chuyen de ltdh hoa

Chuyn Ha hc Nm hc 2011-2012

1

LI M U

Vn gii bi tp ho hc l mt vic lm thng xuyn v quen thuc i vi hc sinh. N

khng ch gip hc sinh cng c phn l thuyt c bn hc m cn pht huy tch cc sng to

trong qu trnh hc tp ca hc sinh.

Tuy nhin vn gii bi tp ho hc thi i hc i vi nhiu hc sinh cn gp nhiu kh

khn, nhiu em cha nh hng, cha tm ra phng php thch hp, cha phn dng nn cn c

mt s kh khn trong khi thi i hc.

Nhm b sung, nng cao k nng gii bi tp ho hc v gip cho hc sinh nhn dng cc

bi ton thi i hc, chng ti su tm, sp xp v gii thiu mt s dng ton thi i hc trong

cc nm qua, ng thi a thm mt s bi tp tham kho gip hc sinh c cc nh hng

gii bi tp trong cc k thi i hc.

Chc chn rng s c nhiu thiu st, knh mong cc ng nghip gp b sung ti liu

cng hon thin hn.

LP BI DNG KIN THC 10 11- 12 V LUYN THI I HCC: Dng Sn, Ha Chu, Ha Vang, Nng

Thc s. Nguyn ThoDownload ti liu, thi, bi ging ti : http://diendan.shpt.info ----T: 0905434602

Ngi thc hin: Nguyn ng Khoa 2

DNG I : VIT NG PHN I.MT S LU

*Xc nh gi tr k da vo cng thc CnH2n+2-2kOz (z0) *Xc nh nhm chc : -OH, -COOH, -CH=O, -COO- *Xc nh gc hirocacbon no, khng no, thm, vng, h *Vit mch C theo th t mch C gim dn. Tm li : T CTTQ k = ? Mch C v nhm chc ng phn (cu to v khng gian)

II.BI TP Cu 1: Tng s hp cht hu c no, n chc, mch h, c cng cng thc phn t C5H10O2, phn ng c vi dung dch NaOH nhng khng c phn ng trng bc l

A. 4. B. 5. C. 8. D. 9. Gii: Ta c k=1 c 1 lin kt phn ng c vi dung dch NaOH Axit hay este no h. Nhng khng c phn ng trng bc Khng phi l este ca axit fomic C-COO-C-C-C C-COO-C(CH3) C C-C-COO-C-C C-C-C-COO-C C-C-C-C-COOH C-C(CH3)-C-COOH C-C-C(CH3)-COOH C-C(CH3)2-COOH

Chn C Cu 2: S ng phn este ng vi cng thc phn t C4H8O2 l

A. 4. B. 6. C. 5. D. 2. Gii: Ta c k=1 este no h. HCOOC-C-C HCOOC(CH3)-C C-COOC-C C-C-COOC

Chn A Cu 3: Cho m gam hn hp gm hai cht hu c n chc, mch h tc dng va vi dung dch cha 11,2 gam KOH, thu c mui ca mt axit cacboxylic v mt ancol X. Cho ton b X tc dng ht vi Na thu c 3,36 lt kh H2 ( ktc). Hai cht hu c l

A. mt este v mt ancol. B. hai axit. C. hai este. D. mt este v mt axit.

KOH11,2

Gii : Tc dng vi KOH to thnh mui axit hu c v mt ancol c este

n = 0,256

mol 2Hancol

3,36n = 2n = 2. 0,322, 4

mol

KOHancoln > n => ban u c ancol.

Vy, hn hp u c 1 este v 1 ancol. Chn A Cu 4 : Cho glixerol phn ng vi hn hp axit bo gm C17H35COOH v C15H31COOH, s loi trieste c to ra ti a l

A. 4. B. 3. C. 6. D. 5. Cu 5 : S ng phn hirocacbon thm ng vi cng thc phn t C8H10 l

A. 4. B. 2. C. 5. D. 3. Cu 6: Cho cc cht sau: CH2=CH-CH2-CH2-CH=CH2, CH2=CH-CH=CH-CH2-CH3, CH3-C(CH3)=CH-CH3, CH2=CH-CH2-CH=CH2. S cht c ng phn hnh hc l

A. 4. B. 2. C. 1. D. 3.




Cu 7: Ba hirocacbon X, Y, Z l ng ng k tip, khi lng phn t ca Z bng 2 ln khi lng phn t ca X. Cc cht X, Y, Z thuc dy ng ng.

A. anken. B. ankin. C. ankaien. D. ankan. Gii : Gi khi lng phn t ca ba hidrocacbon ln lt l MX, MY, MZ. X, Y, Z l ng ng k tip nhau MZ = MX + 28 (1). Theo bi ra ta c: MZ = 2MX (2) T (1) v (2) ta c MX = 28. X l C2H4 => anken

Chn A Cu 8: C bao nhiu ancol bc 2, no, n chc, mch h l ng phn cu to ca nhau m phn

t ca chng c phn trm khi lng cacbon bng 68,18%? A. 5. B. 2. C. 3. D. 4.

Gii : Cng thc tng qut ancol no, n chc : CnH2n+1OH Theo bi ra ta c:

12n 68,18 n = 514n +18 100

Cng thc Ancol l C5H11OH

Cc ng phn bc 2 : C-C-C-C(OH)-C C-C-C(OH)-C-C C-C(CH3)-C(OH)-C Chn C Cu 9: Mt hp cht X cha ba nguyn t C, H, O c t l khi lng mC : mH : mO = 21 : 2 : 4. Hp cht X c cng thc n gin nht trng vi cng thc phn t. S ng phn cu to thuc loi hp cht thm ng vi cng thc phn t ca X l

A. 4. B. 5. C. 6. D. 3. Gii : mC : mH : mO = 21 : 2 : 4 nC : nH : nO = 7 : 8 : 1 CTPT: C7H8O S ng phn thm CH3C6H4OH (3), C6H5OCH3, C6H5CH2OH Chn B. Cu 10: C bao nhiu cht hu c mch h dng iu ch 4-metylpentan-2-ol ch bng phn ng cng H2 (xc tc Ni, to)?

A. 3. B. 5. C. 2. D. 4. Gii : *4-metylpentan-2-ol l: C-C(CH3)-C-C(OH)-C Mch C trong cht ban u l C-C(CH3)-C-C-C

*Cht phn ng vi H2 to ancol bc 2 ch c th l: ancol khng no hay xeton *C=C(CH3)-C-C(OH)-C C-C(CH3)=C-C(OH)-C C-C(CH3)-C-CO-C C=C(CH3)-C-CO-C

Chn D Cu 11 : Vit cc ng phn ng vi cng thc phn t C4H6O2? *Nhn xt : k=2 nn c 2 lin kt hoc 1 lin kt v 1 vng no hoc 2 vng no.

1.ng phn n chc mch h a. Axit CH3-CH=CH-COOH(2) , CH2=CH-CH2-COOH , CH2=C(CH3)-COOH b. Este HCOOCH=CH-CH3(2) , HCOOCH2-CH=CH2 , HCOOC(CH3)=CH2CH3COOCH=CH2 , CH2=CHCOOCH3 2.ng phn n chc mch vng: a. Axit

b. Este




1. ng phn a chc mch h: a. Xeton: CH3-CO-CO-CH3

b.Ete

c.Ancol

d.Andehit 2. ng phn tp chc mch h a. 1-OH; 1-CHO

b. 1-CO-; 1-CHO CH3COCH2CHO c. 1-O-; 1-CHO

CH3OCH=CH-CHO, CH2=CHOCH2-CHO, CH2=CH-CH2OCHO, CH3-CH=CHOCHO

d. 1-CO-; 1-OH HO-CH2-CO-CH=CH2 5. ng phn tp chc mch vng




DNG II : BI TON T CHY I.MT S LU

*t chy CxHy : -CnH2n +2 th n(H2O): n(CO2) > 1 v ngc li, ng thi n(H2O) - n(CO2) =

n(CnH2n+2) -CnH2n th n(H2O): n(CO2) = 1 v ngc li.

-CnH2n -2 th n(H2O): n(CO2) < 1 v n(CO2) - (H2O) = n(CnH2n -2) *t chy CxHyOz cng tng t nh trn *Nu z =1 th n(O) = n(CxHyOz). Ta c th suy ra z =2...

II.BI TP Cu 1: Cho hn hp X gm hai ancol a chc, mch h, thuc cng dy ng ng. t chy hon ton hn hp X, thu c CO2 v H2O c t l mol tng ng l 3 : 4. Hai ancol l

A. C3H5(OH)3 v C4H7(OH)3. B. C2H5OH v C4H9OH. C. C2H4(OH)2 v C4H8(OH)2. D. C2H4(OH)2 v C3H6(OH)2.

Gii: n(H2O) : n(CO2) = 4:3 >1 => ancol no Gi CTC 2 ancol l: OHC xnn 22 S : OHCOOHC nnxnn 2222 )1(

43

1

nn => n =3

X l hn hp ancol a chc p n C

Cu 2: Khi t chy hon ton m gam hn hp hai ancol no, n chc, mch h thu c V lt kh CO2 ( ktc) v a gam H2O. Biu thc lin h gia m, a v V l: A. m = 2a - V/22,4 B. m = 2a - V/11,2 C. m = a + V/5,6 D. m = a - V/5,6

Gii: n(ancol) = n(H2O) - n(CO2) = 4,2218Va

(mol)

m = m(C) + m(H) + m(O) = 12.n(CO2) + 2.n(H2O) +16.n(ancol)

= 12.4,22

V + 2.18a + 16.(

4,2218Va

) = a - 6,5

V (g)

p n D Cu 3: t chy hon ton mt hp cht hu c X, thu c 0,351 gam H2O v 0,4368 lt kh CO2 ( ktc). Bit X c phn ng vi Cu(OH)2 trong mi trng kim khi un nng. Cht X l

A. CH2=CH-CH2-OH. B. C2H5CHO C. CH3COCH3. D. O=CH-CH=O. Gii: X phn ng vi Cu(OH)2/OH- => X c nhm chc CHO

n(CO2) : n(H2O) = 18351,0:

4,224368,0 = 1:1

S nguyn t H = 2C p n B Cu 4: t chy hon ton m gam hn hp 3 ancol n chc, thuc cng dy ng ng, thu c 3,808 lt kh CO2 (ktc) v 5,4 gam H2O. Gi tr ca m l

A. 5,42. B. 5,72. C. 4,72. D. 7,42.

Gii: n(H2O)= 184,5 =0,3 (mol); n(CO2) = 4,22

808,3 =0,17 (mol)




n(H2O) > n(CO2) Hn hp ancol no n(ancol) = n(H2O) - n(CO2) = 0,13 (mol)

m(ancol) = m(C) + m(H) + m(O) = 12.n(CO2) + n(H2O) + 16.n(ancol) = 12.0,17 + 2.0,3 + 16.0,13 = 4,72 (g)

p n C Cu 5: Hn hp X gm axit panmitic, axit stearic v axit linoleic. trung ho m gam X cn 40 ml dung dch NaOH 1M. Mt khc, nu t chy hon ton m gam X th thu c 15,232 lt kh CO2 (ktc) v 11,7 gam H2O. S mol ca axit linoleic trong m gam hn hp X l

A. 0,015. B. 0,010. C. 0,020. D. 0,005. Gii: X: CnH2nO2 (axit panmitic; axit stearic); CmH2m-4O2(axit linoleic)

n(axit linoleic) = 0,5.[n(CO2) n(H2O)] = 0,5( 015,0)187,11

4,22232,15

(mol)

p n A Cu 6: t chy hon ton mt lng hn hp X gm 2 ancol (u no, a chc, mch h, c cng s nhm -OH) cn va V lt kh O2, thu c 11,2 lt kh CO2 v 12,6 gam H2O (cc th tch kh o ktc). Gi tr ca V l

A. 14,56. B. 15,68. C. 11,20. D. 4,48.

Gii: n(H2O) = 186,12 = 0,7 (mol); n(CO2) = 4,22

2,11 =0,5(mol)

n(ancol) = n(H2O) - n(CO2) = 0,2 (mol) Gi ancol l OHC xnn 22

75

1

nn => n = 2,5

x < n , x>1 => x = 2 n(O2) = 0.5.n(O) = 0,5.[n(H2O) + 2.n(CO2) - 2.n(ancol)] = 0,65 (mol) V= 0,65.22,4 = 14,56 (l) p n A

Cu 7: t chy hon ton m gam hn hp X gm ba ancol (n chc, thuc cng dy ng ng) thu c 8,96 lt kh CO2 (ktc) v 11,7 gam H2O. Mt khc, nu un nng m gam X viH2SO4 c th tng khi lng ete ti a thu c l

A. 7,85 gam. B. 7,40 gam. C. 6,50 gam. D. 5,60 gam.

Gii: n(H2O) = 187,11 = 0,65 (mol); n(CO2) = 4,22

96,8 =0,4(mol)

n(ancol) = n(H2O) - n(CO2) = 0,25(mol) m(ancol) = m(C) + m(H) + m(O) = 12.0,4 + 2.0,65 + 16.0,25 = 10,1 (g) Bo ton O => n(H2O p to ete) = n(ancol) = 0,2 (mol) m(ete) = m(ancol) m(H2O) = 10,1 0,25.18 = 5,6 (g)

p n D




DNG III : BI TON POLIME I.MT S LU

*Thng thng khi vit phn ng trng hp v ng trng hp hay trng ngng ngi ta thng vit theo t l 1:1> *Trong gii bi tp c th khng theo t l V d: xC4H6 + C3H3N (C4H6)x(C3H3N)y (C4H6)x(C3H3N)y +x Br2 (C4H6)xBr2x(C3H3N)y *S mt xch = m / M(mi mt xch) *Cht PVC ch cha lin kt n nn tham gia phn ng th, nhng coa su izopren khi tham gia phn ng lu ha li tham gia phn ng th d cn c 1 lin kt i.

