3Fe + 2O2 0t Fe3O42. 4P + 5O2 0t 2P2O53. CH4 + O2 0t CO2 + 2H2O4. CaCO3 0t CaO + CO25. Cu(OH)2 0t CuO + H2O6. Cl2 + H2 askt 2HCl7. SO3 + H2O H2SO48. BaCl2 + H2SO4 BaSO4+ 2HCl9. Ca(OH)2 + Na2CO3 CaCO3 + 2NaOH10.CaO + H2O Ca(OH)2Axit + bazOxit baz + dd axitOxit axit + dd kimOxit axit+ oxit bazDd mui + dd muiDd mui + dd kimMui + dd axitMuiKim loi + phi kim Kim loi + dd axitKim loi + dd mui12.Ba(OH)2 + H2SO4 BaSO4 + 2H2O13.CuO + 2HCl CuCl2 + H2O14.SO2 + 2NaOH Na2SO3 + H2O15.CaO + CO2 CaCO316.BaCl2 + Na2SO4 BaSO4 + 2NaCl17.CuSO4 + 2NaOH Cu(OH)2 + Na2SO418.CaCO3 + 2HCl CaCl2 + CO2+ H2O9Tnh cht ho hc ca kim loiDy hot ng ho hc ca kim loi.K, Na, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au(Khi No May A Zp St Phi Hi Cc Bc Vng) ngha:K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPt+ O2: nhit thng nhit cao Kh phn ng K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtTc dng vi ncKhng tc dng vi nc nhit thngK BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtTc dng vi cc axit thng thng gii phng Hidro Khng tc dng.K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtKim loi ng trc y kim loi ng sau ra khi muiK BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPt+ Axit+ O2+ Phi kim + DD MuiKim loioxitMuiMui + H2Mui + kl 1. 3Fe + 2O2 0t Fe3O42. 2Fe + 3Cl2 0t 2FeCl33. Fe + 2HCl FeCl2 + H24. Fe + CuSO4 FeSO4 + Cu10H2, CO khng kh c oxit kh c oxit cc kim loi ny nhit caoCh :- Cc kim loi ng trc Mg phn ng vi nc nhit thng to thnh dd Kim v gii phng kh Hidro.- Tr Au v Pt, cc kim loi khc u c th tc dng vi HNO3 v H2SO4 c nhng khng gii phng Hidro.So snh tnh cht ho hc ca nhm v st* Ging:- u c cc tnh cht chung ca kim loi.- u khng tc dng vi HNO3 v H2SO4 c ngui* Khc:Tnh cht Al (NTK = 27) Fe (NTK = 56)Tnh chtvt l- Kimloi mutrng, c nh kim, nh, dn in nhit tt.- t0nc = 6600C- Lkimloi nh, ddt mng, do.-Kimloi mutrngxm, c nh kim, dn in nhit km hn Nhm.- t0nc = 15390C- L kim loi nng, do nn d rn.Tc dng viphi kim2Al + 3Cl2 0t 2AlCl32Al + 3S 0t Al2S32Fe + 3Cl2 0t 2FeCl3Fe + S 0t FeSTc dng viaxit2Al + 6HCl 2AlCl3 + 3H2Fe + 2HCl FeCl2 + H2Tc dng vidd mui2Al + 3FeSO4 Al2(SO4)3 + 3FeFe + 2AgNO3Fe(NO3)2+ 2AgTc dng vidd Kim2Al + 2NaOH + H2O 2NaAlO2 + 3H2Khng phn ngHp cht - Al2O3 c tnh lng tnhAl2O3+6HCl2AlCl3+ 3H2OAl2O3+2NaOH2NaAlO2+ H2O- Al(OH)3kt ta dng keo, l hp cht lng tnh- FeO, Fe2O3v Fe3O4u l cc oxit baz- Fe(OH)2mu trng xanh- Fe(OH)3 mu nu 11Kt lun - Nhm l kim loi lng tnh, cthtcdngvi cdd Axit v dd Kim. Trong cc phnngho hc, Nhm th hin ho tr III- St th hin 2 ho tr: II, III+Tcdngvi axit thng thng, vi phi kimyu, vi dd mui: II+ Tc dng viH2SO4c nng, dd HNO3, viphikim mnh: IIIGang v thpGang Thp/N - Gang l hp kim ca St vi Cacbon v1 snguyn t khc nhMn, Si, S (%C=2 5%)- Thp l hp kim ca St vi Cacbonv1snguyn t khc (%C2Al2O3 (r)Phn ng khng c s thay i s oxi ho.BaO (r) + H2O (l) ----> Ba(OH)2 (dd) 2/ Phn ng phn hu.- c im ca phn ng: C th xy ra s thay i s oxi ho hoc khng.V d:Phn ng c s thay i s oxi ho.2KClO3 (r) -------> 2KCl (r)+ 3O2 (k) Phn ng khng c s thay i s oxi ho.CaCO3 (r) ----->CaO (r) + CO2 (k) II/ Phn ng c s thay i s oxi ho.1/ Phn ng th.- c im ca phn ng: Nguyn t ca n cht thay th mt hay nhiu nguyn t ca mt nguyn t trong hp cht.V d:Zn (r) + 2HCl (dd) ----> ZnCl2 (dd) + H2 (k)

2/ Phn ng oxi ho - kh.- c im ca phn ng: Xy ra ng thi s oxi ho v s kh. hay xy ra ng thi s nhng electron v s nhn electron.V d:CuO (r) +H2 (k) ------>Cu (r) +H2O (h) Trong :- H2 l cht kh (Cht nhng e cho cht khc)- CuO l cht oxi ho (Cht nhn e ca cht khc)- T H2 -----> H2O c gi l s oxi ho. (S chim oxi ca cht khc)1- T CuO ----> Cu c gi l s kh. (S nhng oxi cho cht khc)2III/ Phn ng khng c thay i s oxi ho.1/ Phn ng gia axit v baz.- c im ca phn ng: Sn phm thu c l mui v nc.V d:2NaOH (dd) + H2SO4 (dd) ----> Na2SO4 (dd) + 2H2O (l) NaOH (dd) + H2SO4 (dd) ----> NaHSO4 (dd) + H2O (l) Cu(OH)2 (r) + 2HCl (dd) ----> CuCl2 (dd) + 2H2O (l) Trong :Phn ng trung ho (2 cht tham gia trng thi dung dch).- c im ca phn ng: l s tc dng gia axit v baz vi lng va .- Sn phm ca phn ng l mui trung ho v nc.V d: NaOH (dd) + HCl (dd) ----> NaCl (dd) + H2O (l) 2/ Phn ng ga axit v mui.- c im ca phn ng: Sn phm thu c phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.V d:Na2CO3 (r) + 2HCl (dd) ----> 2NaCl (dd) + H2O (l) + CO2 (k) BaCl2 (dd) + H2SO4 (dd) -----> BaSO4 (r) + 2HCl (dd) Lu : BaSO4 l cht khng tan k c trong mi trng axit.3/ Phn ng gia baz v mui.- c im ca phn ng: + Cht tham gia phi trng thi dung dch (tan c trong nc)+ Cht to thnh (Sn phm thu c) phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.+ Ch cc mui kim loi m oxit hay hiroxit c tnh cht lng tnh phn ng vi dung dch baz mnh.V d:2NaOH (dd) + CuCl2 (dd) ---->2NaCl (dd) + Cu(OH)2 (r) Ba(OH)2 (dd) + Na2SO4 (dd) ---> BaSO4 (r) + 2NaOH (dd) NH4Cl (dd) + NaOH (dd) ---> NaCl (dd) + NH3 (k) + H2O (l) AlCl3 (dd) + 3NaOH (dd) ----> 3NaCl (dd) + Al(OH)3 (r) Al(OH)3 (r) + NaOH (dd) ---> NaAlO2 (dd) + H2O (l) 4/ Phn ng gia 2 mui vi nhau.- c im ca phn ng: + Cht tham gia phi trng thi dung dch (tan c trong nc)+ Cht to thnh (Sn phm thu c) phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.V d:NaCl (dd) + AgNO3 (dd) ----> AgCl (r) + NaNO3 (dd) 3BaCl2 (dd) + Na2SO4 (dd) ----> BaSO4 (r) + 2NaCl (dd) 2FeCl3 (dd) + 3H2O (l) + 3Na2CO3 (dd) ----> 2Fe(OH)3 (r) + 3CO2 (k) + 6NaCl (dd) gii thiu 1 sphng php cn bng phng trnh ho hc.1/ Cn bng ph ng trnh theo ph ng php i s. V d: Cn bng phng trnh phn ngP2O5 + H2O -> H3PO4a cc h s x, y, z vo phng trnh ta c:- Cn c vo s nguyn t P ta c: 2x = z(1)- Cn c vo s nguyn t O ta c: 5x + y = z(2)- Cn c vo s nguyn t H ta c: 2y = 3z (3)Thay (1) vo (3) ta c: 2y = 3z = 6x => y = 26x = 3xNu x = 1 th y = 3 v z = 2x = 2.1 = 2=> Phng trnh dng cn bng nhsau: P2O5+ 3H2O -> 2H3PO4 V d: Cn bng phng trnh phn ng.Al + HNO3 (long) ----> Al(NO3)3 + NO + H2OBc 1: t h s bng cc n s a, b, c, d trc cc cht tham gia v cht to thnh (Nu 2 cht m trng nhau th dng 1 n)Ta c.a Al+ b HNO3 ----> a Al(NO3)3 + c NO + b/2 H2O.Bc 2: Lp phng trnh ton hc vi tng loi nguyn t c s thay i v s nguyn t 2 v.Ta nhn thy ch c N v O l c s thay i.N:b = 3a + c (I)O: 3b = 9a + c + b/2 (II)Bc 3: Gii phng trnh ton hc tm h sThay (I) vo (II) ta c.3(3a + c) = 9a + c + b/22c = b/2 ----> b = 4c ---> b = 4 v c = 1. Thay vo (I) ---> a = 1.Bc 4: Thay h s va tm c vo phng trnh v hon thnh phng trnh.Al+4 HNO3 ----> Al(NO3)3 + NO + 2 H2OBc 5: Kim tra li phng trnh va hon thnh.2/ Cn bng theo ph ng php electron .V d:Cu+HNO3 (c) -----> Cu(NO3)2+ NO2 + H2O4Bc 1: Vit PTP xc nh s thay i s oxi ho ca nguyn t.Ban u: Cu0 ----> Cu+ 2 Trong cht sau phn ng Cu(NO3)2 Ban u: N+ 5 (HNO3) ----> N+ 4 Trong cht sau phn ng NO2 Bc 2: Xc nh s oxi ho ca cc nguyn t thay i.Cu0 ----> Cu+ 2N+ 5 ----> N+ 4Bc 3: Vit cc qu trnh oxi ho v qu trnh kh.Cu0 2e ----> Cu+ 2N+ 5 + 1e ----> N+ 4Bc 4: Tm bi chung cn bng s oxi ho.1Cu0 2e ----> Cu+ 22N+ 5 + 1e ----> N+ 4Bc 5: a h s vo phng trnh, kim tra, cn bng phn khng oxi ho - khv hon thnh PTHH. Cu+2HNO3 (c) -----> Cu(NO3)2+ 2NO2 + H2O+ 2HNO3 (c) -----> Cu+4HNO3 (c) -----> Cu(NO3)2+ 2NO2 + 2H2O3/ Cn bng theo ph ng php bn phn ng ( Hay ion - electron)Theo phng php ny th cc bc 1 v 2 ging nh phng php electron.Bc 3: Vit cc bn phn ng oxi ho v bn phn ng kh theo nguyn tc:+ Cc dng oxi ho v dng kh ca cc cht oxi ho, cht kh nu thuc cht in li mnh th vit di dng ion. Cn cht in li yu, khng in li, cht rn, cht kh th vit di dng phn t (hoc nguyn t). i vi bn phn ng oxi ho th vit s e nhn bn tri cn bn phn ng th vit s e cho bn phi.Bc 4: Cn bng s e cho nhn v cng hai bn phn ng ta c phng trnh phn ng dng ion.Mun chuyn phng trnh phn ng dng ion thnh dng phn t ta cng 2 v nhng lng tng ng nh nhau ion tri du (Cation v anion) b tr in tch.Ch : cn bng khi lng ca na phn ng.Mi trng axit hoc trung tnh th ly oxi trong H2O.Bc 5: Hon thnh phng trnh.5Mt s phn ng ho hc thng gp.Cn nm vng iu kin xy ra phn ng trao i trong dung dch.Gm cc phn ng: 1/Axit+ Baz Mui+ H2O2/Axit+ Mui Mui mi+ Axt mi 3/Dung dch Mui + Dung dch Baz Mui mi + Baz mi4/2 Dung dch Mui tc dng vi nhau 2 Mui mi iu kin xy ra phn ng trao i l:Sn phm thu c phi c t nht mt cht khng tan hoc mt cht kh hoc phic H2O v cc cht tham gia phi theo yu cu ca tng phn ng.

Tnh tan ca mt s mui v baz.- Hu ht cc mui clo rua u tan ( tr mui AgCl , PbCl2 )- Tt c cc mui nit rat u tan.- Tt c cc mui ca kim loi kim u tan.- Hu ht cc baz khng tan ( tr cc baz ca kim loi kim, Ba(OH)2 v Ca(OH)2 tan t.* Na2CO3 , NaHCO3 ( K2CO3 , KHCO3 ) v cc mui cacbonat ca Ca, Mg, Ba u tc dng c vi a xt.NaHCO3+ NaHSO4 Na2SO4+H2O+ CO2

Na2CO3 + NaHSO4 Khng xy ra NaHCO3+ NaOH

Na2CO3+H2ONa2CO3 + NaOH Khng xy ra2NaHCO3

Na2CO3+H2O +CO2NaHCO3+ Ba(OH)2

BaCO3+ NaOH+H2O2NaHCO3+ 2KOH

Na2CO3+ K2CO3+2H2O

Na2CO3 + Ba(OH)2 BaCO3 +2NaOHBa(HCO3)2+Ba(OH)2 2BaCO3 + 2H2OCa(HCO3)2+Ba(OH)2 BaCO3 +CaCO3 + 2H2ONaHCO3+ BaCl2

khng xy ra Na2CO3 + BaCl2 BaCO3 +2NaClBa(HCO3)2+BaCl2 khng xy ra Ca(HCO3)2+CaCl2 khng xy ra NaHSO3+ NaHSO4 Na2SO4+H2O+ SO2

Na2SO3 + H2SO4 Na2SO4+ H2O+ SO22NaHSO3+ H2SO4

Na2SO4+2H2O + 2SO2 Na2SO3 + 2NaHSO4 2Na2SO4+ H2O+SO2 2KOH + 2NaHSO4 Na2SO4+K2SO4 + H2O(NH4)2CO3 + 2NaHSO4 Na2SO4+(NH4)2SO4 +H2O+ CO2Fe+CuSO4 FeSO4 + CuCu+ Fe SO4 khng xy ra 6Cu+ Fe2(SO4)3 2FeSO4 + CuSO4

Fe+ Fe2(SO4)3 3FeSO4

2FeCl2 +Cl2 0t 2FeCl3 7Bng tnh tan trong nc ca cc axit baz - muiNhm hiroxit v gc axitHiro v cc kim loiHIKINaIAgIMgIICaIIBaIIZnIIHgIIPbIICuIIFeIIFeIIIAlIII- OHt t - k i t k - k k k k k- Clt/b t t k t t t t t i t t t t- NO3 t/b t t t t t t t t t t t t t- CH3COOt/b t t t t t t t t t t t - t= St/b t t k - t t k k k k k k = SO3 t/b t t k k k k k k k k k - = SO4 t/kbt t i t i k t - k t t t t = CO3 t/b t t k k k k k - k - k - = SiO3 k/kbt t k k k k k k k k = PO4 t/kbt t k k k k k k k k k k kt : hp cht khng tan c trong nc . k: hp cht khng tan i: hp cht t tan.1 b: hp cht bay hi hoc d bi phn hu thnh kh bay ln. kb : hp cht khng bay hi. Vch ngang - " :hp cht khng tn ti hoc b phn hu trong nc.2Mt s PTHH cn lu :V d: Ho tan m( gam ) MxOy vo dung dch axit (HCl, H2SO4, HNO3)Ta c PTHH cn bng nh sau: lu 2y/x l ho tr ca kim loi M MxOy + 2yHCl xMCl2y/x+ yH2O2MxOy+2yH2SO4 xM2(SO4)2y/x+2yH2OMxOy +2yHNO3 xM(NO3)2y/x + yH2OVD: Ho tan m( gam ) kim loi M vo dung dch a xit (HCl, H2SO4)Ta c PTHH cn bng nh sau: lu x l ho tr ca kim loi M 2M + 2xHCl 2MClx +xH2

p dng: Fe +2HCl FeCl2+ H2 2Al +2*3 HCl 2AlCl3 + 3H2 62M+xH2SO4 M2(SO4)x +xH2p dng:Fe+H2SO4 FeSO4+H22Al+3H2SO4 Al2(SO4)3 +3H2Cc phn ng iu ch mt s kim loi: i vi mt s kim loi nh Na, K, Ca, Mg th dng phng php in phn nng chy cc mui Clorua. PTHH chung:2MClx (r ) d p n c 2M(r )+Cl2( k )(i vi cc kim loi ho tr II th nh n gin phn h s) i vi nhm th dng phng php in phn nng chy Al2O3, khi c cht xc tc Criolit(3NaF.AlF3) , PTHH: 2Al2O3 (r ) d p n c4Al ( r ) +3 O2 (k ) i vi cc kim loi nhFe , Pb , Cu th c th dng cc phng php sau:- Dng H2: FexOy +yH2 0t xFe+yH2O ( h )- Dng C:2FexOy+yC(r ) 0t 2xFe+ yCO2 ( k ) - Dng CO: FexOy+yCO (k ) 0t xFe+ yCO2 ( k ) - Dng Al( nhit nhm ): 3FexOy + 2yAl (r ) 0t3xFe + yAl2O3 ( k )- PTP nhit phn st hir xit:4xFe(OH)2y/x+(3x 2y) O2 0t 2xFe2O3 + 4y H2OMt s phn ng nhit phn ca mt s mui 31/ Mui nitrat Nu M l kim loi ng trc Mg (Theo dy hot ng ho hc)2M(NO3)x 2M(NO2)x+xO2(Vi nhng kim loi ho tr II th nh n gin phn h s ) Nu M l kim loi k t Mg n Cu (Theo dy hot ng ho hc)4M(NO3)x 0t2M2Ox + 4xNO2 + xO2

