PHNG PHAP GIAI TOAN HOA HU C

HONG THI VIT TRNG H S PHM H NI 2 H BCH KHOA NNG - 01695316875

PHNG PHP GII TON HA HU C & V CMT S LU V HP CHT HU C

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A-GII NHANH BI TON TRC NGHIM V C

1/ H2SO4 ( 2H+ + SO42- ( H2( HCl ( H+ + Cl-

VD1:Cho 2,81 g hn hp Fe2O3, ZnO, MgO tc dng va vi 500 ml dung dch H2SO4 0,1M.

Khi lng mui sunfat to ra trong dung dch l:

Gii: nH2SO4 =0,05 = n SO42- --->nH += 0,1

2H+ + O2- = H2O

0,1 0,05 mol

m mui = m oxit m O(trong oxit) +m gc axit =2,81 0,05.16 +0,05.96 =6,81 gamVD2:Cho 8 g hn hp bt kim lai Mg va Fe tc dng ht vi dung dch HCl thy thot ra 5,6 lit H2 ktc. Khi lng mui to ra trong dung dch l

Gii: nH2 =0,25 ---> nHCl =nCl = 0,25.2 =0,5. m mui =8 + 0,5.35,5=25,75 gam

VD3Cho 11 gam hn hp 2 kim loi tan hon ton trong HCl d thy c 8,96 lt kh thot ra (kc) v dd X, c cn dd X th khi lng hn hp mui khan thu c l (gam):

Gii: nH2 =0,4 ---> nHCl =nCl- = 0,4.2 =0,8. m mui =kl kim loi +kl ion Cl-=11+0,8.35,3=39,4 gam2/ Axt + Ocid baz ( k c ocid baz khng tan)

VD1: Fe2O3 ( a mol

Phn ng dung dch HCl

FexOy ( b mol

nO2- = 3a+ by ( 2H+ + O2- ( H2O

6a+2yb ( 3a+yb

VD2:Ho tan 2,4 g mt oxit st va 90ml ddHCl 1M. Cng thc ca oxit st ni trn l:

Gi CTPT oxit st l:FexOy( a mol

nHCl =0,09mol

2H+ + O2- ( H2O

0,09 0,045 mol

nO2- =ay = 0,045 (1)

56a + 16ya = 2,4 (2)

xa =0,03 ( x:y =2:3 ( CTPT l Fe2O33/ Axt + Baz ( k c baz khng tan)

VD: Dung dch H2SO4 phn ng vi hn hp: Fe(OH)3 amol, Al(OH)3 bmol, Cu(OH)2 cmol

nOH- = 3a+3b+2c = nH+

4/ Axt + Kim Loi ( Mui v gii phng kh H2

VD: Na( H ( H2

Al ( 3H( 3/2 H2

VD1:Cho 8.3 g hn hp Al,Fe tc dng ht vi dung dch HCl. Sau phn ng khi lng HCl tng thm 7.8 g. Khi lng mi mui to ra trong dung dch v kl mi kim loi trong hh

Gii;n H2 =(8,3-7,8 ):2 =0,25

3/2a+b = 0,25

27a +56 b= 8,3---> a=b= 0,1 mol

VD 2: Cho m gam nhm,Magi, st vo 250 ml dd X cha hh axt HCl 1M,H2SO4 0,5 M, thu c 0,2375 mol kh H2 v dd Y.Tnh pH ca dd Y.

Gii:n H+b=0,25 +0,25.0,5.2 = 0,5

nH+ p = 0,2375.2=0,475

nH+ d =0,025 mol ( CH+=0,1 ( pH =15/ CO, H2 kh oxt kim loi sau Al to Kim loi + CO2 , H2O

VD: Hn hp gm CuO ( amol

Fe2O3 ( bmol + CO ( nO(trong oxt) = a+3b

CO + O ( CO2

a+3b ( a+3b ( a+3b

VD:Mt hn hp X gm Fe2O3, FeO v MgO c khi lng l 4,24 g trong c 1,2 g MgO. Khi cho X phn ng vi CO d (phn ng hon ton), ta c cht rn A v hn hp CO v CO2. Hn hp ny khi qua nc vi trong cho ra 5 g kt ta. Xc nh khi lng Fe2O3, FeO trong hn hp X. Gii: m 2 oxit st l: 4,24 1,2 =3,04 gam---> 160 a +72 b =3,04

n CO2 = n O(trong 2 oxit st) = 0,05 ----> 3a +b = 0,05 ---> a=0,01 ; b= 0,02

6/ Phn ng gia 2 ion ch xy ra khi sn phm c cht kt ta, d bay hi, in li yu.

VD1: Ca2+ + CO32- ( CaCO3( 2H+ + CO32- ( H2O + CO2( 2H+ + S2- ( H2S( Na+ + NO3- x khng xy ra

VD2 : Dung dch cha amol AlCl3, bmol CuCl2, cmol NaCl phn ng dung dch AgNO3 d thu dmol kt ta. Mi lin h a,b,c,d

nCl- = 3a+2b+c

( nAgCl ( = nCl- = nAg+phn ng = 3a+2b+c = d

Ag+ + Cl- ( AgCl(7/ nh lut bo ton khi lng:

mghn hp kim loi + m1 g dung dch HCl thu c m2 g dung dch A, m3 g kh B v m4 g rn khng tan.