II.BI TP Cu 1: Thu phn 1250 gam protein X thu c 425 gam alanin. Nu phn t khi ca X bng100.000 vC th s mt xch alanin c trong phn t X l

A. 328. B. 382. C. 453. D. 479. Gii : T l v khi lng ca alanin trong phn t X l: 425:1250= 0,34 Khi lng alanin trong phn t X l: 100000.0,34 = 34000 (vC) Mt xch Alanin: -NH-CH(CH3)-CO- (M=71vC) S mt xch alanin trong phn t X l: 34000:71=479 Cu 2: Clo ho PVC thu c mt polime cha 63,96% clo v khi lng, trung bnh 1 phn t clo phn ng vi k mt xch trong mch PVC. Gi tr ca k l

A. 4. B. 6. C. 5. D. 3. Gii : C2nH3nCln +Cl2 C2nH3n-1Cln +1 + HCl

T l % v khi lng ca clo: %96,635,345,62

1005,35)1(

nn

n=3 Chn D

Cu 3: Khi lng ca mt on mch t nilon-6,6 l 27346 vC v ca mt on mch t capronl 17176 vC. S lng mt xch trong on mch nilon-6,6 v capron nu trn ln lt l

A. 121 v 114. B. 113 v 114. C. 113 v 152. D. 121 v 152. Gii : T nilon- 6,6: (-NH-[CH2]2CH(COOH)-CH2-NH-CO-[CH2]4-CO-)n (M=242n vC) T capron: (-NH-[CH2]5-CO-)n (M= 113n vC) s mt xch trong nilon-6,6 : 27346: 242= 113 s mt xch trong capron: 17176: 113= 152

Cu 4: C 5,668 g cao su buna-S phn ng va ht vi 3,462 g brom trong CCl4. T l mt xch butaien v stiren trong cao su buna-S l

A. 1 : 3. B. 1 : 2. C. 2 : 3. D. 3 : 5 Gii : (-CH2-CH=CH-CH2-)x(-CH(C6H5)-CH2-)y + xBr2 4CCl

(-CH2-CHBr-CHBr-CH2-)x(-CH(C6H5)-CH2-)y

S mol mt xch butadien trong buna-S bng s mol Br2 bng 160462,3 (mol)

S mol mt xch stiren : 104

54160462,3668,5

(mol)




t l x:y = 160462,3 :

104

54160462,3668,5

=1:2

Cu 5: Khi t chy hon ton mt polime X (to thnh do ng trng hp 2,3-imetyl butaien v acrilo nitin (CH2=CH-CN) vi lng O2 va thy to thnh mt hn hp kh nng p sut xc nh cha 57,69% CO2 v V. T l 2 loi monome l

A. 3/5 B. 3/3 C. 1/3 D. 3/2 Gii : (C6H10)x(C3H3N)y

otO ,2 (6x + 3y) CO2 + (5x + 1,5y) H2O + 0,5N2 T l % v th tch ca CO2 trong hn hp kh: yyxyx

yx5,0)5,15()36(

100)36(

=57,69%

x:y =1:3 Cu 6: Mt loi cao su cha 2% S. Hi c bao nhiu mt xch izopren th c mt cu ni ddiissunfua S-S-?

A.46. B.64. C.80. D.64. Gii: S : (C5H8)n +S2 C5nH8n -2S2

T l % isunfua 6268

10064n

=2% n=46




DNG IV : S DNG PHNG PHP ION TRONG GII BI TP

I.MT S LU

*nh lut bo ton in tch : Trong dung dch tng s in tch m = Tng s in tch dng *Tng khi lng cc cht tan trong dung dch = Tng khi lng ca cation v anion *Khi lng ca mui = Khi lng ca kim loi + khi lng ca caction to mui

II.BI TP Cu 1:Trn 100 ml dung dch (gm Ba(OH)2 0,1M v NaOH 0,1M) vi 400ml dung dch (gm H2SO4 0,0375M v HCl 0,0125M), thu c dung dch X. Gi tr pH ca dung dch X l

A. 7. B. 2. C. 1. D. 6.

Gii : Phn ng: H+ + OH- H2O nOH- = 0,1 0,1 2 + 0,1 0,1=0,03 mol nH+ = 0,4 0,0375 2 + 0,4 0,0125 = 0,035 mol nH+ = nOH- = 0,03 mol nH+ d = 0,035 0,03 = 0,005 mol

[H+] = 0,005/ (0,4 + 0,1) = 0,01 = 10-2 (M) => pHX = 2 Chn p n B

Cu 2:Dung dch X c cha: 0,07 mol Na+; 0,02 mol SO42- v x mol OH- . Dung dch Y c cha ClO4-; NO3-; y mol H+; tng s mol ClO4- v NO3- l 0,04. Trn X v Y c 100 ml dung dch Z. Dung dch Z c pH (b qua s in li ca H2O)

A. 1. B. 12. C. 13. D. 2. Gii : p dng nh lut bo ton in tch cho 2 dung dch X v Y ta c: Dung dch X: 0,07 1 = 0,02 2 + x 1 => x = 0,03 mol Dung dch Y : y 1 = 0,04 1 => y = 0,04 mol Phng trnh : H+ + OH- H2O nH+ = nOH- = 0,03 mol nH+ d = 0,04 0,03 = 0,01 mol [H+] = 0,01/0,1 = 0,1= 10-1 (M) => pHZ = 1

Chn p n A Cu 3: Dung dch X cha cc ion: Fe3+, SO42-, NH4+, Cl-. Chia dung dch X thnh hai phn bng nhau:- Phn mt tc dng vi lng d dung dch NaOH, un nng thu c 0,672 lt kh ( ktc) v 1,07 gam kt ta;

- Phn hai tc dng vi lng d dung dch BaCl2, thu c 4,66 gam kt ta. Tng khi lng cc mui khan thu c khi c cn dung dch X l (qu trnh c cn ch c

nc bay hi) A. 3,73 gam. B. 7,04 gam. C. 7,46 gam. D. 3,52 gam.

Gii : Phn 1 tc dng vi dd NaOH, un nng: NH4+ + OH- NH3 + H2O (1) Fe3+ + 3OH- Fe(OH)3 (2)




n(NH3) = 4,22672,0 = 0, 03 mol. T (1) => n(NH4+) = n(NH3) = 0,03 mol

n(Fe(OH)3) = 1,07/ 107 = 0,01 mol. T (2) => nFe3+ = nFe(OH)3 = 0,01 mol Phn 2: tc dng vi lng d BaCl2

SO42- + Ba2+ BaSO4

nBaSO4 = 23366,4 = 0,02 mol. => nSO42- = n BaSO4= 0,02 mol

Gi x l s ion mol Cl- c trong dd X Theo nh lut bo ton in tch ta c:

30,012 + 0,0312 = 0,0222 + x1 => x = 0,04 mol Khi lng dung dch mui X sau khi c cn l:

0,01256 +0,03218 + 0,02296 + 0,0435,5 = 7,46 g chn p n C

Cu 4: Dung dch X cha cc ion: Ca2+, Na+, HCO3 v Cl, trong s mol ca ion Cl l 0,1. Cho 1/2 dung dch X phn ng vi dung dch NaOH (d), thu c 2 gam kt ta. Cho 1/2 dung dch X cn li phn ng vi dung dch Ca(OH)2 (d), thu c 3 gam kt ta. Mt khc, nu un si n cn dung dch X th thu c m gam cht rn khan. Gi tr ca m l

A. 9,21. B. 9,26. C. 8,79. D. 7,47. Gii : Cho 1/2 dd X tc dng vi dd NaOH d

HCO3- + OH- CO32- + H2O (1) CO32- + Ca2+ CaCO3 (2)

nCaCO3 = 1002 = 0,02 mol. T (1), (2) => nCa2+ = nCaCO3= 0,02 mol

Cho 1/2 dd X cn li tc dng vi Ca(OH)2 d

nCaCO3 = 1003 = 0,03 mol. => nHCO3- = nCaCO3= 0,03 mol

Gi x l s mol ion Na+ c trong dd X p dng nh lut bo ton in tch cho dd X ta c:

0,02 22 +x1 = 10,032 + 0,1 => x =0,08 mol Nu un si dd X th : 2HCO3- CO32- + CO2 + H2O

Ca2+ + CO32 CaCO3 m = m CaCO3 + m Na+ + m Cl- - m CO2 + mCa2+ d

= 0,03100 + 0,0823 + 0,135,5 +(0,04 0,03)40 - 440,03 = 7,47g Chn p n D

Cu 5: Cho 0,3 mol bt Cu v 0,6 mol Fe(NO3)2 vo dung dch cha 0,9 mol H2SO4 (long). Sau khi cc phn ng xy ra hon ton, thu c V lt kh NO (sn phm kh duy nht, ktc). Gi tr ca V l

A. 6,72. B. 8,96. C. 4,48. D. 10,08. Gii : nNO3- = 0,62 = 1,2 mol , nFe2+ = 0,6 mol nH+ = 0,9 2= 1,8 mol Phn ng theo th t : 3Cu + 2NO3- + 8H+ 3 Cu2+ + 2NO + 4 H2O (1) B 0,3 1,2 1,8

P 0,3 0,2 0,8 Sau 0 1,0 1,0 0,2

3Fe2+ + 4H+ + NO3- 3Fe3+ + NO + 2H2O (2) B 0,6 1,0 1,0

P 0,6 0,8 0,2 Sau 0 0,2 0,8 0,2




T (2) => nNO = 31 n Fe2+ =

31 0,6 = 0,2 mol

V NO = (0,2 +0,2 ) 22,4= 8,96 l Chn p n B

DNG V : BI TON V PHN NG GIA HIRO VI HP CHT HU C

I.MT S LU

*S mol gim = s mol H2 phn ng *Tng khi lng trc = Tng khi lng sau *S mol H2 / S mol X = k

II.BI TP Cu 1: un nng hn hp kh gm 0,06 mol C2H2 v 0,04 mol H2 vi xc tc Ni, sau mt thi gian thu c hn hp kh Y. Dn ton b hn hp Y li t t qua bnh ng dung dch brom (d) th cn li 0,448 lt hn hp kh Z ( ktc) c t khi so vi O2 l 0,5. Khi lng bnh dung dch brom tng l

A. 1,64 gam. B. 1,32 gam. C. 1,04 gam. D. 1,20 gam. Gii : m(Y) = m(C2H2) + m(H2) = 0,06.26 + 0,04.2 = 1,64(g)

m(bnh dd Br2 tng) = m(Y) m(Z) = 1,64 - 4,22448.0 .32.0,5 = 1,32(g)

p n B Cu 2: Hn hp kh X gm H2 v C2H4 c t khi so vi He l 3,75. Dn X qua Ni nung nng, thu c hn hp kh Y c t khi so vi He l 5. Hiu sut ca phn ng hiro ho l

A. 25%. B. 50%. C. 20%. D. 40%. Gii : Cch 1: Chn lng cht m(X) = m(Y) = m. Chn 1 mol X m=3,75.4.1=15(g)

nY= 75,04.515

(mol)

n(H2 p) = n(C2H4 p) = n(gim) = 0,25(mol)

H%= %50100.1

2.25,0 p n B

Cch 2: Tng qut

m(X) = m(Y) = m 204.5

154.75,3

n

n

Y

X

m

m

=> nY=0,75nx

n(H2p) = n(C2H4p) = n(gim) = 0,25nX H% = 2.0,25=0,5=50% p n B




Cu 3: Hn hp kh X gm H2 v mt anken c kh nng cng HBr cho sn phm hu c duy nht. T khi ca X so vi H2 bng 9,1. un nng X c xc tc Ni, sau khi phn ng xy ra hon ton, thu c hn hp kh Y khng lm mt mu nc brom; t khi ca Y so vi H2 bng 13. Cng thc cu to ca anken l