(Vi nhng kim loi ho tr II th nh n gin phn h s ) Nu M l kim loi ng sau Cu (Theo dy hot ng ho hc)2M(NO3)x 0t 2M+ 2NO2+ xO2(Vi nhng kim loi ho tr II th nh n gin phn h s)2/ Mui cacbonat- Mui trung ho: M2(CO3)x (r) 0tM2Ox (r) +xCO2(k)(Vi nhng kim loi ho tr II th nh n gin phn h s)- Mui cacbonat axit:2M(HCO3)x(r) 0tM2(CO3)x(r) + xH2O( h )+ xCO2(k)(Vi nhng kim loi ho tr II th nh n gin phn h s)3/ Mui amoniNH4Cl 0tNH3 (k) + HCl ( k )NH4HCO3 0tNH3 (k) + H2O ( h ) +CO2(k)NH4NO3 0tN2O (k) + H2O ( h )NH4NO2 0tN2 (k) + 2H2O ( h )(NH4)2CO3 0t2NH3 (k) + H2O ( h )+ CO2(k)2(NH4)2SO4 0t4NH3 (k) + 2H2O ( h )+2SO2 ( k ) + O2(k) Bi 1: Vit cc phng trnh ho hc biu din cc phn ng ho hc cc th nghim sau:a) Nh vi git axit clohiric vo vi.b) Ho tan canxi oxit vo nc.c) Cho mt t bt iphotpho pentaoxit vo dung dch kali hirxit.d) Nhng mt thanh st vo dung dch ng(II) sunfat.e) Cho mt mu nhm vo dung dch axit sunfuric long.f) Nung mt t st(III) hirxit trong ng nghim.g) Dn kh cacbonic vo dung dch nc vi trong n d.h) Cho mt t natri kim loi vo nc.Bi 2: C nhng baz sau: Fe(OH)3, Ca(OH)2, KOH, Mg(OH)2. Hy cho bit nhng baz no:a) B nhit phn hu?b) Tc dng c vi dung dch H2SO4?4c) i mu dung dch phenolphtalein t khng mu thnh mu hng?Bi 3: Cho cc cht sau: canxi oxit, kh sunfur, axit clohiric, bari hirxit, magi cacbonat, bari clorua, iphotpho penta oxit. Cht no tc dng c vi nhau tng i mt. Hy vit cc phng trnh ho hc ca phn ng.Hngdn:Lpbngthyccccpchttcdngcvi nhau r hn.Bi 4:Choccoxitsau:K2O, SO2, BaO, Fe3O4, N2O5. Vitphng trnh ho hc(nu c) ca cc oxit ny ln lt tc dng vi nc, axit sunfuric, dung dch kali hiroxit.Bi 5: Cho mt lng kh CO d i vo ng thu tinh t nng c cha hn hp bt gm: CuO, K2O, Fe2O3 (u ng thu tinh cn li b hn kn). Vit tt c cc phng trnh ho hc xy ra.Bi 6: Nu hin tng v vit PTHH minh hoa/ Cho Na vo dung dch Al2(SO4)3 b/ Cho K vo dung dch FeSO4 c/ Ho tan Fe3O4 vo dung dch H2SO4 long.d/ Nung nng Al vi Fe2O3 to ra hn hp Al2O3 v FexOy.PTHH tng qut:3x Fe2O3+( 6x 4y ) Al 0t 6 FexOy +( 3x 2y ) Al2O3Bi 7: Cho th nghimMnO2+HCl Kh ANa2SO3 +H2SO4 ( l ) Kh BFeS +HCl Kh CNH4HCO3 + NaOHd Kh DNa2CO3+H2SO4 ( l ) Kh Ea. Hon thnh cc PTHH v xc nh cc kh A, B, C, D, E.b. Cho A tc dng C, B tc dng vi dung dch A, B tc dung vi C, A tc dung dch NaOH iu kin thng, E tc dng dung dch NaOH. Vit cc PTHH xy ra.Bi 8: Nu hin tng xy ra, gii thch v vit PTHH minh ho khi:1/Sc t t n d CO2 vo dung dch nc vi trong; dung dch NaAlO2.2/Cho t t dung dch axit HCl vo dung dch Na2CO3.3/Cho Na vo dung dch MgCl2, NH4Cl. 4/Cho Na vo dung dch CuSO4, Cu(NO3)2.5/Cho Ba vo dung dch Na2CO3, (NH4)2CO3, Na2SO4.6/Cho Fe vo dung dch AgNO3d 7/Cho t t n d dung dch NaOH vo dung dch AlCl3, Al2(SO4)3. 58/Cho Cu ( hoc Fe ) vo dung dch FeCl3.9/Cho t t n d bt Fe vo hn hp dung dch gm AgNO3 v Cu(NO3)2.10/Sc t t NH3 vo dung dch AlCl36Mt s phng php gii ton ho hc thng dng.1. Phng php s hcGii cc php tnh Ho hc cp II ph thng, thng thng s dng phng php s hc: l cc php tnh da vo s ph thuc t l gia cc i lng v cc php tnh phn trm. C s ca cc tnh ton Ho hc l nh lut thnh phn khng i c p dng cho cc php tnh theo CTHH v nh lut bo ton khi lng cc cht p dng cho c php tnh theo PTHH. Trong phng php s hc ngi ta phn bit mt s phng php tnh sau y:a. Phng php t l.im ch yu ca phng php ny l lp c t l thc v sau l p dng cch tnh ton theo tnh cht ca t l thc tc l tnh cc trung t bng tch cc ngoi t.Thd:Tnhkhi lngccbonixitCO2trongc3g cacbon.Bi gii44 ) 2 . 16 ( 122 + CO1mol CO2 = 44gLp t l thc:44g CO2c 12g Cxg 3g C 44 : x= 12 : 3=> x = 11123 . 44Vy, khi lng cacbon ixit l 11gTh d 2: C bao nhiu gam ng iu ch c khi cho tng tc 16g ng sunfat vi mt lng st cn thit.Bi giiPhng trnh Ho hc: CuSO4 + Fe - > FeSO4 + Cu160g64g16g xg=> x = g 4 , 616064 . 16Vy iu ch c 6,4g ng.b. Phng php tnh theo t s hp thc.Dng c bn ca php tnh ny tnh theo PTHH tc l tm khi lng ca mt trong nhng cht tham gia hoc to thnh phn ngtheokhi lngcamttrongnhngchtkhcnhau. Phng 7php tm t s hp thc gia khi lng cc cht trong phn ng c pht biu nh sau:T s khi lng cc cht trong mi phn ng Ho hc th bng t s ca tch cc khi lng mol cc cht vi cc h s trong ph-ng trnh phn ng. C th biu th di dng ton hc nh sau:2 21 121n mn mmmTrong : m1v m2l khi lng cc cht, M1, M2l khi lng mol cc cht cn n1, n2 l h s ca PTHH.Vy khitnh khilng ca mt cht tham gia phn ng Ho hc theo khi lng ca mt cht khc cn s dng nhng t s hp thc tm c theo PTHH nhth no ? minh ho ta xt mt s th d sau:Th d 1: Cn bao nhiugam Ptat n da cho phn ng vi 10g st III clorua ?Bi giiPTHHFeCL3 + 3KOH -> Fe(OH)3 + 3KCL 10g? Tnh t s hp thc gia khi lng Kali hirxit v st II cloruaMKOH = (39 + 16 + 1) = 56gg MFeCL5 , 162 ) 3 . 5 , 35 56 (3 + 5 , 1621685 , 1623 . 563 FeclKOHmm* Tm khi lng KOH: mg gKOH3 , 105 , 162160. 10 Th d 2:Cn bao nhiu gam st III chorua cho tng tc vi kalihirxit thu c 2,5g Kaliclorua?Bi giiPTHHFeCl3 + 3 KOH - >Fe(OH)3 + 3KClTnh t s hp thc gia khi lng FeCl3 v Kalicloruag MFeCL5 , 1623 ; MKCL74,5g5 , 2235 , 1623 . 5 , 745 , 1624 KClFeClmm* Tnh khi lng FeCl3: g MFeCL86 , 15 , 2235 , 162. 5 , 23 c. Phng php tnh theo tha s hp thc.Hng s c tnh ra t t l hp thc gi l tha s hp thc v biu th bng ch ci f. Tha s hp thc c tnh sn v c trong bng tra cu chuyn mn.Vic tnh theo tha s hp thc cng cho cng kt qu nh php tnh theo t s hp thc nhng c tnh n gin hn nh cc bng tra cu c sn.8Th d: Theo th d 2 trn th tha s hp thc l:f = 727 , 05 , 2235 , 162=> 86 , 1 727 , 0 . 5 , 2 . 5 , 23 f MFeCLVy, khi lng FeCl3 l 1,86g2. Phng php i sTrong cc phng php gii cc bi ton Ho hc phng php i s cng thng c s dng. Phng php ny c u im tit kim c thi gian, khi gii cc bi ton tng hp, tng i kh gii bng cc phng php khc. Phng php i s c dng gii cc bi ton Ho hc sau:a. Gii bi ton lp CTHH bng phng php i s.Th d: t chy mt hn hp 300ml hirocacbon v amoniac trongoxi cd. Saukhi chyhonton, thtchkhthucl 1250ml. Sau khi lm ngng t hi nc, th tch gim cn 550ml. Sau khi cho tc dng vi dung dch kim cn 250ml trong c 100ml nit. Th tch ca tt c cc kh o trong iu kin nh nhau. Lp cng thc ca hirocacbonBi giiKhi t chy hn hp hirocacbon v amoniac trong oxi phn ng xy ra theo phng trnh sau:4NH3 + 3O2 -> 2N2 + 6H2O(1)CxHy + (x + )4yO2 -> xCO2 + 2yH2O(2)Theodkinbi ton, saukhi t chyamoniacthto thnh100ml nit. TheoPTHH(1) saukhi t chyhonton amoniac ta thu c th tch nit nh hn 2 ln th tch amoniac trong hn hp ban u, vy th tch amonac khi cha c phn ng l 100. 2 = 200ml. Do th tch hiro ccbon khi cha c phn ng l 300 - 200 = 100ml. Sau khi t chy hn hp to thnh (550 - 250) = 300ml, cacbonnic v (1250 - 550 - 300) = 400ml hi nc.T ta c s phn ng:CxHy + (x + 4y) O2 -> xCO2 + 2yH2O100ml 300ml 400mlTheo nh lut Avogaro, c th thay th t l th tch cc cht kh tham gia v to thnh trong phn ng bng t l s phn t hay s mol ca chng.CxHy + 5O2 -> 3CO2 + 4 H2O=> x = 3; y = 8Vy CTHH ca hydrocacbon l C3H8 9b. Gii bi ton tm thnh phn ca hn hp bng phng php i s.Th d:Ho tan trong nc 0,325g mt hn hp gm 2 mui Natriclorua v Kaliclorua. Thm vo dung dch ny mt dung dch bc Nitrat ly d - Kt ta bc clorua thu c c khi lng l 0,717g. Tnh thnh phn phn trm ca mi cht trong hn hp.Bi giiGi MNaCl l x v mKcl l y ta c phng trnh i s:x + y = 0,35 (1)PTHH: NaCl + AgNO3 -> AgCl + NaNO3KCl + AgNO3 -> AgCl + KNO3 Da vo 2 PTHH ta tm c khi lng ca AgCl trong mi phn ng:mAgCl = x .NaClAgClMM= x . 5 , 58143= x . 2,444mAgCl = y .kclAgClMM= y . 5 , 74143= y . 1,919=> mAgCl = 2,444x + 1,919y = 0,717(2)T (1) v (2) => h phng trnh ' + +7 1 7 , 0 9 1 9 , 1 4 4 4 , 23 2 5 , 0y xy xGii h phng trnh ta c: x = 0,178 y = 0,147=> % NaCl = 325 , 0178 , 0.100% = 54,76%% KCl = 100% - % NaCl = 100% - 54,76% = 45,24%.Vy trong hn hp: NaCl chim 54,76%, KCl chim 45,24%3.Phngphppdngnhlutbotonnguyntv khi lng.a/ Nguyn tc: Trong phn ng ho hc, cc nguyn t v khi lng ca chng -c bo ton.T suy ra:+ Tng khi lng cc cht tham gia phn ng bng tng khi lng cc cht to thnh.+ Tng khi lng cc cht trc phn ng bng tng khi lng cc cht sau phn ng.b/ Phm vi p dng: Trong cc bi ton xy ra nhiu phn ng, lc ny i khi khng cn thit phi vit cc phng trnh phn ng v ch cn lp s 10 phn ng thy mi quan h t l mol gia cc cht cn xc nh v nhng cht m cho.Bi1. Cho mt lung kh clo dtc dng vi9,2g kim loisinh ra 23,4g mui kim loi ho tr I. Hy xc nh kim loi ho tr I v mui kim loi .Hng dn gii:t M l KHHH ca kim loi ho tr I.PTHH:2M+Cl2 2MCl2M(g) (2M + 71)g9,2g23,4gta c:23,4 x 2M =9,2(2M + 71)suy ra: M = 23.Kim loi c khi lng nguyn t bng 23 l Na.Vy mui thu c l: NaClBi 2: Ho tan hon ton 3,22g hn hp X gm Fe, Mg v Zn bng mt lng va dung dch H2SO4 long, thu c 1,344 lit hiro ( ktc) v dung dch cha m gam mui. Tnh m?Hng dn gii:PTHH chung:M + H2SO4 MSO4 +H2 nH2SO4 = nH2= 4 , 22344 , 1 = 0,06 molp dng nh lut BTKL ta c:mMui = mX + m H2SO4- m H2= 3,22 + 98 * 0,06 - 2 * 0,06 = 8,98gBi 3: C 2 l st khi lng bng nhau v bng 11,2g. Mt l cho tc dnght vikh clo, mtl ngmtrongdungdch HCl d.Tnh khi lng st clorua thu c.Hng dn gii:PTHH:2Fe +3Cl2 2FeCl3 (1)Fe + 2HCl FeCl2+H2 (2)Theo phng trnh (1,2) ta c:nFeCl3 =nFe = 562 , 11 = 0,2mol nFeCl2 =nFe = 562 , 11 = 0,2mol S mol mui thu c hai phn ng trn bng nhau nhng khi l-ng mol phn t ca FeCl3 ln hn nn khi lng ln hn.mFeCl2= 127 * 0,2 = 25,4gmFeCl3= 162,5 * 0,2 = 32,5g Bi 4: Ho tan hn hp 2 mui Cacbonnat kim loi ho tr 2 v 3 bng dung dch HCl d thu c dung dch A v 0,672 lt kh (ktc).Hi c cn dung dch A thu c bao nhiu gam mui khc nhau?11Bi gii:Bi 1: Gi 2 kim loi ho tr II v III ln lt l X v Y ta c phng trnh phn ng:XCO3 + 2HCl -> XCl2 + CO2 + H2O(1)Y2(CO3)3 + 6HCl -> 2YCl3 + 3CO2 + 3H2O (2).S mol CO2 thot ra (ktc) phng trnh 1 v 2 l:mol nCO03 , 04 , 22672 , 02 Theo phng trnh phn ng 1 v 2 ta thy s mol CO2 bng s mol H2O.mol n nCO O H03 , 02 2 v mol nHCl006 , 0 2 . 03 , 0 Nh vy khi lng HCl phn ng l:mHCl = 0,06 . 36,5 = 2,19 gamGi x l khi lng mui khan (3 2YCl XClm m+)Theo nh lut bo ton khi lng ta c:10 + 2,19 = x + 44 . 0,03 + 18. 0,03=> x = 10,33 gamBi ton 2: Cho 7,8 gam hn hp kimloi Al v Mg tc dng vi HCl thu c 8,96 lt H2 ( ktc). Hi khi c cn dung dch thu c bao nhiu gam mui khan.Bi gii:Ta c phng trnh phn ng nh sau:Mg + 2HCl -> MgCl2 + H22Al + 6HCl -> 2AlCl3 + 3H2S mol H2 thu c l:mol nH4 , 04 , 2296 , 82 Theo (1, 2) ta thy s mol HCL gp 2 ln s mol H2Nn: S moltham gia phn ng l:n HCl= 2 . 0,4 = 0,8 molS mol (s mol nguyn t) to ra mui cng chnh bng s mol HCl bng 0,8 mol. Vy khi lng Clo tham gia phn ng:mCl = 35,5 . 0,8 = 28,4 gamVy khi lng mui khan thu c l: 7,8 + 28,4 = 36,2 gam4. Phng php tng, gim khi lng.a/ Nguyn tc: So snh khi lng ca cht cn xc nh vi cht m gi thit cho bit lng ca n, t khi lng tng hay gim ny, kt hp vi quan h t l mol gia 2 cht ny m gii quyt yu cu t ra.b/ Phm v s dng: 12 i vi cc bi ton phn ng xy ra thuc phn ng phn hu, phn ng gia kim loi mnh, khng tan trong nc y kim loi yu ra khi dung sch mui phn ng, ...c bit khi cha bit r phn ng xy ra l hon ton hay khng th vic s dng phng php ny cng n gin ho cc bi ton hn.Bi 1: Nhng mt thanh st v mt thanh km vo cng mt cc cha 500 ml dung dch CuSO4. Sau mt thi gian ly hai thanh kim loi ra khi cc th mi thanh c thm Cu bm vo, khi lng dung dch trong cc b gim mt 0,22g. Trong dung dch sau phn ng, nng mol ca ZnSO4 gp 2,5 ln nng mol ca FeSO4. Thm dung dch NaOH dvo cc, lc ly kt ta ri nung ngoi khng kh n khi lng khng i , thu c 14,5g cht rn. S gam Cu bm trn mi thanh kim loi v nng mol ca dung dch CuSO4 ban u l bao nhiu?Hng dn gii:PTHHFe + CuSO4 FeSO4 +Cu( 1 )Zn + CuSO4 ZnSO4 +Cu( 2 )Gi a l s mol ca FeSO4 V th tch dung dch xem nhkhng thay i. Do t l v nng mol ca cc cht trong dung dch cng chnh l t l v s mol.Theo bi ra: CM ZnSO4= 2,5 CM FeSO4Nn ta c: nZnSO4= 2,5 nFeSO4Khi lng thanh st tng: (64 - 56)a = 8a (g)Khi lng thanh km gim: (65 - 64)2,5a = 2,5a (g)Khi lng ca hai thanh kim loi tng: 8a - 2,5a = 5,5a (g)M thc t bi cho l: 0,22gTa c: 5,5a = 0,22 a = 0,04 (mol)Vy khi lng Cu bm trn thanh st l: 64 * 0,04 = 2,56 (g)v khi lng Cu bm trn thanh km l: 64 * 2,5 * 0,04 = 6,4 (g)Dung dch sau phn ng 1 v 2 c: FeSO4, ZnSO4v CuSO4(nu c)Ta c s phn ng:NaOH d t0, kkFeSO4 Fe(OH)2 21Fe2O3a a 2a(mol)mFe2O3