Ta c : m + m1 = m2 + m3 + m4 ( m2 = m + m1 m3 m4

8/ Bo ton in tch:

Trong 1 dung dch : Tng in tch dng = tng in tch m

VD1: Dung dch cha amol Al3+, bmol Ca2+, cmol SO42-, dmol Cl-.

Ta co: 3a + 2b = 2c + d

VD2: mg hn hp Fe, Mg, Zn phn ng dung dch HCl d thu (m+m1) gam mui.

mg hn hp trn phn ng dung dch HCl thu bao nhiu gam mui?

. mmui clorua = mkim loi + mCl- ( mCl- = m1g ( nCl- = mol

. Bo ton in tch: 2Cl- SO42- ( 2.nSO42- = nCl-)

(

. mui sunfat = m + x 96

VD3:Cho m g hn hp Cu, Zn, Fe tc dng vi dung dch HNO3 long, d thu c dung dch A. C cn dung dch A thu c (m+62) gam mui khan. Nung hn hp mui khan trn n khi lng khng i thu c cht rn c khi lng l:

Gii: n NO3- =62:62 = 1mol ---> 2NO3- -------> O2- . n O2- =0,5 mol

1 mol 0.5 mol

m oxit = m kim loi + m O = m + 0,5.16 =( m + 8 ) gam

Vd 4: Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo dung dch HNO3 va , thu c dung dch X ch cha 2 mui sunf v kh NO duy nht. Gi tr a l:

Gii: dd gm:0,12 mol Fe3+, 2a mol Cu2+ ,(0,24+a) mol SO42- .

p dng lbt in tch: 3.0,12 +2.2a =2(0,24 +a)( a=0,06

9/ Bo ton nguyn t :

VD1: Cho 1mol CO2 phn ng 1,2mol NaOH thu mg mui. Tnh m?

. = 1,2 ( sn phm to 2 mui

. Gi CT 2 mui NaHCO3 ( amol BT nguyn t Cacbon: a+b = 1 a= 0,08mol

Na2CO3 ( bmol BT nguyn t Natri: a+2b = 1,2 ( b = 0,02mol

VD2 : Hn hp A gm FeO a mol, Fe2O3 b mol phn ng vi CO t0 cao thu c hn hp B gm: Fe cmol, FeO dmol, Fe2O3 e mol, Fe3O4 f mol. Mi quan h gia a,b,c,d

Ta c : nFe (trong A) = nFe (trong B)

(VD 3:

Hp th hon ton 0,12 mol SO2 vo 2,5 lt dd Ba(OH)2 a mol/l thu c 0,08 mol kt ta. g tr ca a l bao nhiu.

Gii:n BaCO3 =0,08 ( n C cn li to Ba(HCO3)2= 0,04 ( nBa(HCO3)2 =0,02

(n Ba =n Ba(OH)2 =0,08 + 0,02 =0,1 ( CM =0,1/2,5 =0,04 M

VD 4:Ha tan 5,6 gam Fe vo dd H2SO4 d thu dc dd X.dd X phn ng va vi V lt dd KMnO4 0,5 M gi tr ca V l?

Gii:

nFe = nFe2+ =0,1 mol ( nMn2+ = 0,1.1/5=0,02(lBT electron) ( V = 0,02:0.5 =0,04 lt10/ Bo ton Electron :

. Ch s dng i vi phn ng oxi ha kh

. Phng php: + Xc nh cht kh + xc nh cht oxi ha

+ Vit 2 qu trnh + nh lut bo ton Electron : ne cho = ne nhnVD : 0,3 mol FexOy phn ng vi dd HNO3 d thu c 0,1mol kh NO. Xc nh FexOy.

Gii : xFe2y/x ( 3x-2y) ( xFe+3 nFexOy = 0,3 ( nFe2y/x = 0,3x x = 3

0,3x ( 0,3(3x-2y) ( y = 4 hoc x=y=1

N+5 + 3e ( N+2 0,3.(3x 2y) = 0,3 ( 3x 2y = 1

0,30,1

Vy CTPT : Fe3O4 hoc FeO

11/ Xc nh CTPT cht :

VD : 1 oxt ca st c % mFe chim 70%. Xc nh CTPT ca oxt.

Gi CT ca oxt l: FexOy (

EMBED Equation.3

EMBED Equation.3( Fe2O3B. HIDROCACBON:

CT chung: CxHy (x1, y2x+2). Nu l cht kh k thng hoc k chun: x4.

Hoc: CnH2n+2-2k, vi k l s lin kt, k 0.

I- DNG 1: Hn hp gm nhiu hidrocacbon thuc cng mt dy ng ng.

(PP1:Gi CT chung ca cc hidrocacbon (cng dy ng ng nn k ging nhau)

- Vit phng trnh phn ng

- Lp h PT gii

, k.

- Gi CTTQ ca cc hidrocacbon ln lt l ... v s mol ln ln lt l a1,a2.

Ta c: +

+ a1+a2+ =nhhTa c k: n1