A. CH2=CH2. B. CH3-CH=CH-CH3. C. CH2=CH-CH2-CH3. D. CH2=C(CH3)2. Gii : Y khng mt mu dd Br2 => anken b kh ht Cch 1: Chn lng cht Chn 1mol X, m(X) = m(Y) = m, m=9,1.2.1=18,2 (g)

nY= 7,02.132,18 (mol)

n(H2p) = n(anken) = n(gim) = 0,3(mol) n(H2) = 0,7(mol)

M(anken) = 563.0

4,12,18

(g/mol)=> Anken: C4H8

Anken cng HBr to 1 sn phm duy nht p n B

Cch 2: Tng qut

m(X) = m(Y) = m, 262.13

2.182.1,9

n

n

Y

X

m

m

=> nY=0,7nx

n(H2p) = n(anken) = n(gim) = 0,3nX n(H2) = 0,7nx M(anken). 0,3nx + 1,4nX = 18,2nx M(anken) = 56 => Anken: C4H8

Anken cng HBr to 1 sn phm duy nht p n B

Cu 4: Hn hp kh X gm 0,3 mol H2 v 0,1 mol vinylaxetilen. Nung X mt thi gian vi xc tc Ni thu c hn hp kh Y c t khi so vi khng kh l 1. Nu cho ton b Y sc t t vo dung dch brom (d) th c m gam brom tham gia phn ng. Gi tr ca m l

A. 8,0. B. 16,0. C. 32,0. D. 3,2. Gii : m(Y) = m(X) = 0,3.2 + 0,1.52=5,8(g)

nY= 2,01.298,5 (mol) n(H2p) = n(lin kt p) = 0,2(mol)

n(lk d) = n(Br2 p) = 3.0,1 0,2 = 0,1 (mol) m(Br2) = 0,1.160 =16 (g) p n B

Cu 5: Dn V lt ( ktc) hn hp X gm axetilen v hiro i qua ng s ng bt niken nungnng, thu c kh Y. Dn Y vo lng d AgNO3 trong dung dch NH3 thu c 12 gam kt ta. Kh i ra khi dung dch phn ng va vi 16 gam brom v cn li kh Z. t chy hon ton kh Z thu c 2,24 lt kh CO2 ( ktc) v 4,5 gam nc. Gi tr ca V bng

A. 5,60. B. 13,44. C. 8,96. D. 11,2. Gii : X(C2H2,H2)Y(C2H2,H2,C2H4,C2H6)(H2,C2H4,C2H6)(H2,C2H6) CO2+H2O

n(C2Ag2) = 05,024012

(mol) = n(C2H2 d)

n(C2H4) = n(Br2) = 1,016016

(mol) = n(C2H2 p 1) = n(H2 p 1)

n(C2H6) = 0,5n(CO2) = 0,5. 4,2224,2 = 0,05 (mol) = n(C2H2 p 2) = 0,5n(H2 p 2)




n(H2 d) = n(H2O) 3n(C2H6) = 1,005,0.3185,4

(mol)

n(X) = 0,05+0,1+0,1+0,05+0,05.2+0,1=0,5 (mol) V=0,5.22,4=11,2 (l) p n D

Cu 6: un nng V lt hi anehit X vi 3V lt kh H2 (xc tc Ni) n khi phn ng xy ra hon ton ch thu c mt hn hp kh Y c th tch 2V lt (cc th tch kh o cng iu kin nhit , p sut). Ngng t Y thu c cht Z; cho Z tc dng vi Na sinh ra H2 c s mol bng s mol Z phn ng. Cht X l anehit

A. khng no (cha mt ni i C=C), hai chc. B. no, mch h, hai chc.

C.khng no (cha mt ni i C=C), n chc. D. no, mch h, n chc. Gii : Chn 1 mol X

*1 mol andehit X + 3 mol H2 2 mol hn hp Y n (gim) = n(H2 p) = 2 (mol) . Ta c : n(H2)/nX = 2 S lk = 2

*S nhm OH (Z)=2 2)()( 2

Znn H => X l andehit 2 chc c 2 lin kt p n B

Cu 7: Hiro ho hon ton m gam hn hp X gm hai anehit no, n chc, mch h, k tip nhau trong dy ng ng thu c (m + 1) gam hn hp hai ancol. Mt khc, khi t chy hon ton cng m gam X th cn va 17,92 lt kh O2 ( ktc). Gi tr ca m l

A. 10,5. B. 8,8. C. 24,8. D. 17,8.

Gii : m(tng) = m(H2) => n(H2) = 5,021 (mol)

Gi cng thc chung 2 andehit l OHC nn 2OHCOOOHC nnnnn 2222 2

13

Ptp: 1 2

13 n (mol)

Bi ra: 0,5 8,04,22

92,17 (mol)

2

13 n =1,6 => n =1,4 M (X) = 35,6 => m=35,6.0,5=17,8 (g) => p n D

Cu 8: Cho hn hp kh X gm HCHO v H2 i qua ng s ng bt Ni nung nng. Sau khi phn ng xy ra hon ton, thu c hn hp kh Y gm hai cht hu c. t chy ht Y th thu c 11,7 gam H2O v 7,84 lt kh CO2 ( ktc). Phn trm theo th tch ca H2 trong X l

A. 35,00%. B. 65,00%. C. 53,85%. D. 46,15%.

Gii : n(HCHO) = n(CO2) = 35,04,2284,7 (mol)

m(X) = m(C) + m(H) + m(O) = 12.n(CO2) + 2.n(H2O) + 16.n(HCHO)

= 12.0,35 + 2.18

7,11 + 16.0,35 = 11,1 (g)

n(H2) = 3,0230.35,01,11

(mol)

%V(H2) = %n(H2) = %15,46100.35,03,03,0

=> p n D




Cu 9: Cho 0,25 mol mt anehit mch h X phn ng vi lng d dung dch AgNO3 trong NH3, thu c 54 gam Ag. Mt khc, khi cho X phn ng vi H2 d (xc tc Ni, to) th 0,125 mol X phn ng ht vi 0,25 mol H2. Cht X c cng thc ng vi cng thc chung l

A. CnH2n+1CHO (n 0). B. CnH2n-1CHO (n 2). C. CnH2n-3CHO (n 2). D. CnH2n(CHO)2 (n 0).

Gii : S nhm CHO = 1108.25,0.2

54)(.2)(

Xn

Agn S = 2125,025,0

)()( 2

Xnn H

S lk trong gc H-C = 2-1 = 1 => p n B Cu 10: un nng hn hp kh X gm 0,02 mol C2H2 v 0,03 mol H2 trong mt bnh kn (xc tc Ni), thu c hn hp kh Y. Cho Y li t t vo bnh nc brom (d), sau khi kt thc cc phn ng, khi lng bnh tng m gam v c 280 ml hn hp kh Z (ktc) thot ra. T khi caZ so vi H2 l 10,08. Gi tr ca m l

A. 0,585. B. 0,620. C. 0,205. D. 0,328.

Gii : m(Y) = m(X) = 0,02.26 + 0,03.2 = 0,58 (g) m(Z) = 252,02.08,10.4,22

28,0 (g)

m = m(Y) m(Z) = 0,58 0,252 = 0,328 (g) => p n D




DNG VI : DNG KT QU TH I.MT S LU

*Khi gii Bi tp ta nn c ht ton bi (c bit l kt qu) *T kt qu ta c th d don hay em th *Vi bi tp xc nh cng thc phn t hay xc nh thnh phn nh lng

th gip ta gii nhanh hn. *C th dng cc gi tr TB, s mol cc cht hay gii h phng trnh th

II.BI TP Cu 1: Ho tan hon ton 2,44 gam hn hp bt X gm FexOy v Cu bng dung dch H2SO4c nng (d). Sau phn ng thu c 0,504 lt kh SO2 (sn phm kh duy nht, ktc) v dung dch cha 6,6 gam hn hp mui sunfat. Phn trm khi lng ca Cu trong X l

A. 39,34%. B. 65,57%. C. 26,23%. D. 13,11%. Gii : Cch 1: *Nu ch Cu to SO2 th % (m) Cu = 0,0225x64x100/2,44=59,02% khc kt qu => Vy FexOy l FeO hay Fe3O4 *Th vi FeO : 72x + 64y = 2,44 v x + y =0,0225 y =0,01 % (m) Cu = 0,01x64x100/2,44=26,23% => Chn CCch 2: nSO2 = 0,0225 Gi a l s mol H2SO4 tham gia p: S p: X + H2SO4 Mui + SO2 + H2O p dng nh lut bo ton khi lng: 2,44 + 98a = 6,6 + 1,44 + 18a a = 0,7 nSO42- = 0,07- 0,0225 = 0,0475 Gi x, y ln lt l s mol Fe2(SO4)3 v CuSO4 3x + y = 0,0475 x = 0,0125 400x + 160y = 6,6 y = 0,01 m Cu = 0,64 g %mCu = 26,23 % Chn CCch 3: Vit cc phng trnh phn ng ri lp h gii. Cu 2: Hn hp X gm axit Y n chc v axit Z hai chc (Y, Z c cng s nguyn t cacbon).Chia X thnh hai phn bng nhau. Cho phn mt tc dng ht vi Na, sinh ra 4,48 lt kh H2 ( ktc). t chy hon ton phn hai, sinh ra 26,4 gam CO2. Cng thc cu to thu gn v phn trm v khi lng ca Z trong hn hp X ln lt l

A. HOOC-CH2-COOH v 54,88%. B. HOOC-COOH v 60,00%. C. HOOC-COOH v 42,86%. D. HOOC-CH2-COOH v 70,87%.

Gii : Ta c n H2 = 0,2 mol , nCO2 = 0,6 mol Gi a, b l s mol axit n chc v axit 2 chc. n l s nguyn t cacbon trong 2 axit.Ta c: a + 2b = 0,2 v n( a + b ) = 0,6 Da vo kt qu suy ra n = 2 hoc 3 n = 2 => a = 0,2 , b = 0 => loi

Vy 2 axit l HCOO-COOH v CH3COOH mX = 0,2.60 + 0,1.90 = 21g %mZ = 42,86% => Chn C.

Cu 3: Cho hn hp X gm hai hp cht hu c no, n chc tc dng va vi 100 ml dung dch KOH 0,4M, thu c mt mui v 336 ml hi mt ancol ( ktc). Nu t chy hon ton lng hn hp X trn, sau hp th ht sn phm chy vo bnh ng dung dch Ca(OH)2




(d) th khi lng bnh tng 6,82 gam. Cng thc ca hai hp cht hu c trong X l A. HCOOH v HCOOC2H5. B. HCOOH v HCOOC3H7. C. C2H5COOH v C2H5COOCH3. D. CH3COOH v CH3COOC2H5.