=160 x 0,04 x 2a = 3,2 (g) NaOH d t0CuSO4 Cu(OH)2 CuObb b(mol)mCuO= 80b = 14,5 - 3,2 = 11,3 (g) b = 0,14125 (mol)13VynCuSO4 ban u = a+2,5a+b = 0,28125 (mol) CM CuSO4

=5 , 028125 , 0 = 0,5625 MBi 2: Nhng mt thanh st nng 8 gam vo 500 ml dung dch CuSO42M. Sau mt thi gian ly l st ra cn li thy nng 8,8 gam. Xem th tch dung dch khng thay i th nng mol/lit ca CuSO4 trong dung dch sau phn ng l bao nhiu?Hng dn gii:S mol CuSO4 ban u l: 0,5 x 2 = 1 (mol)PTHHFe + CuSO4 FeSO4 +Cu( 1 )1 mol 1 mol56g64g lm thanh st tng thm 64 - 56 = 8 gam M theo bi cho, ta thy khi lng thanh st tng l: 8,8 - 8 = 0,8 gamVy c 88 , 0= 0,1 mol Fe tham gia phn ng, th cng c 0,1 mol CuSO4 tham gia phn ng. S mol CuSO4 cn d : 1 - 0,1 = 0,9 molTa cCM CuSO4 = 5 , 09 , 0 = 1,8 MBi 3: Dn V lit CO2(ktc) vo dung dch cha 3,7 gam Ca(OH)2. Sau phn ng thu c 4 gam kt ta. Tnh V?Hng dn gii:Theo bi ra ta c:S mol ca Ca(OH)2 = 747 , 3 = 0,05 molS mol ca CaCO3 = 1004 = 0,04 molPTHHCO2+Ca(OH)2 CaCO3+H2O- Nu CO2 khng d:Ta c s mol CO2=s mol CaCO3= 0,04 molVy V(ktc) = 0,04 * 22,4 = 0,896 lt- Nu CO2 d: CO2+Ca(OH)2 CaCO3+H2O0,05 0,05 mol 0,05 CO2 + CaCO3+ H2O Ca(HCO3)20,01 (0,05 - 0,04) molVy tng s mol CO2 tham gia phn ng l:0,05 + 0,01 = 0,06 mol V(ktc) = 22,4 * 0,06 = 1,344 lt14Bi 4: Ho tan 20gam hn hp hai mui cacbonat kim loi ho tr 1 v 2 bng dung dch HCl d thu c dung dch X v 4,48 lt kh ( ktc) tnh khi lng mui khan thu c dung dch X.Bi gii: Gi kim loi ho tr 1 v 2 ln lt l A v B ta c ph-ng trnh phn ng sau:A2CO3 + 2HCl -> 2ACl + CO2 + H2O(1)BCO3 + 2HCl -> BCl2 + CO2 + H2O(2)S mol kh CO2 ( ktc) thu c 1 v 2 l:mol nCO2 , 04 , 2248 , 42 Theo (1) v (2) ta nhn thy c 1 mol CO2 bay ra tc l c 1 mol mui cacbonnat chuyn thnh mui Clorua v khi lng tng thm 11 gam (gc CO3 l 60g chuyn thnh gc Cl2 c khi lng 71 gam).Vy c 0,2 mol kh bay ra th khi lng mui tng l:0,2 . 11 = 2,2 gamVy tng khi lng mui Clorua khan thu c l:M(Mui khan) = 20 + 2,2 = 22,2 (gam)Bi 5: Ho tan 10gam hn hp 2 mui Cacbonnat kim loi ho tr 2 v 3 bng dung dch HCl d thu c dung dch A v 0,672 lt kh (ktc).Hi c cn dung dch A thu c bao nhiu gam mui khc nhau?Bi giiMt bi ton ho hc thng l phi c phn ng ho hc xy ra mcphnnghohcthphi vitphngtrnhhohcl iu khng th thiu.Vy ta gi hai kim loi c ho tr 2 v 3 ln lt l X v Y, ta c phn ng:XCO3 + 2HCl -> XCl2 + CO2 + H2O(1)Y2(CO3)3 + 6HCl -> 2YCl3 + 3CO2 + 3H2O (2).S mol cht kh to ra chng trnh (1) v (2) l:4 , 22672 , 02COn = 0,03 molTheo phn ng (1, 2) ta thy c 1 mol CO2bay ra tc l c 1 mol mui Cacbonnat chuyn thnh mui clorua v khi lng tng 71 - 60 = 11 (gam) (; 603g mCO g mCl71 ).S mol kh CO2bay ra l 0,03 mol do khi lng mui khan tng ln: 11 . 0,03 = 0,33 (gam).Vy khi lng mui khan thu c sau khi c cn dung dch.m (mui khan) = 10 + 0,33 = 10,33 (gam).Bi 6: Ho tan 20gam hn hp hai mui cacbonat kim loi ho tr 1 v 2 bng dung dch HCl d thu c dung dch X v 4,48 lt kh ( ktc) tnh khi lng mui khan thu c dung dch X.15Bi gii: Gi kim loi ho tr 1 v 2 ln lt l A v B ta c ph-ng trnh phn ng sau:A2CO3 + 2HCl -> 2ACl + CO2 + H2O(1)BCO3 + 2HCl -> BCl2 + CO2 + H2O(2)S mol kh CO2 ( ktc) thu c 1 v 2 l:mol nCO2 , 04 , 2248 , 42 Theo (1) v (2) ta nhn thy c 1 mol CO2 bay ra tc l c 1 mol mui cacbonnat chuyn thnh mui Clorua v khi lng tng thm 11 gam (gc CO3 l 60g chuyn thnh gc Cl2 c khi lng 71 gam).Vy c 0,2 mol kh bay ra th khi lng mui tng l:0,2 . 11 = 2,2 gamVy tng khi lng mui Clorua khan thu c l:M(Mui khan) = 20 + 2,2 = 22,2 (gam)Bi 1: Nhng mt thanh kim loi M ho tr II vo 0,5 lit dd CuSO4 0,2M. Sau mt thi gian phn ng, khi lng thanh M tng ln 0,40g trong khi nng CuSO4 cn li l 0,1M.a/ Xc nh kim loi M.b/ Ly m(g) kim loi M cho vo 1 lit dd cha AgNO3 v Cu(NO3)2 , nng mi mui l 0,1M. Sau phn ng ta thu c cht rn A khi lng 15,28g v dd B. Tnh m(g)?Hng dn gii: a/ theo bi ra ta c PTHH .M +CuSO4 MSO4+Cu (1)S mol CuSO4 tham gia phn ng (1) l: 0,5 ( 0,2 0,1 ) = 0,05 mol tng khi lng ca M l:mtng = mkl gp -mkl tan = 0,05 (64 M) = 0,40 gii ra: M = 56 , vy M l Feb/ ta ch bit s mol ca AgNO3 v s mol ca Cu(NO3)2. Nhng khng bit s mol ca Fe (cht khFeCu2+ Ag+ (cht oxh mnh)0,1 0,1 ( mol )Ag+ C Tnh oxi ho mnh hn Cu2+ nn mui AgNO3 tham gia phn ng vi Fe trc.PTHH: Fe+2AgNO3 Fe(NO3)2+2Ag (1) Fe +Cu(NO3)2 Fe(NO3)2+Cu (2)Ta c 2 mc so snh:- Nu va xong phn ng (1): Ag kt ta ht, Fe tan ht, Cu(NO3)2 cha phn ng. Cht rn A l Ag th ta c: mA = 0,1 x 108 = 10,8 g- Nu va xong c phn ng (1) v (2) th khi cht rn A gm: 0,1 mol Ag v 0,1 mol Cu mA = 0,1 ( 108 + 64 ) = 17,2 g16theo cho mA = 15,28 g ta c: 10,8 < 15,28 < 17,2 vy AgNO3 phn ng ht, Cu(NO3)2 phn ng mt phn v Fe tan ht.mCu to ra = mA mAg = 15,28 10,80 = 4,48 g. Vy s mol ca Cu = 0,07 mol.Tng s mol Fe tham gia c 2 phn ng l: 0,05 ( p1 ) +0,07 ( p2 ) = 0,12 molKhi lng Fe ban u l: 6,72g5. Phng php lm gim n s.Bi ton 1: (Xt li bi ton nu phng php th nht)Ho tan hn hp 20 gam hai mui cacbonnat kim loi ho tr I v II bng dung dch HCl dthu c dung dch M v 4,48 lt CO2( ktc) tnh khi lng mun to thnh trong dung dch M.Bi giiGi A v B ln lt l kim loi ho tr I v II.Ta c phng trnh phn ng sau:A2CO3 + 2HCl-> 2ACl + H2O + CO2(1)BCO3 + 2HCl -> BCl2 + H2O + CO2(2)S mol kh thu c phn ng (1) v (2) l:mol nCO2 , 04 , 2248 , 43 Gi a v b ln lt l s mol ca A2CO3v BCO3ta c phng trnh i s sau:(2A + 60)a + (B + 60)b = 20 (3)Theo phng trnh phn ng (1) s mol ACl thu c 2a (mol)Theo phng trnh phn ng (2) s mol BCl2 thu c l b (mol)Nu gi s mui khan thu c l x ta c phng trnh:(A + 35.5) 2a + (B + 71)b = x(4)Cng theo phn ng (1, 2) ta c:a + b = ) ( 2 , 02mol nCO(5)T phng trnh (3, 4) (Ly phng trnh (4) tr (5)) ta c: 11 (a + b) = x - 20 (6)Thay a + b t (5) vo (6) ta c:11 . 0,2 = x - 20=> x = 22,2 gamBi ton 2:Ho tan hon ton 5 gam hn hp 2 kim loi bng dung dch HCl thu c dung dch A v kh B, c cn dung dch A thu c 5,71 gam mui khan tnh th tch kh B ktc.Bi gii: Gi X, Y l cc kim loi; m, n l ho tr, x, y l s mol tng ng, s nguyn t khi l P, Q ta c:2X + 2n HCl => 2XCln = nH2(I)2Y + 2m HCl -> 2YClm + mH2(II).17Ta c: xP + y Q = 5 (1)x(P + 35,5n) + y(Q + 35,5m) = 5,71 (2)Ly phng trnh (2) tr phng trnh (1) ta c:x(P + 35,5n) +y(Q + 35,5m)- xP - yQ = 0,71=> 35,5 (nx + my) = 0,71Theo I v II: ) (212my xn nH+ => th tch: V = nx + my = 224 , 0 4 , 22 .2 . 35571 , 0 (lt)186. Phng php dng bi ton cht tng ng.a/ Nguyn tc: Khi trong bi ton xy ra nhiu phn ng nhng cc phn ng cng loi v cng hiu sut th ta thay hn hp nhiu cht thnh 1 cht t-ng ng. Lc lng (s mol, khi lng hay th tch) ca cht t-ng ng bng lng ca hn hp.b/ Phm vi s dng: Trong v c, phng php ny p dng khi hn hp nhiu kim loi hot ng hay nhiu oxit kim loi, hn hp mui cacbonat, ... hoc khi hn hp kim loi phn ng vi nc.Bi 1: Mt hn hp 2 kim loi kim A, B thuc 2 chu k k tip nhau trong bng h thng tun hon c khi lng l 8,5 gam. Hn hp ny tan ht trong nc d cho ra 3,36 lit kh H2 (ktc). Tm hai kim loi A, B v khi lng ca mi kim loi.Hng dn gii:PTHH2A + 2H2O 2AOH+ H2 (1)2B + 2H2O 2BOH+ H2 (2)t a = nA , b = nB ta c: a+b = 2 4 , 2236 , 3 = 0,3 (mol)(I)M trung bnh:M= 3 , 05 , 8 = 28,33Ta thy 23 Bng cch: Chn 1 n s lm chun ri tch cc n s cn li. Nn a v phng trnh ton hc 2 n, trong c 1 n c gii hn (tt nhin nu c 2 n c gii hn th cng tt). Sau c th thit lp bng bin thin hay d vo cc iu kin khc chn cc gi tr hp l.b/ V d:Bi 1: Ho tan 3,06g oxit MxOy bng dung dich HNO3 d sau c cn th thu c 5,22g mui khan. Hy xc nh kim loi M bit n ch c mt ho tr duy nht.22Hng dn gii:PTHH: MxOy+ 2yHNO3 -----> xM(NO3)2y/x +yH2O T PTP ta c t l:y M x1606 , 3+ = y Mx12422 , 5+ ---> M = 68,5.2y/xTrong : t 2y/x = n l ho tr ca kim loi. Vy M = 68,5.n(*)Cho n cc gi tr 1, 2, 3, 4. T (*) ---> M = 137 v n =2 l ph hp.Do M l Ba, ho tr II.Bi 2: A, B l 2 cht kh iu kin thng, A l hp cht ca nguyn t X vi oxi (trong oxi chim 50% khi lng), cn B l hp cht ca nguyn t Y vi hir (trong hiro chim 25% khi l-ng). T khi ca A so vi B bng 4. Xc nh cng thc phn t A, B. Bit trong 1 phn t A ch c mt nguyn t X, 1 phn t B ch c mt nguyn t Y.Hng dn gii:t CTPT A l XOn, MA = X + 16n = 16n + 16n = 32n.t CTPT A l YOm, MB = Y + m = 3m + m = 4m.d = BAMM = mn432 = 4 ---> m = 2n.iu kin tho mn: 0 < n, m < 4, u nguyn v m phi l s chn.Vy m ch c th l 2 hay 4.Nu m = 2 th Y = 6 (loi, khng c nguyn t no tho)Nu m = 4 th Y = 12 (l cacbon) ---> B l CH4v n = 2 th X = 32 (l lu hunh) ---> A l SO29/ Phng php gii hn mt i lng.a/ Nguyn tc p dng:Da vo cc i lng c gii hn, chng hn:KLPTTB ( M ), ho tr trung bnh, s nguyn t trung bnh, ....Hiu sut: 0(%) MA < MR < MB .Vit PTHH xy ra:Theo phng trnh phn ng:nR = 2nH2= 0,2 mol. ----> MR = 6,2 : 0,2 = 31Theo ra: 2 kim loi ny thuc 2 chu k lin tip, nn 2 kim loi l:A l Na(23) v B l K(39)Bi 2: a/ Cho 13,8 gam (A) l mui cacbonat ca kim loi kim vo 110ml dung dch HCl 2M. Sau phn ng thy cn axit trong dung dch thu c v th tch kh thot ra V1 vt qu 2016ml. Vit phng trnh phn ng, tm (A) v tnh V1 (ktc).b/ Ho tan 13,8g (A) trn vo nc. Va khuy va thm tng git dung dch HCl 1M cho ti 180ml dung dch axit, thu c V2 lit kh. Vit phng trnh phn ng xy ra v tnh V2 (ktc).Hng dn:a/ M2CO3 + 2HCl ---> 2MCl + H2O + CO2 Theo PTHH ta c:S mol M2CO3 = s mol CO2 > 2,016 : 22,4 = 0,09 mol---> Khi lng mol M2CO3 < 13,8 : 0,09 = 153,33 (I)Mt khc: S mol M2CO3 phn ng = 1/2 s mol HCl < 1/2. 0,11.2 = 0,11 mol---> Khi lng mol M2CO3 = 13,8 : 0,11 = 125,45 (II)T (I, II) --> 125,45 < M2CO3 < 153,33 ---> 32,5 < M < 46,5 v M l kim loi kim ---> M l Kali (K)Vy s mol CO2 = s mol K2CO3 = 13,8 : 138 = 0,1 mol ---> VCO2 = 2,24 (lit)b/ Gii tng t: ---> V2 = 1,792 (lit)Bi 3: Cho 28,1g qung lmt gm MgCO3; BaCO3 (%MgCO3 = a%) vo dung dch HCl d thu c V (lt) CO2 ( ktc).a/ Xc nh V (lt).Hng dn: a/ Theo bi ra ta c PTHH:MgCO3+2HCl MgCl2 + H2O+CO2(1) x(mol)x(mol)BaCO3 +2HCl BaCl2 + H2O+CO2(2)24 y(mol) y(mol)CO2+Ca(OH)2 CaCO3 + H2O (3) 0,2(mol) 0,2(mol) 0,2(mol) CO2 + CaCO3 +H2O Ca(HCO3)2 (4)Gi s hn hp ch c MgCO3.Vy mBaCO3 = 0 S mol: nMgCO3 = 841 , 28 = 0,3345 (mol)Nu hn hp ch ton l BaCO3 th mMgCO3 = 0S mol: nBaCO3=1971 , 28 =0,143 (mol)Theo PT (1) v (2) ta c s mol CO2 gii phng l: 0,143(mol) nCO2 0,3345 (mol)Vy th tch kh CO2 thu c ktc l: 3,2 (lt) VCO2 7,49 (lt)25Chuyn 2: tan - nng dung dchMt s cng thc tnh cn nh:Cng thc tnh tan: StC 0cht= dmctmm . 100Cng thc tnh nng %:C%= ddctmm . 100%mdd = mdm + mctHoc mdd = Vdd (ml) . D(g/ml)*Mi linhgiatancamtchtvnngphntrm dung dch bo ho ca cht mt nhit xc nh.C 100g dm ho tan c Sg cht tan to thnh (100+S)g dung dch bo ho.Vy:x(g)// y(g) //100g// Cng thc lin h: C% = SS+ 100100 Hoc S=% 100% . 100CCCng thc tnh nng mol/lit:CM = ) () (lit Vmol n = ) () ( . 1000ml Vmol n* Mi lin h gia nng % v nng mol/lit.Cng thc lin h:C% = DM CM10.Hoc CM = MC D % . 10Trong :- mct l khi lng cht tan( n v: gam)- mdm l khi lng dung mi( n v: gam)- mdd l khi lng dung dch( n v: gam)- V l th tch dung dch( n v: lit hoc mililit)- D l khi lng ring ca dung dch( n v: gam/mililit)- M l khi lng mol ca cht( n v: gam)- S l tan ca 1 cht mt nhit xc nh( n v: gam)- C% l nng % ca 1 cht trong dung dch( n v: %)- CMl nng mol/lit ca 1 cht trong dung dch( n v: mol/lit hay M)26Dng 1: Ton tanPhn dng 1: Bi ton lin quan gia tan ca mt cht v nng phn trm dung dch bo ho ca cht .Bi 1: 400C, tan ca K2SO4 l 15. Hy tnh nng phn trm ca dung dch K2SO4 bo ho nhit ny?p s: C% = 13,04%Bi 2: Tnh tan ca Na2SO4 100C v nng phn trm ca dung dch bo ho Na2SO4 nhit ny. Bit rng 100C khi ho tan 7,2g Na2SO4 vo 80g H2O th c dung dch bo ho Na2SO4.p s: S = 9g v C% = 8,257%Phn dng 2: Bi ton tnh lng tinh th ngm nc cn cho thm vo dung dch cho sn.Cch lm:Dng nh lut bo ton khi lng tnh:* Khi lng dung dch to thnh = khi lng tinh th + khi lng dung dch ban u.* Khi lng cht tan trong dung dch to thnh = khi lng cht tan trong tinh th + khi lng cht tan trong dung dch ban u.* Cc bi ton loi ny thng cho tinh th cn ly v dung dch cho sn c cha cng loi cht tan.Bi tp p dng:Bi 1: Tnh lng tinh th CuSO4.5H2O cn dng iu ch 500ml dung dch CuSO4 8%(D = 1,1g/ml).p s: Khi lng tinh th CuSO4.5H2O cn ly l: 68,75gBi 2: iu ch 560g dung dch CuSO416% cn phi ly bao nhiu gamdung dch CuSO48%v bao nhiu gamtinh th CuSO4.5H2O.Hng dn* Cch 1:Trong 560g dung dch CuSO4 16% c cha.mct CuSO4(c trong dd CuSO4 16%) = 10016 . 560 = 252240 = 89,6(g)t mCuSO4.5H2O = x(g)1mol(hay 250g) CuSO4.5H2O cha 160g CuSO4 Vy x(g)// cha 250160 x = 2516 x(g)mdd CuSO4 8% c trong dung dch CuSO4 16% l (560 x) gmct CuSO4(c trong dd CuSO4 8%) l 1008 ). 560 ( x = 252 ). 560 ( x (g)27Ta c phng trnh: 252 ). 560 ( x + 2516 x = 89,6Gii phng trnh c: x = 80.Vy cn ly 80g tinh th CuSO4.5H2O v 480g dd CuSO48% pha ch thnh 560g dd CuSO4 16%.* Cch 2: Gii h phng trnh bc nht 2 n.* Cch 3: Tnh ton theo s ng cho.Lu: Lng CuSO4c thcoi nhdd CuSO464%(vc 250g CuSO4.5H2O th c cha 160g CuSO4). Vy C%(CuSO4) = 250160.100% = 64%.Phndng3:bi tontnhlngchttantchrahay thm vo khi thay i nhit mt dung dch bo ho cho sn.Cch lm:- Bc 1: Tnh khilng cht tan v khilng dung mi c trong dung dch bo ho t1(0c)- Bc 2: t a(g) l khi lng cht tan A cn thm hay tch ra khi dung dch ban u, sau khi thay i nhit t t1(0c) sang t2(0c) vi t1(0c) khc t2(0c).- Bc 3: Tnh khilng cht tan v khilng dung mi c trong dung dch bo ho t2(0c).- Bc4: pdng cng thctnh tan haynng % dung dch bo ho(C% ddbh) tm a.L u : Nu yu cu tnh lng tinh th ngm nc tch ra hay cn thm vo do thay i nhit dung dch bo ho cho sn, bc 2 ta phi t n s l s mol(n)Bi 1: 120C c 1335g dung dch CuSO4bo ho. un nng dung dch ln n 900C. Hi phi thm vo dung dch bao nhiu gam CuSO4 c dung dch bo ho nhit ny.Bit 120C, tan ca CuSO4 l 33,5 v 900C l 80.p s: Khi lng CuSO4 cn thm vo dung dch l 465g.Bi 2: 850C c 1877g dung dch bo ho CuSO4.Lm lnh dung dch xung cn 250C. Hi c bao nhiu gam CuSO4.5H2O tch khi dung dch. Bit tan ca CuSO4 850C l 87,7 v 250C l 40.p s: Lng CuSO4.5H2O tch khi dung dch l: 961,75gBi 3: Cho 0,2 mol CuO tan trong H2SO4 20% un nng, sau lm ngui dung dch n 100C. Tnh khi lng tinh th CuSO4.5H2O tch khi dung dch, bit rng tan ca CuSO4 100Cl 17,4g/100g H2O.p s: Lng CuSO4.5H2O tch khi dung dch l: 30,7g2829Dng 2: Ton nng dung dchBi 1:Cho50ml dungdchHNO340%ckhi lngringl 1,25g/ml. Hy:a/ Tm khi lng dung dch HNO3 40%?b/ Tm khi lng HNO3? c/ Tm nng mol/l ca dung dch HNO3 40%?p s:a/ mdd = 62,5gb/ mHNO3 = 25gc/ CM(HNO3) = 7,94MBi 2:Hy tnh nng mol/l ca dung dch thu c trong mi trng hp sau:a/ Ho tan 20g NaOH vo 250g nc. Cho bit DH2O = 1g/ml, coi nh th tch dung dch khng i.b/ Hotan26,88ltkhhirocloruaHCl (ktc)vo500ml nc thnh dung dch axit HCl. Coi nh th dung dch khng i.c/ Ho tan 28,6g Na2CO3.10H2O vo mt lng nc va thnh 200ml dung dch Na2CO3. p s:a/ CM( NaOH ) = 2Mb/ CM( HCl ) = 2,4Mc/ CM(Na2CO3) = 0,5MBi 3:Cho2,3gNatanhttrong47,8ml ncthucdung dch NaOH v c kh H2 thot ra . Tnh nng % ca dung dch NaOH?p s: C%(NaOH) = 8%30chuyn 3: pha trn dung dchLoi 1: Bi ton pha long hay c dc mt dung dch.a) c im ca bi ton:- Khi phalong, nngdungdchgim. Cncdc, nng dung dch tng.- D phalonghaycc, khi lngchttanlunlun khng thay i.b)Cch lm: C th p dng cng thc pha long hay c cTH1: V khi lng cht tan khng i d pha long hay c c nn. mdd(1).C%(1) = mdd(2).C%(2) TH2: V s mol cht tan khng i d pha long hay c dc nn.Vdd(1). CM (1)=Vdd(2). CM (2) Nu gp bi ton bi ton: Cho thm H2O hay cht tan nguyn cht (A) vo 1 dung dch (A) c nng % cho trc, c th p dng quy tc ng cho gii. Khi c th xem:- H2O l dung dch c nng O%- Cht tan (A) nguyn cht cho thm l dung dch nng 100%+ TH1: Thm H2ODung dch u C1(%)C2(%) - OC2(%) = O Hdau ddmm2.H2O O(%) C1(%) C2(%)+ TH1: Thm cht tan (A) nguyn chtDung dch u C1(%)100 -C2(%) C2(%) = ctAdau ddmm. Cht tan (A)100(%) C1(%) C2(%)Lu : T l hiu s nng nhn c ng bng s phn khi lng dung dch u( hay H2O, hoc cht tan A nguyn cht) cn ly t cng hng ngang.31Bi ton p dng:Bi 1: Phi thm bao nhiu gam H2O vo 200g dung dch KOH 20% c dung dch KOH 16%.p s: mH2O(cn thm) = 50gBi 2: C 30g dung dch NaCl 20%. Tnh nng % dung dch thu c khi:- Pha thm 20g H2O- C c dung dch ch cn 25g.p s: 12% v 24%Bi 3: Tnh s ml H2O cn thm vo 2 lit dung dch NaOH 1M thu c dung dch mi c nng 0,1M.p s: 18 litBi 4: Tnh s ml H2O cn thm vo 250ml dung dch NaOH1,25M to thnh dung dch 0,5M. Gi s s ho tan khng lm thay i ng k th tch dung dch.p s: 375mlBi 5: Tnh s ml dung dch NaOH 2,5%(D = 1,03g/ml) iu ch c t 80ml dung dch NaOH 35%(D = 1,38g/ml).p s: 1500mlBi 6: Lm bay hi 500ml dung dch HNO3 20%(D = 1,20g/ml) ch cn 300g dung dch. Tnh nng % ca dung dch ny.p s: C% = 40%Loi 2:Bi ton ho tan mt ho cht vo nc hay vo mt dung dch cho sn.a/ c im bi ton:- Ho cht em ho tan c th l cht kh, cht lng hay cht rn.- S ho tan c th gy ra hay khng gy ra phn ng ho hc gia cht em ho tan viH2O hoc cht tan trong dung dch cho sn.b/ Cch lm:- Bc 1: Xc nh dung dch sau cng (sau khi ho tan ho cht) c cha cht no:Cn lu xem c phn ng gia cht em ho tan vi H2O hay cht tan trong dung dch cho sn khng? Sn phm phn ng(nu c) gm nhng cht tan no? Nh rng: c bao nhiu loicht tan trong dung dch th c by nhiu nng .32. Nu cht tan c phn ng ho hc vi dung mi, ta phi tnh nng ca sn phm phn ng ch khng c tnh nng ca cht tan .- Bc2:Xcnhlngchttan(khi lnghaysmol)c cha trong dung dch sau cng.. Lng cht tan(sau phn ng nu c) gm: sn phm phn ng v cc cht tc dng cn d.. Lng sn phm phn ng(nu c) tnh theo pttphi da vo cht tc dng ht(lng cho ), tuyt i khng c da vo lng cht tc dng cho d (cn tha sau phn ng)- Bc3: Xcnhlngdungdchmi (khi lnghayth tch). tnh th tch dung dch mi c 2 trng hp (tu theo bi)Nu khng cho bit khi lng ring dung dchmi(Dddm)+ Khi ho tan 1 cht kh hay 1 cht rn vo 1 cht lng c th coi:Th tch dung dch mi = Th tch cht lng+ Khi ho tan 1 cht lng vo 1 cht lng khc, phi gi s s pha trn khng lm thy i ng k th tch cht lng, tnh:Th tch dung dch mi = Tng th tch cc cht lng ban u.Nu cho bit khi lng ring dung dchmi(Dddm)Th tch dung dch mi:Vddm = ddmddmDmmddm: l khi lng dung dch mi+ tnh khi lng dung dch mimddm=Tng khi lng(trc phn ng) khi lng kt ta(hoc kh bay ln) nu c.Bi tp p dng:Bi 1: Cho 14,84g tinh th Na2CO3 vo bnh cha 500ml dung dch HCl 0,4McdungdchB. Tnhnngmol/litccchttrong dung dch B.p s: Nng ca NaCl l: CM = 0,4MNng ca Na2CO3 cn d l: CM = 0,08MBi 2: Ho tan 5,6lit kh HCl ( ktc) vo 0,1lit H2O to thnh dung dch HCl. Tnh nng mol/lit v nng % ca dung dch thu c.p s:- CM = 2,5M33- C% = 8,36%Bi 3: Cho 200g SO3 vo 1 lt dung dch H2SO4 17%(D = 1,12g/ml) c dung dch A. Tnh nng % dung dch A.p s: C% = 32,985%Bi 4: xc nh lng SO3 v lng dung dch H2SO4 49% cn ly pha thnh 450g dung dch H2SO4 83,3%.p s:Khi lng SO3 cn ly l: 210gKhi lng dung dch H2SO4 49% cn ly l 240gBi5: Xc nh khi lng dung dch KOH 7,93% cn ly khi ho tan vo 47g K2O th thu c dung dch 21%.p s: Khi lng dung dch KOH 7,93% cn ly l 352,94gBi 6: Cho 6,9g Na v 9,3g Na2O vo nc, c dung dch A(NaOH 8%). Hi phi ly thm bao nhiu gam NaOH c tinh khit 80%(tan hon ton) cho vo c dung dch 15%?p s: - Khi lng NaOH c tinh khit 80% cn ly l 32,3gLoi 3: Bi ton pha trn hai hay nhiu dung dch.a/ c im bi ton.Khi phatrn2haynhiudungdchvi nhaucthxyrahay khng xy ra phn ng ho hc gia cht tan ca cc dung dch ban u.b/ Cch lm: TH1: Khi trn khng xy ra phn ng ho hc(thng gp bi ton pha trn cc dung dch cha cng loi ho cht)Nguyn tc chung gii l theo phng php i s, lp h 2 phng trnh ton hc (1 theo cht tan v 1 theo dung dch) Cc bc gii:- Bc1: Xcnhdungdchsautrncchachttan no.- Bc2: Xcnhlngchttan(mct)ctrongdungdch mi(ddm)- Bc 3: Xc nh khi lng(mddm) hay th tch(Vddm) dung dch mi.mddm = Tng khi lng( cc dung dch em trn )+ Nu bit khi lng ring dung dch mi(Dddm)34Vddm = ddmddmDm + Nu khng bit khi lng ring dung dch mi: Phi gi s shaohtthtchdosphatrndungdchlkhng ng k, c.Vddm = Tng th tch cc cht lng ban u em trn+ Nu pha trn cc dung dch cng loi cht tan, cng loi nng , c th gii bng quy tc ng cho.m1(g) dd C1(%)C2 C3 C3(%) m2(g) ddC2(%) C3 C1 ( Gi s: C1< C3 < C2 ) v s hao ht th tch do s pha trn cc dd l khng ng k.21mm =1 33 2C CC C + Nu khng bit nng % m li bit nng mol/lit (CM) th p dng s :V1(l) dd C1(M)C2 C3 C3(M) V2(g) ddC2(M) C3 C1 ( Gi s: C1< C3 < C2 )21VV= 1 33 2C CC C + Nu khng bit nng % v nng mol/lit m li bit khi lng ring (D) th p dng s :V1(l) dd D1(g/ml) D2 D3 D3(g/ml) V2(l) dd D2(g/ml)D3 D1 (Gi s: D1< D3 < D2) v s hao ht th tch do s pha trn cc dd l khng ng k.3521VV= 1 33 2D DD D TH2: Khi trn c xy ra phn ng ho hc cng gii qua 3 bc t-ng t bi ton loi 2 (Ho tan mt cht vo mt dung dch cho sn). Tuy nhin, cn lu .- bc1:Phi xcnhcngthcchttanmi, slng cht tan mi. Cn ch kh nng c cht d(do cht tan ban u khng tc dng ht) khi tnh ton.- bc 3: Khi xc nh lng dung dch mi (mddm hay Vddm)Tac: mddm = Tng khi lng cc cht em trng khi lng cht kt ta hoc cht kh xut hin trong phn ng.- Th tch dung dch mi tnh nh trng hp 1 loi bi ton ny.Th d: p dng phng php ng cho.Mt bi ton thng c nhiu cch gii nhng nu bi ton no c th s dng c phng php ng cho gii th s lm bi ton n gin hn rt nhiu.Bi ton 1: Cn bao nhiu gam tinh th CuSO4 . 5H2O ho vo bao nhiu gam dung dch CuSO44% iu ch c 500 gam dung dch CuSO4 8%.Bi gii: Gii Bng phng php thng thng:Khi lng CuSO4 c trong 500g dung dch bng:gam mCuO401008 . 5004 (1)Gi x l khi lng tinh th CuSO4 . 5 H2O cn ly th: (500 - x) l khi lng dung dch CuSO4 4% cn ly:Khi lng CuSO4 c trong tinh th CuSO4 . 5H2O bng:250160 .4xmCuSO(2)Khi lng CuSO4 c trong tinh th CuSO4 4%l:1004 ). 500 (4xmCuSO(3)T (1), (2) v (3) ta c:401004 ). 500 (250) 160 . (+x x=> 0,64x + 20 - 0,04x = 40.Gii ra ta c:X = 33,33g tinh thVy khi lng dung dch CuSO4 4% cn ly l:500 - 33,33 gam = 466,67 gam.+ Gii theo phng php ng choGi x l s gam tinh th CuSO4 . 5 H2O cn ly v (500 - x) l s gam dung dch cn ly ta c s ng cho nh sau:36xx 500 => 141564500 xxGii ra ta tm c: x = 33,33 gam.Bi ton2:Trn500gamdungdchNaOH3%vi 300gam dung dch NaOH 10% th thu c dung dch c nng bao nhiu%.Bi gii: Ta c s ng cho:=> 310300500CCGii ra ta c: C = 5,625%Vy dung dch thu c c nng 5,625%.Bi ton3:Cntrn2dungdchNaOH3%vdungdch NaOH 10% theo t l khilng bao nhiu thu c dung dch NaOH 8%.Bi gii:Gi m1; m2 ln lt l khi lng ca cc dung dch cn ly. Ta c s ng cho sau:=> 3 88 1021mmVy t l khi lng cn ly l: 5221mmBi ton p dng:Bi 1: Cn pha ch theo t l no v khi lng gia 2 dung dch KNO3c nng % tng ng l 45% v 15% c mt dung dch KNO3 c nng 20%.p s: Phi ly 1 phn khi lng dung dch c nng d 45% v 5 phn khi lng dung dch c nng 15% trn vi nhau.Bi 2: Trn V1(l) dung dch A(cha 9,125g HCl) vi V2(l) dung dch B(cha 5,475g HCl) c 2(l) dung dch D.Coi th tch dung dch D = Tng th tch dung dch A v dung dch B.a) Tnh nng mol/lit ca dung dch D.b) Tnhnngmol/lit cadungdchA, dungdchB(Bit hiunngmol/litcadungdchAtrnngmol/lit dung dch B l 0,4mol/l)p s:a) CM(dd D) = 0,2M69 4 - 8 4 8 64 - 8