Gii : Cch 1: Theo kt qu hn hp gm ROOH v RCOOR/ T n(R/OH) = 0,015 mol n(RCOOR/) = 0,015 v n(RCOOH) = 0,25 Mt khc cng thc chung CnH2nO 2 n (CO2 + H2O) n=2,75 loi C Thc hin php th bi phng trnh: (n 0,025 + m 0,015)/0,04 =2,75 => Chn D Cch 2: n KOH = 0,04 mol , n ancol = 0,015 mol Theo bi ra suy ra X gm 1este v 1 axit hu c

Gi CT l :CnH2n+1COOCmH2m+1 v CnH2n+1COOH n(ancol) = n (este) = 0,015mol => n axit = 0,04 0,015 = 0,025 mol m(bnh tng) = m CO2 + m H2O = (m + n + 1).0,015.44 +(n + 1).0,025.44 + (2n + 2m + 2).0,015.9 + (2n + 2).0,025.9 = 6,82 0,93m + 2,48n = 4,34 c cp nghim ph hp n = 1 v m = 2 Vy 2 cht l CH3COOH v CH3COOC2H5. Cu 4: Thu phn hon ton 0,2 mol mt este E cn dng va 100 gam dung dch NaOH 24%, thu c mt ancol v 43,6 gam hn hp mui ca hai axit cacboxylic n chc. Hai axit l

A. HCOOH v C2H5COOH. B. HCOOH v CH3COOH. C. C2H5COOH v C3H7COOH. D. CH3COOH v C2H5COOH.

Gii : Cch 1 : n NaOH = 0,6 mol n NaOH / n este = 3 => este 3 chc CT chung 2 mui RCOONa => R=5,6 = 17/3 => loi C, D

Nhn xt: S mol mui ny gp i mui kia Thc hin php th (1.1 + 29.2)/3 = 19,6 => loi Thc hin php th (1.2 + 29.1)/3 = 10,3 => loi => chn B

Cch 2 : n NaOH = 0,6 mol Ta c : n NaOH / n este = 3 => este 3 chc Gi CT 2 mui ca axit CnH2n+1COONa v CmH2m+1COONa

=>m Mui = 0,2.( 14n + 68) + 0,4.(14m + 68) = 43,6g =>2,8n + 5,6m = 2,8. cp nghim ph hp n = 1, m = 0 Vy 2 axit la HCOOH v CH3COOH => Chn B Cu 3: Thu phn hon ton 444 gam mt lipit thu c 46 gam glixerol v hai loi axit bo. Hai loi axit bo l

A. C15H31COOH v C17H35COOH. B. C17H31COOH v C17H33COOH. C. C17H33COOH v C15H31COOH. D. C17H33COOH v C17H35COOH.

Gii : Ta c : C15H31 = 211 , C17H31 = 235 , C17H33 = 237 , C17H35 = 239. (RCOO)3C3H5 + 3NaOH 3RCOONa + C3H5(OH)3

R = 238,33 = 715/3 Loi B, C. Dng php th : 211.2+239.1 715 Hay 211.1+239.2 715 Chn D




DNG VII : DNG BI TP THEO S CHUYN HA

I.MT S LU

*Theo phng php bo ton ta c th lc b cc trng thi trung gian v gi tr n. *Nu c H% th ban u ta coi nh phn ng hon ton, sau mi a H% vo *H%(chung)=

1001H

1002H

1003H ...100%

II.BI TP Cu 1: Cho s chuyn ha: CH4 C2H2 C2H3Cl PVC. tng hp 250 kg PVC theo s trn th cn V m3 kh thin nhin ( ktc). Gi tr ca V l (bit CH4 chim 80% th tch kh thin nhin v hiu sut ca c qu trnh l 50%).

A. 286,7. B. 448,0. C. 358,4. D. 224,0. Gii : S : 2 CH4 . C2H3Cl

8 kmol 4 kmol S mol CH4 cn dng l : 8.100/50 =16kmol Th tch kh thin nhin cn dng l: 16.22,4.100/80 =448 lt Chn B

Cu 2: Cho m gam tinh bt ln men thnh ancol (ru) etylic vi hiu sut 81%. Ton b lngCO2 sinh ra c hp th hon ton vo dung dch Ca(OH)2, thu c 550 gam kt ta v dung dch X. un k dung dch X thu thm c 100 gam kt ta. Gi tr ca m l

A. 550. B. 650. C. 750. D. 810. Gii : n(CO2) = 5,5 + 2 = 7,5 mol

S : (C6H10O5)n nC6H12O6 2n C2H5OH + 2n CO2 3,75 7,5

Khi lng tinh bt cn dng l 3,75. 163 .100/81 = 750 gam => Chn C Cu 3: Cho s 2 o o+ Cl (1:1) + NaOH, du + HCl6 6 Fe, t t cao,P caoC H X Y Z . Hai cht hu c Y, Z ln lt l: A.C6H5ONa, C6H5OH. B.C6H5OH, C6H5Cl. C.C6H4(OH)2, C6H4Cl2. D.C6H6(OH)6, C6H6Cl6. Gii : *Trong mi trng OH- C6H5OH tn ti dng C6H5ONa

=> Chn A Cu 4: Cho s chuyn ho sau :

0 02Br (1:1mol),Fe,t NaOH(d ),t ,p HCl(d )Toluen X Y Z .

Trong X, Y, Z u l hn hp ca cc cht hu c, Z c thnh phn chnh gm : A. o-bromtoluen v p-bromtoluen. B. m-metylphenol v o-metylphenol. C. benzyl bromua v o-bromtoluen. D. o-metylphenol v p-metylphenol.

Gii : *Trong mi trng OH- hp cht phenol tn ti dng mui. * Nm c qui lut th vng ben zen

Z l o-metylphenol v p-metylphenol.

Cu 5: Cho s chuyn ho: 2 4oH SO + HBr + Mg, etekhantButan - 2 - ol X(anken) Y Z c .

Trong X, Y, Z l sn phm chnh. Cng thc ca Z l




Al + 3Ag+ Al3+ + 3Ag

Ban u 0,1 mol 0,55 mol

P 0,1mol 0,3 mol Cn li 0,25 mol 0,1 mol 0,3 mol

A. CH3-CH(MgBr)-CH2-CH3. B. (CH3)2CH-CH2-MgBr. C. CH3-CH2-CH2 -CH2-MgBr. D. (CH3)3C-MgBr.

Gii : *Nm c qui tc tch Zai xep xc dnh sn phm chnh Chn A.

DNG VIII : BI TON LIN QUAN N DY IN HA CA KIM LOI

I.MT S LU

*Qui tc *Lu : Fe2+ Cu2+ Fe3+ Ag+ NO3-/H+ H2SO4 c Fe Cu Fe2+ Ag NO H2S, SO2, S

*Nu dng oxi ha cng mnh th dng kh cng yu v ngc li. *iu quan trng l bit c th t ca phn ng.

II.BI TP Cu 1: Cho hn hp bt gm 2,7 gam Al v 5,6 gam Fe vo 550 ml dung dch AgNO3 1M. Sau khi cc phn ng xy ra hon ton, thu c m gam cht rn. Gi tr ca m l (bit th t trong dy th in ho: Fe3+/Fe2+ ng trc Ag+/Ag).

A. 54,0. B. 59,4 C. 64,8. D. 32,4. Gii : Ta c nAl = 2,7:27=0,1(mol) , nFe= 5,6 : 56= 0,1(mol) , nAgNO3= 0,55.1= 0,55 (mol)

m=mAg= (0,3+0,2+0,05) x 108= 59,4(g) Cu 2: Cho m1 gam Al vo 100 ml dung dch gm Cu(NO3)2 0,3M v AgNO3 0,3M. Sau khi cc phn ng xy ra hon ton th thu c m2 gam cht rn X. Nu cho m2 gam X tc dng vi lng d dung dch HCl th thu c 0,336 lt kh ( ktc). Gi tr ca m1 v m2 ln lt l

A. 1,08 v 5,43. B. 0,54 v 5,16. C. 1,08 v 5,16. D. 8,10 v 5,43. Gii : n Cu(NO3)2 = nAgNO3 = 0,3.0,1=0,03 (mol) , nH2= 0,336:22,4= 0,015 (mol)

Fe + 2Ag+ Fe2+ + 2Ag


P 0,1mol 0,2 mol Cn li 0,05 mol 0,1 mol 0,2 mol

Fe2+ + Ag+ Fe3+ + Ag


P 0,05mol 0,05 mol Cn li 0,05 mol 0,05mol 0,05mol




Do X tc dng c vi HCl nn X cn cha Al. Vy Al d, Cu(NO3)2 v AgNO3 phn ng ht:

(1)

(2)

(3)

m1=mAl (1,2,3)= (0,2+0,1+0,1) x 27 = 1,08 (g) m2 = mCu + m Ag+ mAl(3)= 0,03x64 + 0,03x108 + 0,01x27= 5,43(g) Cu 3: Tin hnh hai th nghim sau: - Th nghim 1: Cho m gam bt Fe (d) vo V1 lt dung dch Cu(NO3)2 1M;. - Th nghim 2: Cho m gam bt Fe (d) vo V2 lt dung dch AgNO3 0,1M. Sau khi cc phn ng xy ra hon ton, khi lng cht rn thu c hai th nghim u bng nhau. Gi tr ca V1 so vi V2 l

A. V1 = V2. B. V1 = 2 V2. C. V1 = 5V2. D. V1 = 10 V2. Gii : nCu(NO3)2= 1.V1= V1 (mol) , nAgNO3 = 0,1.V2= 0,1V2(mol)

Khi lng cht rn thu c sau 2 phn ng bng nhau => 8V1=8V2 V1=8V2 Cu 4: Cho hn hp gm 1,2 mol Mg v x mol Zn vo dung dch cha 2 mol Cu2+ v 1 mol Ag+n khi cc phn ng xy ra hon ton, thu c mt dung dch cha ba ion kim loi. Trong cc gi tr sau y, gi tr no ca x tho mn trng hp trn?

A. 1,8. B. 1,2. C. 2,0. D. 1,5. Gii : Sp xp theo tng dn tnh oxi ha: Mg2+


Do khi lng kim loi thu c bng trc phn ng => 8a-b=0 8a=b

%m(Zn)=856165

100165

= 12,67%

Cu 6: Cho 6,72 gam Fe vo dung dch cha 0,3 mol H2SO4 c, nng (gi thit SO2 l sn phm kh duy nht). Sau khi phn ng xy ra hon ton, thu c.

A. 0,03 mol Fe2(SO4)3 v 0,06 mol FeSO4. B. 0,05 mol Fe2(SO4)3 v 0,02 mol Fe d. C. 0,02 mol Fe2(SO4)3 v 0,08 mol FeSO4. D. 0,12 mol FeSO4.

Gii : nFe= 6,72:56=0,12 mol

Cu 7: Th tch dung dch HNO3 1M (long) t nht cn dng ho tan hon ton mt hn hp gm 0,15 mol Fe v 0,15 mol Cu l (bit phn ng to cht kh duy nht l NO).

A. 0,6 lt. B. 1,0 lt. C. 1,2 lt. D. 0,8. Gii : Qu trnh cho e

Qu trinh nhn e

s mol HNO3 cn dng l t nht th a43 =0,3+0,3=0.6 a= 0,8 (mol)

Vdd HNO3 = 0,8/1=0,8 (lt) Cu 8: Cho hn hp gm 1,12 gam Fe v 1,92 gam Cu vo 400 ml dung dch cha hn hp gmH2SO4 0,5M v NaNO3 0,2M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v kh NO (sn phm kh duy nht). Cho V ml dung dch NaOH 1M vo dung dch X th lng kt ta thu c l ln nht. Gi tr ti thiu ca V l

A. 240. B. 400. C. 120. D. 360. Gii : nH+=2nH2SO4=0,5x0,4=0,4 mol , nNO3-= nNaNO3= 0,2x0,4=0,08 mol , nFe=1,12:56=0,02 mol , nCu=1,92:64=0,03 mol

2Fe + 6H2SO4 Fe2(SO4)3 + 3SO2 + 6H2O Ban u 0,12 0,3 P ng 0,1 0,3 Cn li 0,02 0,05

Fe2(SO4)3 + Fe 3FeSO4 Ban u 0,05mol 0,02 mol P ng 0,02mol 0,02 mol Cn li 0,03 mol 0,06mol

Fe Fe2+ + 2e

0,15 mol 0,3 mol

Cu Cu2+ + 2e

0,15 mol 0,3 mol

4H+ + NO3- + 3e NO + 2H2O

a mol a43




Vdd NaOH= (0,24+0,06+0,06)x1000/1=360ml Cu 9: Cho 6,72 gam Fe vo 400 ml dung dch HNO3 1M, n khi phn ng xy ra hon ton, thu c kh NO (sn phm kh duy nht) v dung dch X. Dung dch X c th ho tan ti a m gam Cu. Gi tr ca m l

A. 3,84. B. 3,20. C. 1,92. D. 0,64. Gii : nHNO3= 1x0,4=0,4 mol , nFe = 6,72:56= 0,12 mol

m = 0,03x64=1,92 (g) Cu 10: Cho 2,24 gam bt st vo 200 ml dung dch cha hn hp gm AgNO3 0,1M v Cu(NO3)2 0,5M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v m gam cht rn Y. Gi tr ca m l

A. 4,08. B. 0,64. C. 2,16. D. 2,80. Gii : nFe= 2,24:56=0,04 (mol) , nCu2+=0,5x0,2=0,1(mol) , nAg+=0,1x0,2=0,02(mol)

H+ + OH- H2O 0,24 mol 0,24 mol

Fe + 4H+ + NO3- Fe3+ +NO +H2O

Ban u 0,02 0,4 0,08 P ng 0,02 0,08 0,02 Cn li 0,32 0,06 0,02

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO +h2O

Ban u 0,03 0,4 0,06 P ng 0,03 0,08 0,02 Cn li 0,24 0,04 0,03

Fe3+ + 3OH- Fe(OH)3 0,02 mol 0,06 mol

Cu2+ + 2OH- Cu(OH)2 0,03 mol 0,06 mol

Fe + 4H+ + NO3- Fe3+ + NO +H2O

Ban u 0,12 0,4 0,08 P ng 0,1 0,4 0,02 Cn li 0,02 0,02 0,06 0,1

2Fe3+ + Fe 3Fe2+

Ban u 0,1 mol 0,02 mol P ng 0,04 mol 0,02 mol Cn li 0,06 mol

2Fe3+ + Cu 2Fe2+ + Cu2+

0,06 mol 0,03 mol

Fe + 2Ag+ Fe2+ + Ag


P 0,01mol 0,02 mol Cn li 0,03 mol 0,02 mol 0,01mol 0,02mol




m Y=mCu +mAg = 0,03x64+0,02x108=4,08 (g)

Fe + Cu2+ Fe2+ + Cu


P 0,03mol 0,03 mol Cn li 0,03 mol 0,07 mol 0,03mol 0,03mol




DNG IX : DNG TON T L I.MT S LU

*Mt s bi tp ta khng xc nh c s mol c th tng cht m ch tm c t l mol cc cht. *T t l ta c th chn lng cht *Ty tng trng hp nhng nn t cho cht c t l nh hn l x...

II.BI TP Cu 1: Cho hn hp gm Na v Al c t l s mol tng ng l 1 : 2 vo nc (d). Sau khi cc

phn ng xy ra hon ton, thu c 8,96 lt kh H2 ( ktc) v m gam cht rn khng tan. Gi tr ca m l

A. 5,4. B. 43,2. C. 7,8. D. 10,8. Gii : Gi a l nNa => nAl = 2a Do sau p/ cn cht rn khng tan nn Al dPhn ng : Na + H2O NaOH + H2

a a a/2 NaOH + Al + H2O NaAlO2 + 3/2 H2 a a 3a/2

a/2 + 3a/2 = 8.96/22.4 a = 0.2 nAl d = 2a a = 0.2 (mol) m = 5.4 (g) chn A Cu 2: Hn hp X gm Na v Al. Cho m gam X vo mt lng d nc th thot ra V lt kh. Nu cng cho m gam X vo dung dch NaOH (d) th c 1,75V lt kh. Thnh phn phn trm theo khi lng ca Na trong X l (bit cc th tch kh o trong cng iu kin).

A. 29,87%. B. 39,87%. C. 77,31%. D. 49,87%. Gii : Gi a,b ln lt l s mol ca Na v Al

Do khi tc dng vi nc ch to V(l) H2 cn khi tc dng vi NaOH th to 1,75V(l) H2 nn Al d trng hp u. Do t l s mol = t l th tch nn c th coi nH2 ln lt l V v 1,17V. Na + H2O NaOH + H2 (1) NaOH + Al + H2O NaAlO2 + 3/2 H2 (2) Ta c : (x/2 +3x/2) : (x/2+3y/2) = 1/1,75 x/y=1/2 % (m)Na = 29,87 chn A Cu 3: Trn dung dch cha a mol AlCl3 vi dung dch cha b mol NaOH. thu c kt ta th cn c t l.

A. a : b > 1 : 4 B. a : b < 1 : 4. C. a : b = 1 : 4. D. a : b = 1 : 5. Gii : Do t l trong phn ng to kt ta gia AlCl3 v NaOH l 1:3 v t l trong phn ng ha tan kt ta l 1:1 nn thu c kt ta th cn c t l a:b Chn A Cu 4: Ha tan hon ton 12 gam hn hp Fe, Cu (t l mol 1:1) bng axit HNO3, thu c V lt (ktc) hn hp kh X (gm NO v NO2) v dung dch Y (ch cha hai mui v axit d). T khi ca X i vi H2 bng 19. Gi tr ca V l

A. 2,24. B. 3,36. C. 5,60. D. 4,48. Gii : Gi a, b ln lt l s mol ca Zn v Cu Theo bi ra ta c h: a : b = 1:1 v 56a + 64 b = 12 a = 0.1 v b = 0.1 gi x, y ln lt l s mol NO v NO2 pt bo ton e: Cu Cu2+ + 2e N+5 + 3e N+2 0.1 0.2 x 3x




Fe Fe3+ +3e N+5 + 1e N+4 0.1 0.3 y y ta c h: 3x + y = 0.5 v 30x + 46y = 19*2(x+y) x = y = 0.125(mol)

V = 2x0,125x22,4 = 5,6(l) chn C Cu 5: Cho 19,3 gam hn hp bt gm Zn v Cu c t l mol tng ng l 1 : 2 vo dung dch cha 0,2 mol Fe2(SO4)3. Sau khi cc phn ng xy ra hon ton, thu c m gam kim loi. Gi tr ca m l

A. 12,80. B. 12,00. C. 6,40. D. 16,53. Gii : n(Fe3+) = 0.4 mol. Gi a, b ln lt l s mol ca Zn v Cu Gii h tng t cu 4 ta c a = 0.1 v b = 0.2 Cc phn ng: Zn + 2Fe3+ Zn2+ + 2Fe2+ 0.1 0,2 Cu + 2Fe3+ Cu2+ + 2Fe2+ (0.4-0,2) nCu d = 0,1 m = 6,4 chn C Cu 6: Hn hp kh X gm N2 v H2 c t khi so vi He bng 1,8. un nng X mt thi gian trong bnh kn (c bt Fe lm xc tc), thu c hn hp kh Y c t khi so vi He bng 2. Hiusut ca phn ng tng hp NH3 l

A. 25%. B. 50%. C. 36%. D. 40%. Gii : Chn nhh = 1mol v gi a,b ln lt l s mol ca N2 v H2. Ta c h: a + b =1 v 28a + 2b = 1,8 . 4 a = 0,2 v b = 0,8 gi x l s mol N2 phn ng : N2 + 3H2 2NH3

x 3x 2x Ta c : n(sau) = 1-2x

p dng LBTKL. mhhsau = mhhtrc = 1,8*4 = 7,2 7,2/(2*4) = 1- 2x x = 0,05(mol)

H = 0,05*100/0,2 =25% chn A Cu 7: Cho x mol Fe tan hon ton trong dung dch cha y mol H2SO4 (t l x : y = 2 : 5), thu c mt sn phm kh duy nht v dung dch ch cha mui sunfat. S mol electron do lng Fe trn nhng khi b ho tan l

A. 2x. B. 3x. C. 2y. D. y. Gii :

Fe Fe3+ + 3e 4H+ + SO42- + 2e SO2 + 2H2O y/3 2(y-y/2)/3 y 2y y/2 y y/2

p/ Fe + Fe3+ Fe2+, ch xy ra s cho nhn e gia Fe v Fe3+ (c to ra t x mol Fe ban u) coi nh khng c s nhng e ca Fe ban u. Vy s mol e m Fe nhng l y chn D




DNG X : DNG TON QUI V CHT, CHN CHT I.MT S LU

*Chn lng ban u l 100 gam, a gam.., hay 1 mol *Qui nhiu cht v 1 hay 2 cht. *Chn lng trong mt hn hp xc nh...

II.BI TP Cu 1: ho tan hon ton 2,32 gam hn hp gm FeO, Fe3O4 v Fe2O3 (trong s mol FeO bng s mol Fe2O3), cn dng va V lt dung dch HCl 1M. Gi tr ca V l

A. 0,16. B. 0,23. C. 0,08. D. 0,18. Gii : Do nFeO = nFe2O3 nn c th coi hn hp ch gm Fe3O4, n Fe3O4 = 0.01 mol Fe3O4 + 8HCl FeCl2 + 2FeCl3 +4H2O 0.01 0.08 V = 0.08 (l) chn C

Cu 2: Cho 9,12 gam hn hp gm FeO, Fe3O4 v Fe2O3 tc dng vi dung dch HCl (d). Sau khi cc phn ng xy ra hon ton, c dung dch Y; c cn Y thu c 7,62 gam FeCl2 v m gam FeCl3. Gi tr ca m l

A. 7,80. B. 8,75. C. 6,50. D. 9,75. Gii : C th coi Fe3O4 l hn hp ca FeO v Fe2O3. 7.62g FeCl2 l do FeO to ra nFeO = nFeCl2 = 0.06 mol nFe2O3 = 0.03 mol nFeCl3 = 2nFe2O3 = 0.06 mol m = 9.75 (g) chn D Cu 3:T khi hi ca hn hp H2, CO, CH4 so vi H2 bng 7,8. t chy hon ton mt th tch hn hp ny cn 1,4 th tch O2. Thnh phn % ( th tch ) ca mi kh trong hn hp u l:

A.20; 50; 30. B.33,33; 66,67; 16,67. C.20; 20; 60. D.10; 80; 10. Gii : Gi a, b, c ln lt l nH2, CO, CH4 trong 1mol hn hp Phn ng H2 + O2 H2O a a/2 CO + O2 CO2 b b/2 CH4 +2 O2 CO2 + 2H2O c 2c Ta c h: a + b + c = 1 v 2a + 28b + 16c = 7,8.2 v a/2 + b/2 + 2c = 1,4 Gii h: a = 0.2, b = 0.2, c = 0.6 chn C Cu 4: Hn hp kh gm N2 v H2 c t l th tch tng ng l 1:3 cha trong bnh, xc tc. Sau phn ng thy th tch kh i ra gim 1/10 so vi ban u. Thnh phn % th tch ca N2, H2 v NH3 trong hn hp sau phn ng ln lt l:

A.20 ; 60 ; 20. B.22,22 ; 66,67; 11,11. C.30; 60; 10. D.33,33; 50; 16,67. Gii : Chn nhh = 4 mol th nN2 = 1 mol v nH2 = 3 mol Gi a l nN2 p/ N2 + 3H2 2NH3 a 3a 2a

ngim = 2a = 0,4 a = 0.2%VN2 = 22,22 %VH2 = 66,67 %VNH3 = 11,11 chn C Cu 5: Ha tan han ton 30,4 gam hn hp: Cu, CuS, S, Cu2S bng lng HNO3 d, thy thot ra 20,16 lt NO duy nht(ktc) v dung dch Y. Thm Ba(OH)2 d vo Y th khi lng kt ta thu c l:

A.81,55 gam. B.104,2 gam. C.110,95 gam. D. 115,85 gam. Gii : Coi hn hp ban u gm Cu v S. Gi a, b ln lt l Cu v S trong 30,4 gam hn hp




Qu trnh cho nhn e: Cu Cu2+ + 2e N+5 + 3e N+2 a 2a 2,7 0,9 S S+6 + 6e b 6b Ta c h : 64a + 32b = 30,4 v 2a + 6b = 2,7 Gii h : a = 0,3 v b = 0,35

Kt ta thu c sau phn ng vi Ba(OH)2 l BaSO4 v Cu(OH)2 Khi lng kt ta = 0,3.98 + 0,35.233 = 110,95 chn C Cu 6:Hn hp A cha mt s hirocacbon th kh, B l khng kh. Trn A v B theo t l th tch l 1:15 thu c hn hp kh D. Cho D vo bnh kn th tch khn i V. Nhit v p sut trong bnh l t0C v P. Sau khi t chy trong bnh ch c N2, CO2 v hi nc vi VCO2:VH2O=7:4. a nhit bnh v t0C, p sut l P1. Th:

A.P1=47P/48. B.P1=P. C.P1=16P/17. D.P1=3P/5. Gii : Gi cng thc chung ca cc hidrocacbon l CxHy v chn nhh = 16mol Trong : nCxHy = 1mol, nO2 = 3mol, nN2 = 12mol Phn ng : CxHy + (x + y/4)O2 x CO2 + y/2H2O 1mol 3mol x y/2 Ta c: x + y/4 = 3 v 4x = 7y/2. Gii h: x = 7/3 v y = 8/3 nCO2 + nH2O = 7/3 + 4/3 = 11/3(mol) Ta c: nP1 = n1P P1 = n1P/n = (11/3 + 12)P/16 = 47P/48 Chn A Cu 7: Hn hp gm hirocacbon X v oxi c t l s mol tng ng l 1:10. t chy hon ton hn hp trn thu c hn hp kh Y. Cho Y qua dung dch H2SO4 c, thu c hn hp kh Z c t khi i vi hiro bng 19. Cng thc phn t ca X l

A. C3H6. B. C3H8. C. C4H8. D. C3H4. Gii : Chn nhh = 11mol nCxHy = 1 v nO2 = 10.

CxHy + (x + y/4)O2 x CO2 + y/2H2O 1mol (x + y/4 ) x y/2 -> 44x + 32(10-x-y/4) = 19*2(10-y/4) 12x + 3y/2 = 60 8x + y = 40 Chn x= 4, y = 8 C4H8 Chn C Cu 8: t chy hon ton 20,0 ml hn hp X gm C3H6, CH4, CO (th tch CO gp hai ln th tch CH4), thu c 24,0 ml CO2 (cc th tch kh o cng iu kin nhit v p sut). T khi ca X so vi kh hiro l