3 10 - C%

10C% C% - 3% 500:300:3 10 - 8

10 8 8 - 3 m1m237b) t nng mol/l ca dung dch A l x, dung dch B l y ta c:x y = 0,4 (I)V th tch: Vdd D = Vdd A + Vdd B = x25 , 0 + y15 , 0 = 2 (II)Gii h phng trnh ta c: x = 0,5M, y = 0,1MVy nng mol/l ca dung dch A l 0,5M v ca dung dch B l 0,1M.Bi 3: Hi phi ly 2 dung dch NaOH 15% v 27,5% mi dung dch baonhiugamtrnvonhauc500ml dungdchNaOH 21,5%, D = 1,23g/ml?ps:DungdchNaOH27,5%cnlyl319,8gvdungdch NaOH 15% cn ly l 295,2gBi 4:Trnln150ml dungdchH2SO42Mvo200gdungdch H2SO45M( D=1,29g/ml ). Tnhnngmol/l cadungdch H2SO4 nhn c.p s: Nng H2SO4 sau khi trn l 3,5MBi 5: Trn 1/3 (l) dung dch HCl (dd A) vi 2/3 (l) dung dch HCl (dd B) c 1(l) dung dch HCl mi (dd C). Ly 1/10 (l) dd C tc dng vi dung dch AgNO3 d th thu c 8,61g kt ta.a) Tnh nng mol/l ca dd C.b) Tnh nng mol/l ca dd A v dd B. Bit nng mol/l dd A = 4 nng d mol/l dd B.p s: Nng mol/l ca dd B l 0,3M v ca dd A l 1,2M.Bi 6: Trn 200ml dung dch HNO3 (dd X) vi 300ml dung dch HNO3 (dd Y) c dung dch (Z). Bit rng dung dch (Z) tc dng va vi 7g CaCO3.a) Tnh nng mol/l ca dung dch (Z).b) Ngi ta c th iu ch dung dch (X) t dung dch (Y) bng cch thm H2O vo dung dch (Y) theo t l th tch: VH2O: Vdd(Y) = 3:1.Tnh nng mol/l dung dch (X) v dung dch (Y)? Bit s pha trn khng lm thay i ng k th tch dung dch.p s:a) CMdd(Z) = 0,28Mb) Nng mol/l ca dung dch (X) l 0,1M v ca dung dch (Y) l 0,4M.Bi7:trungho 50ml dungdch NaOH 1,2Mcn V(ml) dung dch H2SO4 30% (D = 1,222g/ml). Tnh V?p s: Th tch dung dch H2SO4 30% cn ly l 8,02 ml.38Bi 8:Cho 25gdung dch NaOH4% tcdng vi51gdungdch H2SO4 0,2M, c khi lng ring D = 1,02 g/ml. Tnh nng % cc cht sau phn ng.p s:- Nng % ca dung dch Na2SO4 l 1,87%- Nng % ca dung dch NaOH (d) l 0,26%Bi 9:Trn ln 100mldung dch NaHSO41M vi 100mldung dch NaOH 2M c dung dch A.a) Vit phng trnh ho hc xy ra.b) CcndungdchAththuchnhpnhngchtno? Tnh khi lng ca mi cht.p s: b) Khi lng cc cht sau khi c cn.- Khi lng mui Na2SO4 l 14,2g- Khi lng NaOH(cn d) l 4 gBi 10:Khi trungho100ml dungdchca2axitH2SO4vHCl bngdungdchNaOH, ri ccnththuc13,2gmui khan. Bit rng c trung ho 10 ml dung dch 2 axit ny th cn va 40ml dung dch NaOH 0,5M. Tnh nng mol/l ca mi axit trong dung dch ban u.ps: Nng mol/lca axit H2SO4l0,6Mv ca axit HCl l 0,8MBi 11: Tnhnngmol/l cadungdchH2SO4vdungdch NaOH bit rng:C 30mldung dch H2SO4c trung ho ht bi20mldung dch NaOH v 10ml dung dch KOH 2M.Ngcli:30ml dungdchNaOHctrunghohtbi 20ml dung dch H2SO4 v 5ml dung dch HCl 1M.ps:Nngmol/l caddH2SO4l0,7MvcaddNaOHl 1,1M.Hng dn gii bi ton nng bng phng php i s:Thd:TnhnngbanucadungdchH2SO4v dung dch NaOH bit rng:- Nu 3 lt dung dch NaOH vo 2 lt dung dch H2SO4 th sau phn ng dung dch c tnh kim vi nng 0,1M.- Nu 2 lt dung dch NaOH vo 3 lt dung dch H2SO4 th sau phn ng dung dch c tnh axit vi nng 0,2M.Bi giiPTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2OGi nng dung dch xt l x v nng dung dch axit l y th:* Trong trng hp th nht lng kim cn li trong dung dch l 390,1 . 5 = 0,5mol.Lng kim tham gia phn ng l: 3x - 0,5 (mol)Lng axt b trung ho l: 2y (mol)Theo PTP s mol xt ln hn 2 ln H2SO4Vy 3x - 0,5 = 2y.2 = 4y hay 3x - 4y = 0,5 (1)* Trong trng hp th 2 th lng a xt d l 0,2.5 = 1molLng a xt b trung ho l 3y - 1 (mol)Lng xt tham gia phn ng l 2x (mol). Cng lp lun nh trn ta -c:3y - 1 = 21. 2x = x hay 3y - x = 1 (2)T (1) v (2) ta c h phng trnh bc nht:' 1 35 , 0 4 3x yy xGii h phng trnh ny ta c x = 1,1 v y = 0,7.Vy, nng ban u ca dung dch H2SO4 l 0,7M ca dung dch NaOH l 1,1M.Bi 12: Tnhnngmol/l cadungdchNaOHvdungdch H2SO4. Bit nu ly 60ml dung dch NaOH th trung ho hon ton 20ml dung dch H2SO4. Nu ly 20ml dung dch H2SO4 tc dng vi 2,5g CaCO3 th mun trung ho lng axit cn d phi dng ht 10ml dung dch NaOH trn.ps:Nngmol/l caddH2SO4l1,5MvcaddNaOHl 1,0M.Bi 13: Tnhnngmol/l cadungdchHNO3vdungdch KOH. Bit- 20ml dung dch HNO3c trung ho ht bi 60ml dung dch KOH.- 20mldungdchHNO3saukhitc dng ht vi 2gCuO th c trung ho ht bi 10ml dung dch KOH.p s: Nng ca dung dch HNO3 l 3M v ca dung dch KOH l 1M.Bi 14: C 2 dung dch H2SO4 l A v B.a) Nu 2 dung dch A v B c trn ln theo t l khi lng 7:3 th thu c dung dch C c nng 29%. Tnh nng % ca dd A v dd B. Bit nng dd B bng 2,5 ln nng dd A.b) Ly 50ml dd C (D = 1,27g/ml) cho phn ng vi 200ml dd BaCl2 1M. Tnh khi lng kt ta v nng mol/l ca dd E cn li saukhi tchhtkt ta, gisthtchddthayi khng ng k.40Hng dn:a/ Gi s c 100g dd C. c 100g dd C ny cn em trn 70g dd A nng x% v 30g dd B nng y%. V nng % dd C l 29% nn ta c phng trnh:mH2SO4(trong dd C) = 10070 x + 10030 y = 29 (I)Theo bi ra th: y = 2,5x (II)Gii h (I, II) c: x% = 20% v y% = 50%b/ nH2SO4( trong 50ml dd C ) = Mm Cdd100%. = 98 . 100) 27 , 1 . 50 ( 29 = 0,1879 molnBaCl2 = 0,2 mol > nH2SO4. Vy axit phn ng htmBaSO4 = 0,1879 . 233 = 43,78gDung dch cn li sau khi tch ht kt ta c cha 0,3758 mol HCl v 0,2 0,1879 = 0,0121 mol BaCl2 cn d.Vy nng ca dd HCl l 1,5M v ca dd BaCl2 l 0,0484MBi 15: Trn dd A cha NaOH v dd B cha Ba(OH)2theo th tch bng nhau c dd C. Trung ho 100ml dd C cn ht 35ml dd H2SO4 2M v thu c 9,32g kt ta. Tnh nng mol/l ca cc dd A v B. Cn trn bao nhiu ml dd B vi 20ml dd A ho tan va ht 1,08g bt Al.p s:nH2SO4 = 0,07 mol; nNaOH = 0,06 mol; nBa(OH)2 = 0,04 mol.CM(NaOH) = 1,2M; CM(Ba(OH)2) = 0,8M.Cn trn 20ml dd NaOH v 10ml dd Ba(OH)2 ho tan ht 1,08g bt nhm.41Chuyn 4: Xc nh cng thc ho hcPhng php 1: Xc nh cng thc ho hc da trn biu thc i s. *Cch gii:- Bc 1: t cng thc tng qut.- Bc 2: Lp phng trnh(T biu thc i s)- Bc 3: Gii phng trnh -> Kt lun Cc biu thc i s thng gp.- Cho bit % ca mt nguyn t.- Cho bit t l khi lng hoc t l %(theo khi lng cc nguyn t). Cc cng thc bin i.- Cng thc tnh % ca nguyn t trong hp cht.CTTQ AxBy AxBy%A = AxByAMx M ..100%--> BA%% = y Mx MBA..- Cng thc tnh khi lng ca nguyn t trong hp cht.CTTQ AxBy AxBy mA = nA xB y.MA.x--> BAmm = y Mx MBA..Lu :- xc nh nguyn t kim loi hoc phi kim trong hp cht c th phi lp bng xt ho tr ng vi nguyn t khi ca kim loi hoc phi kim .- Ho tr ca kim loi (n): 1n4, vi n nguyn. Ring kim loi Fe phi xt thm ho tr 8/3.- Ho tr ca phi kim (n): 1 n 7, vi n nguyn.- Trongoxitcaphi kimthsnguyntphi kimtrong oxit khng qu 2 nguyn t.Bi tp p dng:Bi 1: Mt oxit nit(A) c cng thc NOx v c %N = 30,43%. Tm cng thc ca (A).p s: NO2Bi 2: Mt oxit st c %Fe = 72,41%. Tm cng thc ca oxit.p s: Fe3O4 Bi 3: Mt oxit ca kim loi M c %M = 63,218. Tm cng thc oxit.p s: MnO242Bi 4: Mt qung st c cha 46,67% Fe, cn li l S.a) Tm cng thc qung.b) T qung trn hy iu ch 2 kh c tnh kh.p s: a) FeS2b) H2S v SO2.Bi 5: Oxit ng c cng thc CuxOyv c mCu: mO= 4 : 1.Tm cng thc oxit.p s: CuOBi 6: Oxit ca kim loi M. Tm cng thc ca oxit trong 2 trng hp sau:a) mM : mO = 9 : 8b) %M : %O = 7 : 3p s: a) Al2O3 b) Fe2O3 Bi 7: Mt oxit (A) ca nit c t khi hi ca A so vi khng kh l 1,59. Tm cng thc oxit A.p s: NO2Bi 8: Mt oxit ca phi kim (X) c t khi hi ca (X) so vi hiro bng 22. Tm cng thc (X).p s: TH1: CO2TH2: N2OPhng php 2: Xc nh cng thc da trn phn ng. Cch gii:- Bc 1: t CTTQ- Bc 2: Vit PTHH.- Bc 3: Lp phng trnh ton hc da vo cc n s theo cch t.- Bc 4: Gii phng trnh ton hc. Mt s gi :- Vi cc bi ton c mt phn ng, khi lp phng trnh ta nn p dng nh lut t l.- Tng qut:C PTHH: aA+ bB-------> qC+ pD(1)Chun b: a b.MB q.22,4 cho: nA p nB p VC (l ) ktcTheo(1) ta c:43pu Ana. = pu BBmM b.. = CVq 4 , 22 .Bi tp p dng:Bi 1: t chy hon ton 1gam nguyn t R. Cn 0,7 lit oxi(ktc), thu c hp cht X. Tm cng thc R, X.p s: R l S v X l SO2 Bi 2: Kh ht 3,48 gam mt oxit ca kim loi R cn 1,344 lit H2 (ktc). Tm cng thc oxit.- y l phn ng nhit luyn.- Tng qut:Oxit kim loi A + (H2, CO, Al, C) ---> Kim loi A + (H2O, CO2, Al2O3, CO hoc CO2)- iu kin: Kim loi A l kim loi ng sau nhm.p s: Fe3O4Bi 3: Nung ht 9,4 gam M(NO3)nthu c 4 gam M2On. Tm cng thc mui nitratHng dn: - Phn ng nhit phn mui nitrat.- Cng thc chung: -----M: ng trc Mg---> M(NO2)n (r) + O2(k)M(NO3)3(r) -----t0------ -----M: ( t Mg --> Cu)---> M2On (r) + O2(k) + NO2(k) -----M: ng sau Cu------> M(r) + O2(k) + NO2(k) p s: Cu(NO3)2. Bi 4: Nung ht 3,6 gam M(NO3)n thu c 1,6 gam cht rn khng tan trong nc. Tm cng thc mui nitrat em nung.Hng dn: Theo ra, cht rn c th l kim loi hoc oxit kim loi. Gii bi ton theo 2 trng hp.Ch : TH: Rn l oxit kim loi.Phn ng: 2M(NO3)n (r) ----t----> M2Om (r) + 2nO2(k) + 22 m n O2(k) Hoc 4M(NO3)n (r) ----t----> 2M2Om (r) + 4nO2(k) + (2n m)O2(k)iu kin: 1nm3, vi n, m nguyn dng.(n, m l ho tr ca M )p s: Fe(NO3)2Bi 5: t chy hon ton 6,8 gam mt hp cht v c A ch thu -c 4,48 lt SO2(ktc) v 3,6 gam H2O. Tm cng thc ca cht A.p s: H2S44Bi 6: Ho tan hon ton 7,2g mt kim loi (A) ho tr II bng dung dch HCl, thu c 6,72 lit H2 (ktc). Tm kim loi A.p s: A l MgBi 7: Cho 12,8g mt kim loi R ho tr II tc dng vi clo va th thu c 27g mui clorua. Tm kim loi R.p s: R l CuBi 8: Cho 10g st clorua(cha bit ho tr ca st ) tc dng vi dungdchAgNO3ththuc22,6gAgCl(r)(khngtan). Hyxc nh cng thc ca mui st clorua.p s: FeCl2Bi 9: Ho tan hon ton 7,56g mt kim loi R cha r ho tr vo dung dch axit HCl, th thu c 9,408 lit H2 (ktc). Tm kim loi R.p s: R l AlBi 10: Ho tan hon ton 8,9g hn hp 2 kim loi A v B c cng ho tr II v c t l mol l 1 : 1 bng dung dch HCl dng dthu c 4,48 lit H2(ktc).Hi A, B l cc kim loi no trong s cc kim loi sau y: ( Mg, Ca, Ba, Fe, Zn, Be )p s:A v B l Mg v Zn.Bi 11: Ho tan hon ton 5,6g mt kim loi ho tr II bng dd HCl thu c 2,24 lit H2(ktc). Tm kim loi trn.p s: FeBi 12: Cho 4,48g mt oxit ca kim loi ho tr tc dng ht 7,84g axit H2SO4. Xc nh cng thc ca oxit trn.p s: CaOBi 13: ho tan 9,6g mt hn hp ng mol (cng s mol) ca 2 oxit kim loi c ho tr II cn 14,6g axit HCl. Xc nh cng thc ca 2 oxit trn. Bit kim loi ho tr II c th l Be, Mg, Ca, Fe, Zn, Ba.p s: MgO v CaOBi 14: Ho tan hon ton 6,5g mt kim loi A cha r ho tr vo dung dch HCl th thu c 2,24 lit H2(ktc). Tm kim loi A.p s: A l ZnBi 15: C mt oxit st cha r cng thc, chia oxit ny lm 2 phn bng nhau.a/ ho tan ht phn 1 cn dng 150ml dung dch HCl 1,5M.b/ Cho lung kh H2 d i qua phn 2 nung nng, phn ng xong thu c 4,2g st.45Tm cng thc ca oxit st ni trn.p s: Fe2O3 Bi 16: Kh hon ton 4,06g mt oxit kim loi bng CO nhit cao thnh kim loi. Dn ton b kh sinh ra vo bnh ng nc vi trong d, thy to thnh 7g kt ta. Nu ly lng kim loi sinh ra ho tan ht vo dung dch HCl dth thu c 1,176 lit kh H2(ktc). Xc nh cng thc oxit kim loi.Hng dn:Gi cng thc oxit l MxOy = amol. Ta c a(Mx +16y) = 4,06MxOy + yCO -----> xM + yCO2aayaxay(mol)CO2 + Ca(OH)2----> CaCO3+ H2Oayayay (mol)Ta c ay = s mol CaCO3 = 0,07 mol.---> Khi lng kim loi = M.ax = 2,94g.2M + 2nHCl ----> 2MCln+ nH2ax 0,5nax (molTa c: 0,5nax = 1,176 :22,4=0,0525molhaynax=0,105Lptl: naxMax0525 , 094 , 2=28.Vy M = 28n ---> Ch c gi tr n = 2 v M = 56 l ph hp.Vy M l Fe. Thay n = 2 ---> ax = 0,0525. Ta c: ayax= 07 , 00525 , 0 = 43 = yx----> x = 3 v y = 4. Vy cng thc oxit l Fe3O4.Chuyn 5: Bi ton v oxit v hn hp oxit Tnh cht:- Oxit baz tc dng vi dung dch axit.- Oxit axit tc dng vi dung dch baz.- Oxitlngtnhvatcdngvi dungdchaxit, vatc dng dung dch baz.- Oxit trung tnh: Khng tc dng c vi dung dch axit v dung dch baz.Cch lm:- Bc 1: t CTTQ- Bc 2: Vit PTHH.- Bc 3: Lp phng trnh ton hc da vo cc n s theo cch t.- Bc 4: Gii phng trnh ton hc.- Bc 5: Tnh ton theo yu cu ca bi.46-A - Ton oxit bazBi tp p dng:Bi 1: Cho 4,48g mt oxit ca kim loi ho tr tc dng ht 7,84g axit H2SO4. Xc nh cng thc ca oxit trn.p s: CaOBi 2: Ho tan hon ton 1 gam oxit ca kim loi R cn dng 25ml dung dch hn hp gm axit H2SO4 0,25M v axit HCl 1M. Tm cng thc ca oxit trn.p s: Fe2O3Bi 3: C mt oxit st cha r cng thc, chia oxit ny lm 2 phn bng nhau.a/ ho tan ht phn 1 cn dng150ml dung dch HCl 1,5M.b/ Cho lung kh H2 d i qua phn 2 nung nng, phn ng xong thu c 4,2g st.Tm cng thc ca oxit st ni trn.p s: Fe2O3Bi 4: Hotanhonton20,4goxitkimloi A, hotrIII trong 300ml dungdchaxitH2SO4ththuc68,4gmui khan. Tm cng thc ca oxit trn.p s:Bi 5: ho tan hon ton 64g oxit ca kim loi ho tr III cn va 800ml dung dch axit HNO3 3M. Tm cng thc ca oxit trn.p s:Bi 6: Khi ho tan mt lng ca mt oxit kim loi ho tr II vo mt l-ng va dung dch axit H2SO44,9%, ngi ta thu c mt dung dch mui c nng 5,78%. Xc nh cng thc ca oxit trn.Hng dn:t cng thc ca oxit l ROPTHH: RO+ H2SO4 ----> RSO4 + H2O(MR + 16) 98g(MR + 96)gGi s ho tan 1 mol (hay MR + 16)g ROKhi lng dd RSO4(5,87%) = (MR + 16) + (98 : 4,9).100 = MR + 2016C% = 201696++RRMM.100% = 5,87%Gii phng trnh ta c: MR = 24, kim loi ho tr II l Mg.p s: MgO47Bi 7: Ho tan hon ton mt oxit kim loi ho tr II bng dung dch H2SO414% va th thu c mt dung dch mui c nng 16,2%. Xc nh cng thc ca oxit trn.p s: MgOB - bi ton v oxit axitBi tp1: ChottkhCO2(SO2)vodungdchNaOH(hoc KOH) th c cc PTHH xy ra:CO2 + 2NaOH Na2CO3+H2O ( 1 )Sau khi s mol CO2 = s mol NaOH th c phn ng. CO2+ NaOH NaHCO3