A. 22,2. B. 25,8 C. 11,1. D. 12,9. Gii : Chn nhh = 20 mol v goi a, b ln lt l nC3H6 v CH4 nCO = 2b Ta c h: a + 3b = 20 v 3a + b + 2b = 24 Gii h ta c : a=2 v b= 6 -> nCO = 12mol Vy dhh/H2 = 12.9 chn D Cu 9: Hn hp X c t khi so vi H2 l 21,2 gm propan, propen v propin. Khi t chy hon ton 0,1 mol X, tng khi lng ca CO2 v H2O thu c l

A. 18,60 gam. B. 18,96 gam. C. 20,40 gam. D. 16,80 gam. Gii : mhh = 0.1*2*21.2 = 4.24 (g) , nCO2 = nC(hh) = 3nhh = 0.3 (mol) ,nH2O = nH(hh)/2 = 0.32

mCO2 + mH2O = 18.96(g) => chn B Cu 10: Cho hn hp X gm CH4, C2H4 v C2H2. Ly 8,6 gam X tc dng ht vi dung dch brom (d) th khi lng brom phn ng l 48 gam. Mt khc, nu cho 13,44 lt ( ktc) hn hp kh X tc dng vi lng d dung dch AgNO3 trong NH3, thu c 36 gam kt ta. Phn trm th tch ca CH4 c trong X l

A. 25%. B. 20%. C. 50%. D. 40%.




Gii : Goi a, b, c ln lt l s mol ca CH4, C2H4 v C2H2 trong 8,6 gam hn hp. Ta c nBr2 = 0.3 mol Cc phn ng v s : C2H4 + Br2 C2H4Br2 (1)C2H2 + 2Br2 C2H2Br4 (2)C2H2 C2Ag2 (3) b b c 2c c c C (13,44/22,4) = 0,6 mol hn hp th to ra (36/240)=0.15 mol kt ta

Vy (a + b + c)mol 0.25 (a + b + c) mol Ta c h 16a + 28b + 26c = 8,6 v b + 2c = 0.3 v c = 0.25(a + b +c) Gii h : a = 0,2, b = 0.1, c = 0,1 %(V)CH4 = 50% Chn C




DNG XI : BI TON NG DNG I.MT S LU

*Xut pht t mt bi ton c bn ta pht trin thnh bi ton ng dng *i ngc kt qu thnh gi thit hay ngc li *Thay mt cht bng nhiu cht...

II.BI TP Cu 1: Dung dch X cha hn hp gm Na2CO3 1,5M v KHCO3 1M. Nh t t tng git cho

n ht 200 ml dung dch HCl 1M vo V ml dung dch X, sinh ra 1,12 lt kh ( ktc). Gi tr ca V l

A. 100. B. 200. C. 150. D. 250. Gii : Ta c : n HCl = 0,2 mol P theo th t : CO32- + H +HCO3- (1) HCO3- + H+ CO2 + H2O (2) 0,15 0,15 0,05 0,05 V = 0,15 :1,5 = 0,1 lt Chn A. Cu 2: Nh t t tng git n ht 30 ml dung dch HCl 1M vo 100 ml dung dch cha Na2CO3 0,2M v NaHCO3 0,2M, sau phn ng thu c s mol CO2 l

A. 0,020. B. 0,030. C. 0,015. D. 0,010. Gii : Ta c nHCl = 0,03 mol , n Na2CO3 = 0.02 mol , n NaHCO3 = 0,2 mol

CO32- + H + HCO3- 0,02 0,02

HCO3- + H+ CO2 + H2O 0,01 0,01 0,01 Chn D Cu 3: Nung 2,23 gam hn hp X gm cc kim loi Fe, Al, Zn, Mg trong oxi, sau mt thi gian thu c 2,71 gam hn hp Y. Ho tan hon ton Y vo dung dch HCl (d), thu c a mol H2. S mol HCl phn ng l

A. 0,12. B. 0,06. C. 0,06 + 2a. D. 0,18. Gii : Ta c : nO = (2,71 2,23)/16 = 0,03 S : O2- + 2H+ H2O

0,03 0,06 n H2 = a n H+ = 2a. Vy s mol HCl p = n H+ = 2a + 0,06 chn C Cu 4: Nung 2,23 gam hn hp X gm cc kim loi Fe, Al, Zn, Mg trong oxi, sau mt thi gian thu c 2,71 gam hn hp Y. Ho tan hon ton Y vo dung dch HNO3 (d), thu c 0,672 lt kh NO (sn phm kh duy nht, ktc). S mol HNO3 phn ng l

A. 0,12. B. 0,14. C. 0,16. D. 0,18. Gii : Cch 1: NO3- + 4H+ + 3e NO + 2H2O

0,12 0,03 O2 + 4e 2O2- 0,015 0,03 S : O2- + 2H+ H2O

0,03 0,06 n = 0,06 + 0,12 = 0,18

Cch 2: M Mn+ + ne 0,15/n 0,15




N+5 + 3e N+2 0,09 0,03

O2 + 4e 2O2- 0,015 0,06 n HNO3 = n.n M(NO3)n + n NO = 0,18 mol Cch 3: n O = (2,71 2,23)/ 16 = 0,03 mol Gi a l s mol HNO3 tham gia phn ng. Hn hp kim loi + 2O + HNO3 hn hp mui + NO + H2O

0,03 3a 3(a 0,03) 0,03 0,5a p dng nh lut bo ton nguyn t oxi ta c: 0,03 + 3a = 3(a 0,03) + 0,03 + 0,5a a = 0,18 mol. Cch 4: Gi s mol HNO3 tham gia phn ng l a Y + HNO3 Mui + NO + H2O 2,71 63a 2,33+62(a-0,03) 30.0,03 9a p dng nh lut bo ton khi lng ta c : 2,71 + 63a = 2,33 + 62( a 0,03 ) + 30.0,03 + 9a a = 0,18. chon D

Cu 5: Cho mt mu hp kim Na-Ba tc dng vi nc (d), thu c dung dch X v 3,36 lt H2 ( ktc). Th tch dung dch axit H2SO4 2M cn dng trung ho dung dch X l

A. 150ml. B. 75ml. C. 60ml. D. 30ml. Gii : 2Na + 2H2O 2NaOH + H2 Ba + H2O Ba(OH)2 + H2 Theo pt ta c n OH- = 2n H2 = 0,3 mol OH- + H+ H2O 0,3 0,3 n H2SO4 = 0,3/2 = 0,15. vy V = 0,15/2 = 0.075lt Chn B Cu 6: Ho tan hon ton 8,94 gam hn hp gm Na, K v Ba vo nc, thu c dung dch X v 2,688 lt kh H2 (ktc). Dung dch Y gm HCl v H2SO4, t l mol tng ng l 4 : 1. Trung ho dung dch X bi dung dch Y, tng khi lng cc mui c to ra l

A. 13,70 gam. B. 12,78 gam. C. 18,46 gam. D. 14,62 gam. Gii : Ta c n OH- = 2n H2 = 0,24 mol , n H+ trung ha = 0,24 mol n HCl phn ng = 0,24.4/6 = 0,16

n H2SO4 = ( 0,24.2)/(2.6 ) = 0,04 m Mui = m Cation + m Anion = 8,94 + 0,16.35,5 + 0,04.96 = 18,46 Chn C. Cu 7: Cho 0,448 lt kh CO2 ( ktc) hp th ht vo 100 ml dung dch cha hn hp NaOH 0,06M v Ba(OH)2 0,12M, thu c m gam kt ta. Gi tr ca m l

A. 1,182. B. 3,940. C. 1,970. D. 2,364. Gii : Phn ng : CO2 + OH- HCO3-

CO2 +2 OH- CO32- +H2O H : a + b = 0,02 mol v a + 2b = 0,03 a =0,01 v b = 0,01 m(BaCO3) = 1,97 gam Chn C Cu 8: t chy hon ton m gam FeS2 bng mt lng O2 va , thu c kh X. Hp th ht X vo 1 lt dung dch cha Ba(OH)2 0,15M v KOH 0,1M, thu c dung dch Y v 21,7 gam kt ta. Cho Y vo dung dch NaOH, thy xut hin thm kt ta. Gi tr ca m l

A. 23,2. B. 12,6. C. 18,0. D. 24,0. Gii : n Ba(OH)2 = 0,15 , n KOH = 0,1 mol, n BaSO3 = 27,1/217 = 0,1 mol Phn ng to mui trung ha v mui axit. Phn ng : SO2 + OH- HSO3-




SO2 +2 OH- SO32- +H2O Ba2+ + SO32- BaSO3 Ta c : b = 0,1 mol v a + 2 . 0,1 = 0,04 a = 0,2 n(SO2) = 0,3 mol Phn ng : 4FeS2 + 11O2 2Fe2O3 + 8SO2 m = 0,3/2 . 120 = 18 gam

DNG XII : DNG BI TP S DNG NGUYN L NGC

I.MT S LU

* Axit Baz + H+ ; Nu axit cng mnh th baz cng yu v ngc li.

* Oxh + ne Kh ; Nu dng oxi ha cng mnh th dng kh cng yu v ngc li.

* Axit c th tc dng vi Baz, cht kh c th tc dng vi cht oxi ha...

II.BI TP Cu 1: Cho bit cc phn ng xy ra sau:

2FeBr2 + Br2 2FeBr3. 2NaBr + Cl2 2NaCl + Br2. Pht biu ng l:

A. Tnh kh ca Cl- mnh hn Br-. B. Tnh kh ca Br- mnh hn ca Fe2+. C. Tnh oxi ha ca Br2 mnh hn ca Cl2. D. Tnh oxi ha ca Cl2 mnh hn ca Fe3+.

Gii : 2FeBr2 + Br2 2FeBr3 => tnh oxi ho ca Br2 mnh hn ca Fe2+ 2NaBr + Cl2 2NaCl + Br2 => tnh oxi ho ca Cl2 mnh hn ca Br2 Tnh oxi ho ca Cl2 mnh hn ca Fe2+ => chn D Cu 2: Trong s cc dung dch: Na2CO3, KCl, CH3COONa, NH4Cl, NaHSO4, C6H5ONa, nhng dung dch c pH > 7 l

A. KCl, C6H5ONa, CH3COONa. B. NH4Cl, CH3COONa, NaHSO4. C. Na2CO3, NH4Cl, KCl. D. Na2CO3, C6H5ONa,

CH3COONa. Gii : Nhng dung dch c pH >7 l: Na2CO3, C6H5ONa, CH3COONa v (1)Na2CO3 2Na+ + CO32- CO32- + HOH HCO32- + OH- (2) CH3COONa CH3COO- + Na+ CH3COO- + HOH CH3COOH + OH- (3) C6H5ONa C6H5O- + Na+ C6H5O- + HOH C6H5OH + OH- Cc anion OH- c gii phng, nn mi trng c pH > 7,0. chn p n D Cu 3: Dy cc ion xp theo chiu gim dn tnh oxi ho l (bit trong dy in ha, cp Fe3+/Fe2+ng trc cp Ag+/Ag):

A. Ag+, Fe3+, Cu2+, Fe2+. B. Fe3+, Ag+, Cu2+, Fe2+.

C. Fe3+, Fe2+, Cu2+, Ag+. D. Ag+, Cu2+, Fe3+, Fe2+.




Gii : V cp Fe3+/Fe2+ ng trc cp Ag+/Ag => tnh oxi ho ca Ag+ > Fe3+ (1) Tnh kh ca Cu > Fe2+ => tnh oxi ho ca Cu2+< Fe3+ (2) t (1),(2) ta c tnh oxi ho ca Ag+ > Fe3+> Cu2+> Fe2+ => chn p n A Cu 4: Cho cc phn ng xy ra sau y: (1)AgNO3 + Fe(NO3)2 Fe(NO3)3 + Ag. (2)Mn + 2HCl MnCl2 + H2.Dy cc ion c sp xp theo chiu tng dn tnh oxi ho l A.Ag+, Fe3+, H+, Mn2+. B.Ag+, Mn2+, H+, Fe3+. C.Mn2+, H+, Ag+, Fe3+. D.Mn2+, H+, Fe3+, Ag+.

Gii : (1) AgNO3 + Fe (NO3)2 Fe(NO3)3 + Ag => tnh oxi ho ca Ag+ > Fe3+ ( 2) Mn + 2HCl MnCl2 + H2 => tnh oxi ho ca Mn2+ < H+ Tnh oxi ho ca H+ < Fe3+ tnh oxi ho ca cc ion c sp xp theo chiu tng dn nh sau:

Mn2+ < H+< Fe3+< Ag+ => chn p n D

Cu 5: Cho cc ion kim loi: Zn2+, Sn2+, Ni2+, Fe2+, Pb2+. Th t tnh oxi ho gim dn l A. Zn2+>Sn2+ > Ni2+ > Fe2+> Pb2+. B. Pb2+ > Sn2+ > Ni2+>Fe2+ > Zn2+. C. Pb2+ > Sn2+ > Fe2+> Ni2+ > Zn2+. D. Sn2+ > Ni2+ > Zn2+> Pb2+ > Fe2+.