( 2 ) H ng gii:xt t l s mol vit PTHH xy ra. tT= 2CONaOHnn - Nu T 1 th ch c phn ng ( 2 ) v c th d CO2.- Nu T 2 th ch c phn ng ( 1 ) v c th d NaOH.- Nu 1VCO2=1,568litv%VCO2= 15,68%Bi 3: Dn V lit CO2(ktc) vo 200ml dung dch Ca(OH)2 1M, thu -c 10g kt ta. Tnh v.p s:TH1: CO2 ht v Ca(OH)2 d. ---> VCO2 = 2,24 lit.TH2: CO2 d v Ca(OH)2 ht ---->VCO2 = 6,72 lit.Bi 4: Cho m(g) kh CO2 sc vo 100ml dung dch Ca(OH)2 0,05M, thu c 0,1g cht khng tan. Tnh m.p s:49TH1: CO2 ht v Ca(OH)2 d. ---> mCO2 = 0,044gTH2: CO2 d v Ca(OH)2 ht ---->mCO2 = 0,396gBi 5: Phi t bao nhiu gam cacbon khi cho kh CO2to ra trong phn ng trn tc dng vi 3,4 lit dung dch NaOH 0,5M ta c 2 mui vi mui hiro cacbonat c nng mol bng 1,4 ln nng mol ca mui trung ho.p s:Vthtchdungdchkhngthayi nntlvnng cng chnh l t l v s mol. ---> mC = 14,4g.Bi 6: Cho4,48litCO2(ktc)i qua190,48ml dungdchNaOH 0,02% c khi lng ring l 1,05g/ml. Hy cho bit mui no c to thnh v khi lng lf bao nhiu gam.p s: Khi lng NaHCO3 to thnh l: 0,001.84 = 0,084gBi7: Thi2,464 lit kh CO2vo mt dung dch NaOH th c 9,46g hn hp 2 mui Na2CO3 v NaHCO3. Hy xc nh thnh phn khi lngcahnhp2mui . Numunchthucmui NaHCO3 th cn thm bao nhiu lt kh cacbonic na.p s: 8,4g NaHCO3 v 1,06g Na2CO3. Cn thm 0,224 lit CO2.Bi 8: t chy 12g C v cho ton b kh CO2 to ra tc dng vi mt dung dch NaOH 0,5M. Vi th tch no ca dung dch NaOH 0,5M th xy ra cc trng hp sau:a/ Ch thu c mui NaHCO3(khng d CO2)?b/ Ch thu c mui Na2CO3(khng d NaOH)?c/ Thu c c 2 muivinng molca NaHCO3bng 1,5 ln nng mol ca Na2CO3?Trong trng hp ny phi tip tc thm bao nhiu lit dung dch NaOH 0,5M na c 2 mui c cng nng mol.p s:a/ nNaOH = nCO2 = 1mol ---> Vdd NaOH 0,5M = 2 lit.b/ nNaOH = 2nCO2= 2mol ---> Vdd NaOH 0,5M = 4 lit.c/ t a, b ln lt l s mol ca mui NaHCO3 v Na2CO3.Theo PTHH ta c:nCO2 = a + b = 1mol (I)V nng mol NaHCO3 bng 1,5 ln nng mol Na2CO3 nn.Va =1,5Vb ---> a = 1,5b (II)Gii h phng trnh (I, II) ta c: a = 0,6 mol, b = 0,4 molnNaOH = a + 2b = 0,6 + 2 x 0,4 = 1,4 mol ---> Vdd NaOH 0,5M = 2,8 lit.Gi x l s mol NaOH cn thm v khi ch xy ra phn ng.NaHCO3+NaOH --->Na2CO3+ H2O x(mol)x(mol) x(mol)50nNaHCO3 (cn li) = (0,6 x) molnNa2CO3 (sau cng) = (0,4 + x) molV bi cho nng mol 2 mui bng nhau nn s mol 2 mui phi bng nhau.(0,6 x) = (0,4 + x) ---> x = 0,1 mol NaOHVy s lit dung dch NaOH cn thm l: Vdd NaOH 0,5M = 0,2 lit.Bi 9: Sc x(lit) CO2 (ktc) vo 400ml dung dch Ba(OH)2 0,5M th thu c 4,925g kt ta. Tnh x.p s:TH1: CO2 ht v Ca(OH)2 d. ---> VCO2 = 0,56 lit.TH2: CO2 d v Ca(OH)2 ht ---->VCO2 = 8,4 lit.C - Ton hn hp oxit.Cc bi ton vn dng s mol trung bnh v xc nh khong s mol ca cht. 1/ i vi cht kh. (hn hp gm c 2 kh)Khi lng trung bnh ca 1 lit hn hp kh ktc:MTB= VV M V M4 , 222 1 2 1+Khi lng trung bnh ca 1 mol hn hp kh ktc:MTB= VV M V M2 2 1 1+Hoc: MTB=nn n M n M ) (1 2 1 1 +(n l tng s mol kh trong hn hp)Hoc: MTB=1) 1 (1 2 1 1x M x M + (x1l % ca kh th nht)Hoc:MTB=dhh/kh x . Mx2/ i vi cht rn, lng. MTB ca hh=hhhhnmTnh cht 1: MTB ca hhc gi tr ph thuc vo thnh phn v lng cc cht thnh phn trong hn hp.Tnh cht 2: MTB ca hh lun nm trong khong khi lng mol phn t ca cc cht thnh phn nh nht v ln nht.Mmin FeCl2 + H2Cu+HCl ----> Khng phn ng.Cng thc 2: Kim loi phn ng vi axit loi 2:Kim loi + Axit loi 2 ----->Mui + H2O + Sn phm kh.c im:- Phn ng xy ra vi tt c cc kim loi (tr Au, Pt).- Mui c ho tr cao nht(i vi kim loi a ho tr)Bi tp p dng:Bi 1: Ho tan ht 25,2g kim loi R trong dung dch axit HCl, sau phn ng thu c 1,008 lit H2 (ktc). Xc nh kim loi R.p s:Bi2: Ho tan hon ton 6,5g mt kim loi A cha r ho tr vo dung dch axit HCl, th thu c 2,24 lit H2(ktc). Xc nh kim loi A.p s: A l Zn.Bi 3: Cho 10g mt hn hp gm Fe v Cu tc dng vi dung dch axit HCl, th thu c 3,36 lit kh H2 (ktc). Xc nh thnh phn % v khi lng ca mi kim loi trong hn hp u.p s: % Fe = 84%,% Cu = 16%.Bi 4:Cho1hnhpgmAl vAgphnngvi dungdchaxit H2SO4 thu c 5,6 lt H2 (ktc). Sau phn ng th cn 3g mt cht rn khng tan. Xc nh thnh phn % theo khi lng cu mi kim loi trong hn hp ban u.p s: % Al = 60% v % Ag = 40%.Bi 5: Cho 5,6g Fe tc dng vi 500ml dung dch HNO30,8M. Sau phn ng thu c V(lit) hn hp kh A gm N2O v NO2 c t khi so vi H2 l 22,25 v dd B.a/ Tnh V (ktc)?b/ Tnh nng mol/l ca cc cht c trong dung dch B.Hng dn:Theo bi ra ta c:nFe = 5,6 : 56 = 0,1 molnHNO3 = 0,5 . 0,8 = 0,4 molMhh kh = 22,25 . 2 = 44,5t x, y ln lt l s mol ca kh N2O v NO2.PTHH xy ra:8Fe + 30HNO3 ----> 8Fe(NO3)3 + 3N2O + 15H2O (1) 8mol3mol8x/3 xFe + 6HNO3 ----->Fe(NO3)3 + 3NO2 + 3H2O (2) 551mol3moly/3yT l th tch cc kh trn l:Gi a l thnh phn % theo th tch ca kh N2O.Vy (1 a) l thnh phn % ca kh NO2. Ta c: 44a + 46(1 a) = 44,5 a = 0,75 hay % ca kh N2O l 75% v ca kh NO2l 25%T phng trnh phn ng kt hp vi t l th tch ta c: x = 3y (I)---> y = 0,012 v x = 0,036 8x/3 + y/3 = 0,1 (II) Vy th tch ca cc kh thu c ktc l:VN2O = 0,81(lit) v VNO2= 0,27(lit)Theo phng trnh th:Smol HNO3(phn ng)=10nN2O+2nNO2=10.0,036+2.0,012= 0,384 molS mol HNO3 (cn d) = 0,4 0,384 = 0,016 molS mol Fe(NO3)3 = nFe = 0,1 molVy nng cc cht trong dung dch l:CM(Fe(NO3)3) = 0,2MCM(HNO3)d = 0,032MBi 6: ho tan 4,48g Fe phi dng bao nhiu ml dung dch hn hp HCl 0,5M v H2SO4 0,75M.Hng dn: Gi s phi dng V(lit) dung dch hn hp gm HCl 0,5M v H2SO4 0,75MS mol HCl = 0,5V (mol)S mol H2SO4 = 0,75V (mol)S mol Fe = 0,08 molPTHH xy ra:Fe + 2HCl ---> FeCl2 + H2Fe + H2SO4 ---> FeSO4 + H2Theo phng trnh ta c: 0,25V + 0,75V = 0,08---> V = 0,08 : 1 = 0,08 (lit)Bi 7: ho tan 4,8g Mg phi dng bao nhiu ml dung dch hn hp HCl 1,5M v H2SO4 0,5M.a/ Tnh th tch dung dch hn hp axit trn cn dng.b/ Tnh th tch H2 thu c sau phn ng ktc.p s:a/ Vhh dd axit = 160ml.b/ Th tch kh H2 l 4,48 lit.56Bi 8: Ho tan 2,8g mt kim loi ho tr (II) bng mt hn hp gm 80ml dung dch axit H2SO4 0,5M v 200ml dung dch axit HCl 0,2M. Dung dch thu c c tnh axit v mun trung ho phi dng 1ml dung dch NaOH 0,2M. Xc nh kim loi ho trII em phn ng.Hng dn:Theo bi ra ta c:S mol ca H2SO4 l 0,04 molS mol ca HCl l 0,04 molS mol ca NaOH l 0,02 molt R l KHHH ca kim loi ho tr IIa, b l s mol ca kim loi R tc dng vi axit H2SO4 v HCl.Vit cc PTHH xy ra.Sau khi kim loi tc dng vi kim loi R. S mol ca cc axit cn li l:S mol ca H2SO4 = 0,04 a (mol)S mol ca HCl = 0,04 2b(mol)Vit cc PTHH trung ho:T PTP ta c: S mol NaOH phn ng l: (0,04 2b) + 2(0,04 a) = 0,02---> (a + b) = 0,1 : 2 = 0,05Vy s mol kim loi R = (a + b) = 0,05 mol---> MR = 2,8 : 0,05 = 56 v R c ho tr II ---> R l Fe.Bi 9: Chia 7,22g hn hp A gm Fe v R (R l kim loi c ho tr khng i) thnh 2 phn bng nhau:- Phn 1: Phn ng vi dung dch HCl d, thu c 2,128 lit H2(ktc)- Phn 2: Phn ng vi HNO3, thu c 1,972 lit NO(ktc)a/ Xc nh kim loi R.b/ Tnh thnh phn % theo khi lng mi kim loi trong hn hp A.Hng dn:a/ Gi 2x, 2y (mol) l s mol Fe, R c trong hn hp A --> S mol Fe, R trong 1/2 hn hp A l x, y.Vit cc PTHH xy ra:Lp cc phng trnh ton hc;mhh A = 56.2x + 2y.MR(I)nH2= x + ny/2 = 0,095(II)nNO = x + ny/3 = 0,08(III)Gii h phng trnh ta c: MR = 9n (vi n l ho tr ca R)Lp bng: Vi n = 3 th MR = 27 l ph hp. Vy R l nhm(Al)b/ %Fe = 46,54% v %Al = 53,46%.Chuyn 7: 57axit tc dng vi baz(Bi ton hn hp axit tc dng vi hn hp baz)* Axit n: HCl, HBr, HI, HNO3. Ta cnH+ = nA xit * Axit a: H2SO4, H3PO4, H2SO3.Ta c nH+ =2nA xit hoc nH+ = 3nA xit* Baz n: KOH, NaOH, LiOH.Ta c nOH =2nBaZ * Baz a: Ba(OH)2, Ca(OH)2.Ta c nOH =2nBaZ PTHH ca phn ng trung ho: H++OH - H2O*L u : trong mt hn hp m c nhiu phn ng xy ra th phn ng trung ho c u tin xy ra trc.Cch lm:- Vit cc PTHH xy ra.- t n s nu bi ton l hn hp.- Lp phng trnh ton hc- Gii phng trnh ton hc, tm n.- Tnh ton theo yu cu ca bi.Lu :- Khi gpdungdchhnhpccaxittcdngvi hnhpcc baz th dng phng php t cng thc tng ng cho axit v baz.- t th tch dung dch cn tm l V(lit)- Tm V cn nh: nHX = nMOH.Bi tp:Cho t t dung dch H2SO4 vo dung dch NaOH th c cc phn ng xy ra:Phn ng u tin to ra mui trung ho trc.H2SO4+ 2NaOH Na2SO4 + H2O( 1 ) Sau khi s mol H2SO4 = s mol NaOH th c phn ngH2SO4+NaOH NaHSO4+H2O ( 2 ) H ng gii:xt t l s mol vit PTHH xy ra.tT= 4 2SO HNaOHnn - Nu T 1 th ch c phn ng (2) v c th d H2SO4.- Nu T 2 th ch c phn ng (1) v c th d NaOH.- Nu 1 MA = 23(Na) v MB = 39(K)mNaOH = 2,4g v mKOH = 5,6g.62Chuyn 8:axit tc dng vi mui1/ Phn loi axitGm 3 loi axit tc dng vi mui.a/ Axit loi 1:- Thng gp l HCl, H2SO4long, HBr,..- Phn ng xy ra theo c ch trao i.b/ Axit loi 2:- L cc axit c tnh oxi ho mnh: HNO3, H2SO4c.- Phn ng xy ra theo c ch phn ng oxi ho kh.c/ Axit loi 3:- L cc axit c tnh kh.- Thng gp l HCl, HI, H2S.- Phn ng xy ra theo c ch phn ng oxi ho kh.2/ Cng thc phn ng.a/ Cng thc 1:Mui + Axit ---> Mui mi + Axit mi.iu kin: Sn phm phi c:- Kt ta.- Hoc c cht bay hi(kh).- Hoc cht in li yu hn.c bit: Cc mui sunfua ca kim loi k t Pb tr v sau khng phn ng vi axit loi 1.V d: Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2 (k)