Gii : Theo chiu dy in ho ca kim loi ta c, i t tri sang phi tnh oxi ho ca cc ion kim loi tng dn => th t cc ion theo chiu gim dn nh sau : Pb2+> Sn2+>Ni2+>Fe2+>Zn2+ => chn p n B Cu 6: Cho phn ng ha hc: Fe + CuSO4 FeSO4 + Cu. Trong phn ng trn xy ra.

A. s oxi ha Fe v s kh Cu2+.. B. s oxi ha Fe v s oxi ha Cu. C. s kh Fe2+ v s oxi ha Cu. D. s kh Fe2+ v s kh Cu2+.

Gii : Fe + CuSO4 FeSO4 + Cu Phn ng xy ra do s oxi ho Fe v s kh Cu2+ => chn p n A Cu 7: Th t mt s cp oxi ho - kh trong dy in ho nh sau: Fe2+Fe; Cu2+/Cu; Fe3+/Fe2+. Cp cht khng phn ng vi nhau l

A. Fe v dung dch CuCl2. B. dung dch FeCl2 v dung dch CuCl2.

C. Cu v dung dch FeCl3. D. Fe v dung dch FeCl3. Gii : Theo cp oxi ho - kh trong dy in ho : Fe2+/Fe, Cu2+/Cu, Fe3+/Fe2+ nn cc cp cht phn ng vi nhau l :

Fe + CuCl2 FeCl2 + Cu Cu + 2FeCl3 CuCl2 + 2FeCl2 Fe + 2FeCl3 3FeCl2

Cp cht khng phn ng l dung dch FeCl2 v dd CuCl2 => chn p n B Cu 8: Cp cht khng xy ra phn ng ho hc l

A. Cu + dung dch FeCl2 B. Fe + dung dch FeCl3.

C. Cu + dung dch FeCl3. D. Fe + dung dch HCl. Gii : Ta c cc phng trnh phn ng sau: Cu + 2FeCl3 CuCl2+ 2FeCl2 Fe + 2FeCl3 3FeCl2 Fe +2 HCl FeCl2 + H2 cp cht khng phn ng l: Cu + dd FeCl2 => chn p n A

Cu 9: Mnh khng ng l: A. Fe2+ oxi ha c Cu. B. Tnh oxi ha ca cc ion tng theo th t: Fe2+, H+, Cu2+, Ag+.




C. Fe3+ c tnh oxi ha mnh hn Cu2+. D. Fe kh c Cu2+ trong dung dch.

Gii : Mnh khng ng l : A : Fe2+ oxi ho c Cu Cu 10: Cho dy cc cht: phenol, anilin, phenylamoni clorua, natri phenolat, etanol. S cht trong dy phn ng c vi dung dch NaOH l

A. 1. B. 3. C. 2.. D. 4. Gii : Cc phng trnh phn ng xy ra l: C6H5OH + NaOH C6H5ONa + H2O C6H5NH3Cl + NaOH C6H5NH2 + NaCl + H2O s cp cht phn ng l 2 => chn p n C

Cu 11: Cho dy cc cht: C6H5OH (phenol), C6H5NH2 (anilin), H2NCH2COOH, CH3CH2COOH, CH3CH2CH2NH2. S cht trong dy tc dng c vi dung dch HCl l

A. 4. B. 3. C. 2. D. 5. Gii : Cc phng trnh phn ng xy ra l: C6H5NH2 + HCl C6H5NH3Cl H2NCH2COOH + HCl ClH3NCH2COOH CH3CH2CH2NH2 + HCl CH3CH2CH2NH3Cl => s cp cht phn ng c l 3 => chn p n B




THI TH I HC LN III NM 2010 TRNG THPT CHUYN L QU N Mn thi: HO HC ( thi c 06 trang) Thi gian lm bi: 90 pht, khng k thi gian pht .

H, tn th sinh:.......................................................................... M thi 135 S bo danh:...............................................................................

Cho bit khi lng nguyn t (theo vC) ca cc nguyn t: H = 1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; S = 32; Cl = 35,5; K = 39; Ca = 40; Mn = 55; Fe = 56; Cu = 64; Zn = 65; Cl=35,5; Br = 80; Ag = 108; Ba = 137; Pb = 207.

I. PHN CHUNG CHO TT C TH SINH (40 cu t cu 1 n cu 40) Cu 1: S ng phn l hp cht thm c cng thc phn t C8H10O tc dng c vi NaOH l

A. 9. B. 6. C. 7. D. 8. Cu 2: Cho cc phn ng: (I) Fe + HCl ; (II) Fe3O4 + H2SO4 (c) ;

(III)KMnO4 + HCl ; (IV) FeS2 + H2SO4 (long) ; (IV) Al + H2SO4 (long) ; S phn ng m H+ ng vai tr l cht oxi ho l: A . 3 B . 2 C . 4 D . 1

Cu 3: Hn hp X c C2H5OH, C2H5COOH, CH3CHO trong C2H5OH chim 50% theo s mol. t chy m gam hn hp X thu c 3,06 gam H2O v 3,136 lt CO2 (ktc). Mt khc 13,2 gam hn hp X thc hin phn ng trng bc thy c p gam Ag kt ta. Gi tr ca p l

A. 9,72. B. 8,64. C. 10,8. D. 2,16. Cu 4: Cc nhn nh sau: 1)Axit hu c l axit axetic. 2)Gim n l dung dch axit axetic c nng t 2% - 5%. 3)Khi cho 1 mol axit hu c (X) tc dng vi Na d, s mol H2 sinh ra bng s mol X v khi t chy axit X th thu c n(H2O) : nCO2 =1. Vy X l axit no n chc no. 4)Khi t chy hirocacbon no th ta c n(H2O) : nCO2 >1. Cc nhn nh sai l:

A. 1, 2, 3, 4. B. 2, 3, 4. C. 1, 2, 3. D. 1, 3, 4. Cu 5: Cho s phn ng sau X + H2SO4 (c, nng) Fe2(SO4)3 + SO2 + H2O

S cht X c th thc hin phn ng trn l A. 4. B. 6. C. 5. D. 7.

Cu 6: C 4 dung dch long ca cc mui: BaCl2, ZnCl2, FeCl2, FeCl3. Khi sc kh H2S d vo cc dung dch mui trn th s trng hp c phn ng to kt ta l

A. 3. B. 2. C. 1. D. 4. Cu 7: Chia m gam hn hp M gm hai ancol n chc k tip trong dy ng ng thnh 2 phn bng nhau. Cho phn mt tc dng vi Na (d), c 3,36 lt kh hiro (ktc). Cho phn hai phn ng hon ton vi CuO nhit cao, c hn hp M1 cha hai anehit (ancol ch bin thnh anehit). Ton b lng M1 phn ng ht vi AgNO3 trong NH3, thu c 86,4 gam Ag. Gi tr ca m l

A. 24,8 gam B. 30,4 gam C. 15,2 gam D. 45,6 gam Cu 8: Chia 156,8 gam hn hp L gm FeO, Fe3O4, Fe2O3 thnh hai phn bng nhau. Cho phn th nht tc dng ht vi dung dch HCl d c 155,4 gam mui khan. Phn th hai tc dng va vi dung dch M l hn hp HCl, H2SO4 long thu c 167,9 gam mui khan. S mol ca HCl trong dung dch M l

A. 1,75 mol B. 1,50 mol C. 1,80 mol D. 1,00 mol Cu 9: Hn hp bt X gm BaCO3, Fe(OH)2, Al(OH)3, CuO, MgCO3. Nung X trong khng kh n khi lng khng i c hn hp rn A1. Cho A1 vo nc d khuy u c dung dch B cha 2 cht tan v phn khng tan C1. Cho kh CO d qua bnh cha C1 nung nng c hn hp rn E (Cho cc phn ng xy ra hon ton). E cha ti a:




A. 1 n cht v 2 hp cht. B. 3 n cht. C. 2 n cht v 2 hp cht. D. 2 n cht v 1 hp cht.

Cu 10: Cho a gam Na vo 160 ml dung dch gm Fe2(SO4)3 0,125M v Al2(SO4)3 0,25M. Tch kt ta ri nung n khi lng khng i th thu c 5,24 gam cht rn. Gi tr a no sau y l ph hp?

A. 9,43. B. 11,5. C. 9,2. D. 10,35. Cu 11: Cho hn hp Na, Al, Fe, FeCO3, Fe3O4 tc dng vi dung dch NaOH d, lc ly kt ta ri chia lm 2 phn. Phn 1 em tc dng vi dung dch HNO3 long d. Phn 2 em tc dng vi dung dch HCl d. S phn ng oxi ha kh xy ra l:

A. 5. B. 6. C. 8. D. 7. Cu 12: Cho s dng: X Y Z. Cho cc cht sau y: etilen, etyl clorua, ancol etylic. S s nhiu nht th hin mi quan h gia cc cht trn l

A. 3 B. 4 C. 5 D. 6 Cu 13: Cho cc cht sau: propyl clorua, anlyl clorua, phenyl clorua, natri phenolat, anilin, mui natri ca axit amino axetic, ancol benzylic. S cht tc dng c vi dung dch NaOH long khi un nng l

A. 3. B. 4. C. 2. D. 1. Cu 14: Mt hn hp X gm Na, Al v Fe (vi t l mol Na v Al tng ng l 5:4) tc dng vi H2O d th thu c V lt kh, dung dch Y v cht rn Z. Cho Z tc dng vi dung dch H2SO4 long d th thu c 0,25V lt kh (cc kh o cng iu kin). Thnh phn % theo khi lng ca Fe trong hn hp X l

A. 14,4% B. 33,43% C. 20,07% D. 34,8%. Cu 15: Oxi ho 25,6 gam CH3OH (c xc tc) thu c hn hp sn phm X. Chia X thnh hai phn bng nhau Phn 1 tc dng vi AgNO3 d trong NH3 un nng thu c m gam Ag. Phn 2 tc dng va vi 100ml dung dch KOH 1M. Hiu sut qu trnh oxi ho CH3OH l 75%. Gi tr ca m l

A. 64,8. B. 32,4. C. 129,6. D. 108. Cu 16: t chy hon ton 0,1 mol hp cht A ch thu c 0,3 mol CO2, 0,45 mol H2O, 0,05 mol nit (ktc), bit A tc dng vi dung dch NaOH to kh c mi khai. S ng phn ca A l

A. 2. B. 5. C. 3. D. 4. Cu 17: Cho V lt hn hp kh A gm clo v oxi tc dng va ht vi hn hp B gm 0,2 mol Al v 0,1 mol Mg th thu c 25,2 gam hn hp mui clorua v oxit ca 2 kim loi. S mol ca Cl2 c trong V lt hn hp kh A l

A. 0,15. B. 0,3. C. 0,2. D. 0,25. Cu 18: Cho tng cht: Fe, FeO, Fe(OH)2, Fe(OH)3, Fe3O4, Fe2O3, Fe(NO3)2, Fe(NO3)3, FeSO4, Fe2(SO4)3, FeCO3 ln lt phn ng vi HNO3 c, nng, d. S phn ng thuc loi phn ng oxi ha kh l

A. 7 B. 8 C. 10 D. 9 Cu 19: Cho bt Fe vo dung dch NaNO3 v H2SO4. n phn ng hon thu c dung dch A, hn hp kh X gm NO v H2 c v cht rn khng tan. Bit dung dch A khng cha mui amoni. Trong dung dch A cha cc mui:

A. FeSO4, Fe(NO3)2, Na2SO4, NaNO3. B. FeSO4, Fe2(SO4)3, NaNO3, Na2SO4. C. FeSO4, Na2SO4. D. FeSO4, Fe(NO3)2, Na2SO4.

Cu 20: Hn hp X c 2 hirocacbon ng ng lin tip, c t khi hi so vi H2 bng 15,8. Ly 6,32 gam X li vo 200 gam dung dch cha xc tc thch hp th thu c dung dch Z v thy thot ra 2,688 lt kh kh Y iu kin tiu chun c t khi hi so vi H2 bng 16,5. Bit rng dung dch Z cha anehit vi nng C%. Gi tr ca C% l:

A. 1,305% B. 1,407% C. 1,043% D. 1,208% Cu 21: C cc nhn nh sau y:

1)Nguyn tc sn xut gang l kh st bng CO nhit cao.