BaCl2 + H2SO4 ---> BaSO4(r) + 2HClb/ Cng thc 2:Mui + Axit loi 2 ---> Mui + H2O + sn phm kh.iu kin:- Mui phi c tnh kh.- Mui sinh ra sau phn ng th nguyn t kim loi trong mui phi c ho tr cao nht.Ch : C 2 nhm mui em phn ng.- Vi cc mui: CO32-, NO3-, SO42-, Cl- .+ iu kin: Kim loi trong mui phi l kim loi a ho tr v ho tr ca kim loi trong mui trc phi ng khng cao nht.- Vi cc mui: SO32-, S2-, S2-.+ Phn ng lun xy ra theo cng thc trn vi tt c cc kim loi.c/ Cng thc 3:Thng gp vi cc mui st(III). Phn ng xy ra theo quy tc 2.(l phn ng oxi ho kh)2FeCl3 + H2S ---> 2FeCl2 + S(r) + 2HCl.Ch :63Bi tp: Cho t t dung dch HCl voNa2CO3 (hoc K2CO3) th c cc PTHH sau:Giai on 1 Ch c phn ng.Na2CO3 + HCl NaHCO3+ NaCl( 1 ) x (mol) x molx molGiai on 2Ch c phn ng NaHCO3+HCl d NaCl+ H2O+ CO2

( 2 ) x x x molHoc ch c mt phn ng khi s mol HCl = 2 ln s mol Na2CO3.Na2CO3+ 2HCl 2NaCl +H2O+CO2 ( 3 )i vi K2CO3cng tng t.H ng gii:xt t l s mol vit PTHH xy ra tT= 3 2CO NaHClnn - Nu T 1 th ch c phn ng (1) v c th d Na2CO3.- Nu T 2 th ch c phn ng (3) v c th d HCl.- Nu 1 2MCl + H2O + CO2 Theo PTHH ta c:S mol M2CO3 = s mol CO2 > 2,016 : 22,4 = 0,09 mol---> Khi lng mol M2CO3 < 13,8 : 0,09 = 153,33 (I)Mt khc: S mol M2CO3 phn ng = 1/2 s mol HCl < 1/2. 0,11.2 = 0,11 mol---> Khi lng mol M2CO3 = 13,8 : 0,11 = 125,45 (II)T (I, II) --> 125,45 < M2CO3 < 153,33 ---> 32,5 < M < 46,5 v M l kim loi kim 66---> M l Kali (K)Vy s mol CO2 = s mol K2CO3 = 13,8 : 138 = 0,1 mol ---> VCO2 = 2,24 (lit)b/ Gii tng t: ---> V2 = 1,792 (lit)Bi 3: Ho tan CaCO3 vo 100ml hn hp dung dch gm axit HCl v axit H2SO4 th thu c dung dch A v 5,6 lit kh B (ktc), c cn dung dch A th thu c 32,7g mui khan. a/ Tnh nng mol/l mi axit trong hn hp dung dch ban u.b/ Tnh khi lng CaCO3 dng.Bi 4: Cho 4,2g mui cacbonat ca kim loi ho tr II. Ho tan vo dung dch HCl d, th c kh thot ra. Ton b lng kh c hp th vo 100ml dung dch Ba(OH)2 0,46M thu c 8,274g kt ta. Tm cng thc ca mui v kim loi ho tr II.p s:- TH1 khi Ba(OH)2 d, th cng thc ca mui l: CaCO3 v kim loi ho tr II l Ca.- TH2 khi Ba(OH)2 thiu, th cng thc ca mui l MgCO3 v kim loi ho tr II l Mg.Bi 5: Cho 1,16g mui cacbonat ca kim loi R tc dng ht vi HNO3, thu c 0,448 lit hn hp G gm 2 kh c t khi hi so vi hiro bng 22,5. Xc nh cng thc mui (bit th tch cc kh o ktc).Hng dn:Hn hp G gm c kh CO2 v kh cn li l kh X.C dhh G/ H2= 22,5 --> MTB ca hh G = 22,5 . 2 = 45M MCO2= 44 < 45 ---> Mkh X > 45. nhn thy trong cc kh ch c NO2 v SO2 c khi lng phn t ln hn 45. Trong trng hp ny kh X ch c th l NO2.t a, b ln lt l s mol ca CO2 v NO2.Ta c hnhh G = a + b = 0,02a = 0,01 MTB hh G = b ab a++46 44 = 45 b = 0,01PTHH: R2(CO3)n + (4m 2n)HNO3 ---> 2R(NO3)m + (2m 2n)NO2 + nCO2 + (2m n)H2O.2MR + 60n2m 2n 1,16g0,01 molTheo PTHH ta c: 16 , 160 2 n MR + = 01 , 02 2 n m----> MR = 116m 146nLp bng: iu kin 1 n m 4 n 1 2 2 3 367m 3 2 3 3 4MR56Ch c cp nghim n = 2, m = 3 --> MR = 56 l ph hp. Vy R l FeCTHH: FeCO3Bi 6: Cho 5,25g mui cacbonat ca kim loi M tc dng ht vi HNO3, thu c 0,336 lit kh NO v V lit CO2. Xc nh cng thc mui v tnh V. (bit th tch cc kh c o ktc)p s: Gii tng t bi 3 ---> CTHH l FeCO3 Bi 7: Ho tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dch HCl d thu c 0,672 lt kh CO2 (ktc). Tnh thnh phn % s mol mi mui trong hn hp.Bi giiCc PTHH xy ra:CaCO3 + 2HCl CaCl2 + CO2 + H2O (1)MgCO3 + 2HCl MgCl2 + CO2 + H2O (2)T (1) v (2) nhh = nCO2 = 4 , 22672 , 0 = 0,03 (mol)Gi x l thnh phn % s mol ca CaCO3 trong hn hp th (1 - x) l thnh phn % s mol ca MgCO3.Ta cM2 mui = 100x + 84(1 - x) = 03 , 084 , 2 x = 0,67 % s mol CaCO3 = 67% ; % s mol MgCO3 = 100 - 67 = 33%.Bi 8: Ho tan 174 gam hn hp gm 2 mui cacbonat v sunfit ca cng mt kim loi kim vo dung dch HCl d. Ton b kh thot ra c hp th ti thiu bi 500 ml dung dch KOH 3M.a/ Xc nh kim loi kim.b/ Xc nh % s mol mi mui trong hn hp ban u.Bi giicc PTHH xy ra:M2CO3 + 2HCl 2MCl + CO2 + H2O (1)M2SO3 + 2HCl 2MCl + SO2 + H2O (2)Ton b kh CO2 v SO2 hp th mt lng ti thiu KOH sn phm l mui axit.CO2 + KOH KHCO3 (3)SO2 + KOH KHSO3 (4)T (1), (2), (3) v (4)suy ra: n 2 mui = n 2 kh = nKOH = 10003 . 500 = 1,5 (mol)M2 mui = 5 , 1174 = 116 (g/mol) 2M + 60 Ba(AlO2)2 + 3BaCl2 + 4H2OBi tp: Cho t t dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) vo dung dch Al2(SO4)3th c cc PTHH sau.6NaOH +Al2(SO4)3 2Al(OH)3+3Na2SO4 ( 1 )NaOHd + Al(OH)3 NaAlO2 +2H2O( 2 )8NaOH+Al2(SO4)3 2NaAlO2+3Na2SO4 + 4H2O ( 3 )V:3Ba(OH)2+ Al2(SO4)3 2Al(OH)3 +3BaSO4

( 1 )Ba(OH)2 d+2Al(OH)3 Ba(AlO2)2 +4H2O( 2 )4Ba(OH)2+Al2(SO4)3 Ba(AlO2)2 + 3BaSO4+ 4H2O( 3 ) Ngc li: Cho t t dung dch Al2(SO4)3 vo dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) th c PTHH no xy ra?Al2(SO4)3+ 8NaOH 2NaAlO2+3Na2SO4 +4H2O (3 )/Al2(SO4)3+4Ba(OH)2 Ba(AlO2)2 + 3BaSO4+ 4H2O(3 )// Mt s phn ng c bit:NaHSO4 (dd) + NaAlO2 + H2O Al(OH)3 + Na2SO4 NaAlO2 + HCl + H2O Al(OH)3 + NaClNaAlO2 + CO2 + H2O Al(OH)3 + NaHCO3 Bi tp p dng: Bi 1: Cho 200 ml dd gm MgCl2 0,3M; AlCl3 0,45; HCl

0,55M tc dng hon ton vi V(lt) dd C cha NaOH 0,02 M v Ba(OH)2 0,01 M. Hy tnh th tich V(lt) cn dng thu c kt ta ln nht v 70lng kt ta nh nht. Tnh lng kt ta . (gi s khi Mg(OH)2 kt ta ht th Al(OH)3 tan trong kim khng ng k)H ng dn gi i :nHCl = 0,11mol ; nMgCl2 = 0,06 mol ; nAlCl3 = 0,09 mol.Tng s mol OH- =0,04 V(*)Cc PTHH xy ra:H+ +OH- H2O (1)Mg2++OH- Mg(OH)2 (2)Al3+ +3OH- Al(OH)3 (3)Al(OH)3 +OH- AlO2- +2H2O (4)Tr ng hp 1 : c kt ta ln nht th ch c cc phn ng (1,2,3 ).Vy tng s mol OH- dng l: 0,11 + 0,06 x 2 + 0,09 x 3 = 0,5 mol (**)T (*) v (**) ta c Th tch dd cn dng l: V = 0,5 : 0,04 = 12,5 (lit)mKt ta = 0,06 x 58 +0,09 x 78 = 10,5 gTr ng hp 2 : c kt ta nh nht th ngoi cc p (1, 2, 3) th cn c p (4) na.Khi lng Al(OH)3 tan ht ch cn li Mg(OH)2, cht rn cn li l: 0,06 x 58 = 3,48 gV lng OH- cn dng thm cho p (4) l 0,09 mol.Vy tng s mol OH- tham gia p l: 0,5 + 0,09 = 0,59 molTh tch dd C cn dng l: 0,59/ 0,04 = 14,75 (lit)Bi 2: Cho 200ml dung dch NaOH vo 200g dung dch Al2(SO4)3 1,71%. Sau phn ng thu c 0,78g kt ta. Tnh nng mol/l ca dung dch NaOH tham gia phn ng.p s:TH1: NaOH thiuS mol NaOH = 3s mol Al(OH)3 = 3. 0,01 = 0,03 mol ---> CM NaOH = 0,15MTH2: NaOH d ---> CM NaOH = 0,35MBi 3: Cho 400ml dung dch NaOH 1M vo 160ml dung dch hn hp cha Fe2(SO4)3 0,125M v Al2(SO4)3 0,25M. Sau phn ng tch kt ta em nung n khi lng khng i c cht rn C.a/ Tnh mrn C.b/ Tnh nng mol/l ca mui to thnh trong dung dch.p s:a/ mrn C = 0,02 . 160 + 0,02 . 102 = 5,24gb/ Nng ca Na2SO4 = 0,18 : 0,56 = 0,32M v nng ca NaAlO2 = 0,07M71Bi 4: Cho 200g dung dch Ba(OH)2 17,1% vo 500g dung dch hn hp (NH4)2SO4 1,32% v CuSO4 2%. Sau khi kt thc tt c cc phn ng ta thu c kh A, kt ta B v dung dch C.a/ Tnh th tch kh A (ktc)b/ Ly kt ta B ra sch v nung nhit cao n khi lng khng i th c bao nhiu gam rn?c/ Tnh nng % ca cc cht trong C.p s:a/ Kh A l NH3 c th tch l 2,24 litb/ Khi lng BaSO4 = 0,1125 . 233 = 26,2g v mCuO = 0,0625 . 80 = 5gc/ Khi lng Ba(OH)2 d = 0,0875 . 171 = 14,96gmdd = Tng khi lng cc cht em trn - mkt ta - mkhmdd = 500 + 200 26,21 6,12 1,7 = 666gNng % ca dung dch Ba(OH)2 = 2,25%Bi 5: Cho mt mu Na vo 200ml dung dch AlCl3 thu c 2,8 lit kh (ktc) v mt kt ta A. Nung A n khi lng khng i thu c 2,55 gam cht rn. Tnh nng mol/l ca dung dch AlCl3 .Hng dn:mrn: Al2O3 --> s mol ca Al2O3 = 0,025 mol ---> s mol Al(OH)3 = 0,05 mols mol NaOH = 2s mol H2 = 0,25 mol.TH1: NaOH thiu, ch c phn ng.3NaOH + AlCl3 ---> Al(OH)3 + 3NaClKhng xy ra v s mol Al(OH)3 to ra trong phn ng > s mol Al(OH)3 cho.TH2: NaOH d, c 2 phn ng xy ra.3NaOH + AlCl3 ---> Al(OH)3 + 3NaCl0,150,050,05mol 4NaOH+ AlCl3 ---> NaAlO2 + 3NaCl + H2O(0,25 0,15)0,025Tng s mol AlCl3 phn ng 2 phng trnh l 0,075 mol ----> Nng ca AlCl3 = 0,375MBi 6: Cho 200ml dung dch NaOH x(M) tc dng vi 120 ml dung dch AlCl3 1M, sau cng thu c 7,8g kt ta. Tnh tr s x?p s:- TH1: Nng AlCl3 = 1,5M- TH2: Nng AlCl3 = 1,9MBi 7: Cho 9,2g Na vo 160ml dung dch A c khi lng ring 1,25g/ml cha Fe2(SO4)3 0,125M v Al2(SO4)3 0,25M. Sau khi phn ng kt thc ngi ta tch kt ta v em nung nng n khi lng khng i thu c cht rn.a/ Tnh khi lng cht rn thu c.72b/ Tnh nng % ca dung dch mui thu c.p s:a/ mFe2O3 = 3,2g v mAl2O3 = 2,04g.b/ Nng % ca cc dung dch l: C%(Na2SO4) = 12,71% v C%(NaAlO2) = 1,63%73Chuyn 10: Hai dung dch mui tc dng vi nhau.Cng thc 1:Mui + Mui ---> 2 Mui miiu kin:- Mui phn ng: tan hoc tan t trong nc.- Sn phm phi c cht:+ Kt ta.+ Hoc bay hi+ Hoc cht in li yu. H2OV d: BaCl2 + Na2SO4 ---> BaSO4 + 2NaClCng thc 2:Cc mui ca kim loi nhm, km, st(III) ---> Gi chung l mui APhn ng vi cc mui c cha cc gc axit: CO3, HCO3, SO3, HSO3, S, HS, AlO2 ---> Gi chung l mui B.Phn ng xy ra theo quy lut:Mui A + H2O ----> Hiroxit (r) + AxitAxit + Mui B ----> Mui mi + Axit mi.V d: FeCl3 phn ng vi dung dch Na2CO3 2FeCl3 + 6H2O -----> 2Fe(OH)3 + 6HCl6HCl + 3Na2CO3 ---> 6NaCl+ 3CO2 + 3H2OPT tng hp:2FeCl3 + 3H2O + 3Na2CO3 ---> 2Fe(OH)3 + 3CO2 + 6NaCl.Cng thc 3:Xy ra khi gp st, phn ng xy ra theo quy tc 2.V d: AgNO3 + Fe(NO3)2 ---> Fe(NO3)3 + Ag.Bi 1: Cho 0,1mol FeCl3 tc dng ht vi dung dch Na2CO3 d, thu -c cht kh B v kt ta C. em nung C n khi lng khng i thu c cht rn D. Tnh th tch kh B (ktc) v khi lng cht rn D.p s: - Th tch kh CO2 l 3,36 lit - Rn D l Fe2O3 c khi lng l 8gBi 2: Trn 100g dung dch AgNO3 17% vi 200g dung dch Fe(NO3)2 18% thu c dung dch A c khi lng ring (D = 1,446g/ml). Tnh nng mol/l ca dung dch A.p s:- Dung dch A gm Fe(NO3)2 0,1 mol v Fe(NO3)3 0,1 mol.74- Nng mol/l ca cc cht l: CM(Fe(NO3)2) = CM(Fe(NO3)3) = 0,5MBi 3: Cho 500ml dung dch A gm BaCl2 v MgCl2 phn ng vi 120ml dung dch Na2SO4 0,5M d, thu c 11,65g kt ta. em phn dung dch c cn thu c 16,77g hn hp mui khan. Xc nh nng mol/l cc cht trong dung dch.Hng dn:Phn ng ca dung dch A vi dung dch Na2SO4.BaCl2 + Na2SO4 ----> BaSO4 + 2NaCl0,05 0,05 0,05 0,1 molTheo (1) s mol BaCl2 trng dd A l 0,05 mol v s mol NaCl = 0,1 mol.S mol Na2SO4 cn d l 0,06 0,05 = 0,01 molS mol MgCl2 = 955 , 58 . 1 , 0 142 . 01 , 0 77 , 16 = 0,1 mol.Vy trong 500ml dd A c 0,05 mol BaCl2 v 0,1 mol MgCl2.---> Nng ca BaCl2 = 0,1M v nng ca MgCl2 = 0,2M.Bi 4: Cho 31,84g hn hp NaX, NaY (X, Y l 2 halogen 2 chu k lin tip) vo dung dch AgNO3 d, thu c 57,34g kt ta. Tm cng thc ca NaX, NaY v tnh thnh phn % theo khi lng ca mi mui.Hng dn;* TH1: X l Flo(F) --> Y l Cl. Vy kt ta l AgCl.Hn hp 2 mui cn tm l NaF v NaClPTHH: NaCl + AgNO3 ---> AgCl + NaNO3 Theo PT (1) th nNaCl = nAgCl = 0,4 mol ---> %NaCl = 73,49% v %NaF = 26,51%.* TH2: X khng phi l Flo(F).Gi Na Xl cng thc i din cho 2 mui.PTHH:Na X +AgNO3 --->Ag X + NaNO3 (23 +X ) (108 +X )31,84g57,34gTheo PT(2) ta c: 31,84X 23 + = 34 , 57108 X + --->X= 83,13Vy hn hp 2 mui cn tm l NaBr v NaI ---> %NaBr = 90,58% v %NaI = 9,42%Bi 5: Dung dch A cha 7,2g XSO4 v Y2(SO4)3. Cho dung dch Pb(NO3)2 tc dng vi dung dch A (va ), thu c 15,15g kt ta v dung dch B.a/ Xc nh khi lng mui c trong dung dch B.b/ Tnh X, Y bit t l s mol XSO4 v Y2(SO4)3 trong dung dch A l 2 : 1 v t l khi lng mol nguyn t ca X v Y l 8 : 7.Hng dn:75PTHH xy ra:XSO4 + Pb(NO3)2 ---> PbSO4 + X(NO3)2 xx x molY2(SO4)3 + 3Pb(NO3)2 ---> 3PbSO4 + 2Y(NO3)3 y 3y 2yTheo PT (1, 2) v cho ta c:mhh mui = (X+96)x + (2Y+3.96)y = 7,2 (I) ---> X.x + 2Y.y = 2,4Tng khi lng kt ta l 15,15g --> S mol PbSO4 = x + 3y = 15,15/303 = 0,05 mol Giih ta c: mmui trong dd B = 8,6g(c th p dng nh lut bo ton khi lng)Theo ra v kt qu ca cu a ta c:x : y = 2 : 1X : Y = 8 : 7x + 3y = 0,05X.x + 2.Y.y = 2,4---> X l Cu v Y l FeVy 2 mui cn tm l CuSO4 v Fe2(SO4)3.Bi 6: C 1 lit dung dch hn hp gm Na2CO3 0,1M v (NH4)2CO3 0,25M. Cho 43g hn hp BaCl2 v CaCl2 vo dung dch trn. Sau khi cc phn ng kt thc thu c 39,7g kt ta A v dung dch B.a/ Chng minh mui cacbonat cn d.b/ Tnh thnh phn % theo khi lng cc cht trong A.c/ Cho dung dch HCl d vo dung dch B. Sau phn ng c cn dung dch v nung cht rn cn li ti khi lng khng i thu c rn X. Tnh thnh phn % theo khi lng rn X.Hng dn: chng minh mui cacbonat d, ta chng minh mmui phn ng < mmui ban u Ta c: S mol Na2CO3 = 0,1 mol v s mol (NH4)2CO3 = 0,25 mol.Tng s mol CO3 ban u = 0,35 molPhn ng to kt ta:BaCl2 + CO3 ----> BaCO3 + 2ClCaCl2 + CO3 ---> CaCO3 + 2ClTheo PTHH ta thy: Tng s mol CO3 phn ng = (43 39,7) : 11 = 0,3 mol.Vy s mol CO3 phn ng < s mol CO3 ban u.---> s mol CO3 db/ V CO3 d nn 2 mui CaCl2 v BaCl2 phn ng ht.mmui kt ta = 197x + 100y = 39,7Tng s mol Cl phn ng = x + y = 0,3----> x = 0,1 v y = 0,2Kt ta A c thnh phn: %BaCO3 = 49,62% v %CaCO3 = 50,38%c/ Cht rn X ch c NaCl. ---> %NaCl = 100%.7677Chuyn 11: bi ton hn hp kim loi.Thng gp di dng kim loi phn ng vi axit, baz, mui v vi nc.Dy hot ng ho hc ca kim loi.K, Na, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au(Khi No May A Zp St Phi Hi Cc Bc Vng) ngha:K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPt+ O2: nhit thng nhit cao Kh phn ng K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtTc dng vi ncKhng tc dng vi nc nhit thngK BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtTc dng vi cc axit thng thng gii phng Hidro Khng tc dng.K BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtKim loi ng trc y kim loi ng sau ra khi muiK BaCaNa Mg Al ZnFeNiSnPbH CuAgHgAuPtH2, CO khng kh c oxit kh c oxit cc kim loi ny nhit caoCh :- Cc kim loi ng trc Mg phn ng vi nc nhit thng to thnh dd Kim v gii phng kh Hidro.- Tr Au v Pt, cc kim loi khc u c th tc dng vi HNO3 v H2SO4 c nhng khng gii phng Hidro.78 ngha ca dy hot ng ho hcKNa Ba CaMgAlZnFeNiSn PbH CuHg AgPt Au- Dy c sp xp theo chiu gim dn tnh hot ng ho hc (t tri sang phi)- Mt s kim loi va tc dng c vi axit v vi nc:K, Na, Ba, CaKim loi + H2O ----> Dung dch baz + H2 - Kim loi va tc dng vi axit, va tc dng vi baz:(Be), Al, Zn, Cr 2A + 2(4 n)NaOH + 2(n 2)H2O ---> 2Na4 nAO2 + nH2 V d: 2Al + 2NaOH + 2H2O ----> 2NaAlO2+ 3H22Al + Ba(OH)2 + 2H2O ----> Ba(AlO2)2 + 3H2Zn + 2NaOH --->Na2ZnO2 + H2Zn + Ba(OH)2 ---> BaZnO2 + H2- Kim loi ng trc H tc dng vi dung dch axit HCl, H2SO4 long to mui v gii phng H2.Kim loi + Axit ----> Mui + H2Lu : Kim loi trong mui c ho tr thp (i vi kim loi a ho tr)- K t Mg tr i kim loi ng trc y c kim loi ng sau ra khi mui ca chng. theo quy tc:Cht kh mnh + cht oxi ha mnh cht oxi ho yu + cht kh yu.Lu : nhng kim loi u dy (kim loi tc dng c vi nc) th khng tun theo quy tc trn m n xy ra theo cc bc sau: Kim loi kim (hoc kim th)+H2O Dung dch baz +H2 Sau : Dung dch baz+dung dch mui Mui mi + Baz mi(*)iu kin(*): Cht to thnh phi c t nht 1 cht kt ta (khng tan).VD: cho Ba vo dung dch CuSO4. Trc tin:Ba+ 2H2O Ba(OH)2 +H2

Ba(OH)2+CuSO4 Cu(OH)2 + BaSO4 c bit: Cu + 2FeCl3 ---> CuCl2 + 2FeCl2 Cu + Fe2(SO4)3 ---> CuSO4 + 2FeSO4 Cc bi ton vn dng s mol trung bnhv xc nh khong s mol ca cht.1/ i vi cht kh. (hn hp gm c 2 kh)79Khi lng trung bnh ca 1 lit hn hp kh ktc:MTB= VV M V M4 , 222 1 2 1+Khi lng trung bnh ca 1 mol hn hp kh ktc:MTB= VV M V M2 2 1 1+Hoc: MTB=nn n M n M ) (1 2 1 1 +(n l tng s mol kh trong hn hp)Hoc: MTB=1) 1 (1 2 1 1x M x M + (x1l % ca kh th nht)Hoc:MTB=dhh/kh x . Mx2/ i vi cht rn, lng. MTB ca hh=hhhhnmTnh cht 1: MTB ca hhc gi tr ph thuc vo thnh phn v lng cc cht thnh phn trong hn hp.Tnh cht 2: MTB ca hh lun nm trong khong khi lng mol phn t ca cc cht thnh phn nh nht v ln nht.Mmin x = 0.Vy y = 9,86 : 65 = 0,1517 molGi s hn hp ch Mg ---> y = 0Vy x = 9,86 : 24 = 0,4108 mol0,1517 < nhh kim loi < 0,4108V x > 0 v y > 0 nn s mol axit tham gia phn ng vi kim loi l:0,3034 < 2x + 2y < 0,8216 nhn thy lng axit dng < 0,86 mol.Vy axit d --> Do Zn v Mg phn ng ht.Sau khi ho tan ht trong dung dch c.x mol MgX2 ; y mol ZnX2 ; 0,86 2(x + y) mol HX v 0,43 mol SO4.Cho dung dch tc dng vi dung dch baz. HX+ ROH ---> RX + H2O.0,86 2(x + y) 0,86 2(x + y) mol MgX2+2ROH ---->Mg(OH)2+2RX x 2xx molZnX2+ 2ROH----> Zn(OH)2+ 2RXy2y y molTa c nROH phn ng = 0,86 2(x + y) + 2x + 2y = 0,86 molVy nROH d = 0,96 0,86 = 0,1molTip tc c phn ng xy ra:Zn(OH)2 +2ROH ----> R2ZnO2 + 2H2Ob:y 0,1 molP: y12y1 molcn: y y10,1 2y1 mol( iu kin: y y1)Phn ng to kt ta. Ba(OH