2)Nguyn tc sn xut thp l kh cc tp cht trong gang. 3)Tnh cht ha hc ca Fe2+ l tnh kh. 4)Nc cng l nc c cha ion Ca2+, Mg2+ di dng mui Cl-, HCO3-, SO42-. S nhn nh ng l A. 2. B. 3. C. 0. D. 1.

Cu 22: C cc nhn nh sau: 1)Cu hnh electron ca ion X2+ l 1s22s22p63s23p63d6. Trong bng tun hon cc nguyn t ho

hc, nguyn t X thuc chu k 4, nhm VIIIB. 2)Cc ion v nguyn t: Ne , Na+ , F c im chung l c cng s electron. 3)Khi t chy ancol no th ta c n(H2O) : n(CO2)>1. 4)Dy gm cc nguyn t c sp xp theo chiu gim dn bn knh nguyn t t tri sang phi l

K, Mg, Si, N. 5)Tnh baz ca dy cc hiroxit: NaOH, Mg(OH)2, Al(OH)3 gim dn. Cho: N (Z = 7), F (Z=9), Ne (Z=10), Na (Z=11), Mg (Z=12), Al (Z=13), K (Z = 19), Si (Z = 14). S nhn nh ng: A. 3. B. 5. C. 4. D. 2.

Cu 23: Hn hp X gm 3 ancol n chc A, B, C trong B, C l 2 ancol ng phn. t chy hon ton 0,08 mol X thu c 3,96 gam H2O v 3,136 lt kh CO2 (ktc). S mol ancol A bng 5/3 tng s mol 2 ancol (B + C). Khi lng ca B, C trong hn hp l:

A. 3,6 gam B. 0,9 gam C. 1,8 gam D. 2,22 gam Cu 24: Cho m in ca cc nguyn t nh sau: O(3,44), Cl(3,16), Mg(1,31), C(2,55), H(2,2)

Trong cc phn t: MgO, CO2, CH4, Cl2O. S cht c kiu lin kt cng ha tr c cc l A. 2. B. 3. C. 1. D. 4.

Cu 25: t chy hon ton 0,1 mol ancol no, mch h X cn va 5,6 lt oxi (ktc). X cng vi axit HOOCC6H4COOH l 2 monome c dng iu ch polime, lm nguyn liu sn xut t:

A. Nilon-6,6. B. Lapsan. C. Capron. D. Enang Cu 26: Nung m gam hn hp X gm FeS v FeS2 trong mt bnh kn cha khng kh (gm 20% th tch O2 v 80% th tch N2) n khi phn ng xy ra hon ton, thu c cht rn v hn hp kh Y c thnh phn th tch: N2 = 84,77%; SO2 = 10,6% cn li l O2. Thnh phn % theo khi lng ca FeS trong X l

A. 68,75% B. 59,46% C. 26,83% D. 42,3% Cu 27: Cho 32,8 gam hn hp gm axit glutamic v tyrosin (t l mol 1:1) tc dng vi 500ml dung dch NaOH 1M, phn ng hon ton thu c dung dch Y. C cn cn thn dung dch Y thu c m gam cht rn khan. Gi tr ca m l

A. 49,2 B. 52,8 C. 43,8 D. 45,6 Cu 28: Cho cc cht v ion sau y: NO2-, Br2, SO2, N2, H2O2, HCl, S. S cht v ion c c tnh oxi ha v tnh kh l

A. 7. B. 4. C. 6. D. 5. Cu 29: Cho qu tm vo ln lt cc dung dch: CH3COOK, FeCl3, NH4NO3, K2S, Zn(NO3)2, Na2CO3. S dung dch lm i mu giy qu l

A. 4 B. 6 C. 5 D. 3 Cu 30: Trong s cc loi t sau: t tm, t visco, t nilon-6,6, t axetat, t capron, t enang. Nhng loi t no thuc loi t nhn to l:

A. T nilon 6,6 v t capron. B. T tm v t enang. C. T visco v t nilon-6,6. D. T visco v t axetat.

Cu 31: Cho phn ng sau: 2SO2(k) + O2(k) 2SO3(k) ;H < 0 cn bng trn chuyn dch theo chiu thun th: (1): tng tng nhit , (2): tng p sut, (3): h

nhit , (4): dng xc tc l V2O5, (5): Gim nng SO3. Bin php ng l:




A. 1, 2, 5. B. 2, 3, 5. C. 1, 2, 3, 4, 5. D. 2, 3, 4, 5. Cu 32: T etilen v benzen, cht v c v iu kin phn ng c . Tng s phn ng t nht c th iu ch c polibutaien, polistiren, poli (butaien - stiren), poli (vinyl clorua) l

A. 11. B. 12. C. 9. D. 10. Cu 33: X phng ha hon ton 100gam cht bo c ch s axit bng 7 cn a gam dung dch Na0H 25%, thu c 9,43gam glyxerol v b gam mui natri. Gi tr ca a, b ln lt l:

A. 49,2 v 103,37 B. 51,2 v 103,145 C. 51,2 v 103,37 D. 49,2 v 103,145 Cu 34: Tch s ion ca nc mt s nhit nh sau: 200C l 7,00.10-15, 250C l 1,00.10-14, 300C l 1,50.10-14. S in ly ca nc l

A. thu nhit B. ta nhit hay thu nhit ty theo iu kin phn ng C. ta nhit D. khng xc nh ta nhit hay thu nhit

Cu 35: t chy hon ton mt lng hn hp hai este n chc no, mch h cn 3,976 lt oxi (ktc) thu c 6,38 gam CO2. Cho lng este ny tc dng va vi KOH thu c hn hp hai ancol k tip v 3,92 gam mui ca mt axit hu c. Cng thc ca hai cht hu c trong hn hp u l:

A. HCOOC3H7 v HCOOC2H5 B. CH3COOCH3 v CH3COOC2H5 C. C2H5COOC2H5 v C2H5COOCH3 D. CH3COOC2H5 v CH3COOC3H7

Cu 36: Xt cc cht: imetylete (1), ancol metylic (2), ancol etylic (3), axit axetic (4), axeton (5). Cc cht trn c xp theo nhit si tng dn (t tri sang phi) l:

A. 1, 5, 2, 3, 4 B. 2, 3, 4, 5, 1 C. 5, 1, 2, 3, 4 D. 1, 2, 3, 4, 5 Cu 37: Cho lung kh CO d i qua hn hp BaO, Al2O3 v FeO t nng thu c cht rn X1. Ho tan cht rn X1 vo nc thu c dung dch Y1 v cht rn E1. Sc kh CO2 d vo dung dch Y1 thu c kt ta F1. Ho tan E1 vo dung dch NaOH d thy b tan mt phn v cn cht rn G1. Cho G1vo dung dch AgNO3 d (Coi CO2 khng phn ng vi nc). Tng s phn ng xy ra l

A. 7. B. 8. C. 6. D. 9. Cu 38: Cho 25,65 gam mui gm H2NCH2COONa v H2NCH2CH2COONa tc dng va vi 250 ml dung dch H2SO41M. Sau phn ng c cn dung dch th khi lng mui do H2NCH2COONa to thnh l:

A. 29,25 gam B. 18,6 gam C. 37,9 gam D. 12,4 gam Cu 39: Nung m gam hn hp X gm KClO3 v KMnO4 thu c cht rn Y v O2. Bit KClO3 phn hy hon ton, cn KMnO4 ch b phn hy mt phn. Trong Y c 0,894 gam KCl chim 8,132% theo khi lng. Trn lng O2 trn vi khng kh theo t l th tch 2OV : KKV =1:3 trong mt bnh kn ta thu c hn hp kh Z. Cho vo bnh 0,528 gam cacbon ri t chy ht cacbon, phn ng hon ton, thu c hn hp kh T gm 3 kh O2, N2, CO2, trong CO2 chim 22,92% th tch. Gi tr m (gam) l

A. 12,59 B. 12,53 C. 12,70 D. 12,91 Cu 40: Cht hu c X mch h, khng cha lin kt (-O-O-) v c cng thc phn t l C3H6On. Bit X ch cha mt loi nhm chc. S ng phn cu to c th c ca X l

A. 4. B. 6. C. 7. D. 5.

II. PHN RING [10 cu] Th sinh ch c lm mt trong hai phn (phn A hoc B) A.Theo chng trnh Chun (10 cu, t cu 41 n cu 50) Cu 41: Dng mt lng dung dch H2SO4 nng 20%, un nng ha tan va a mol CuO. Sau phn ng lm ngui dung dch n 1000C th khi lng tinh th CuSO4.5H2O tch ra khi dung dch l 30,7 gam. Bit rng tan ca dung dch CuSO4 1000C l 17,4 gam. Gi tr ca a l

A. 0,1. B. 0,15. C. 0,2. D. 0,25.




Cu 42: C dung dch X gm (KI v mt t h tinh bt). Cho ln lt tng cht sau: O3, Cl2, H2O2, FeCl3, AgNO3 tc dng vi dung dch X. S cht lm dung dch X chuyn sang mu xanh l

A. 4 cht B. 5 cht C. 3 cht D. 2 cht Cu 43: Ha tan hon ton 3,76 gam hn hp: S, FeS, FeS2 trong HNO3 d c 0,48 mol NO2 v dung dch X. Cho dung dch Ba(OH)2 d vo X, lc kt ta nung n khi lng khng i th khi lng cht rn thu c l:

A. 17,545 gam B. 18,355 gam C. 15,145 gam D. 2,4 gam Cu 44: Ngi ta c th iu ch cao su Buna t g theo s sau:

Xenluloz %35 glucoz %80 C2H5OH %60 Buta-1,3-ien TH Cao su Buna

Khi lng xenluloz cn sn xut 1 tn cao su Buna l A. 25,625 tn. B. 37,875 tn. C. 5,806 tn. D. 17,857 tn.

Cu 45: Hn hp A gm 0,1 mol anehit metacrylic v 0,3 mol kh hiro. Nung nng hn hp A mt thi gian, c mt cht xc tc Ni, thu c hn hp hi B gm hn hp cc ancol, cc anehit v hiro. T khi hi ca B so vi He bng 95/12. Hiu sut ca phn ng hiro ha anehit metacrylic l:

A. 100% B. 70% C. 65% D. 80% Cu 46: Cho 6,0 gam Mg tc dng ht vi dung dch HCl 18,25% (va ). Sau phn ng thu c dung dch mui A v hiro that ra. Bit phn ng xy ra hon ton. Vy nng phn trm ca dung dch mui s l:

A. 22,41% B. 22,51% C. 42,79% D. 42,41% Cu 47: C 5 kh ng ring bit trong 5 l l Cl2, O2, HCl, O3, SO2. Hy chn trnh t tin hnh no trong cc trnh t sau phn bit cc kh:

A. Qu tm m, dung dch KI/h tinh bt, Cu un nng. B. Dung dch AgNO3, dung dch KI/h tinh bt, dng u que m cn tn . C. Nhn bit mu ca kh, dung dch AgNO3, dung dch KI, dng u que m cn tn . D. Dung dch H2S, dung dch AgNO3, dung dch KI.

Cu 48: T butan, cht v c v iu kin phn ng c . S phng trnh phn ng ti thiu iu ch glixerol l

A. 5. B. 4. C. 7. D. 6. Cu 49: Tm pht biu sai trong cc pht biu sau :

A. Nhit si ca ankanol cao hn so vi ankanal c phn t khi tng ng. B. Phenol l cht rn kt tinh iu kin thng. C. Metylamin l cht lng c mi khai, tng t nh amoniac. D. Etylamin d tan trong H2O.

Cu 50: Cho cc cht sau: C2H5OH, C6H5OH, C6H5NH2, dung dch C6H5ONa, dung dch NaOH, dung dch CH3COOH, dung dch HCl. Cho tng cp cht tc dng vi nhau c xc tc, s cp cht c phn ng xy ra l

A. 12 B. 8 C. 9 D. 10

B.Theo chng trnh Nng cao (10 cu, t cu 51 n cu 60) Cu 51: Cho NH3 d vo ln lt cc dung dch sau: CrCl3, CuCl2, ZnCl2, AgNO3, NiCl2. S trng hp kt ta hnh thnh b tan l

A. 4. B. 3. C. 1. D. 5. Cu 52: Mt bnh kn dung tch 1 lt cha 1,5 mol H2 v 1,0 mol N2 (c xc tc v nhit thch hp). trng thi cn bng c 0,2 mol NH3 to thnh. Mun hiu sut t 25 % cn phi thm vo bnh bao nhiu mol N2?

A. 1,5 B. 2,25 C. 0,83 D. 1,71




Cu 53: Cho 12gam hn hp Fe v Cu vo 200ml dung dch HNO3 2M, thu c mt cht kh (sn phm kh duy nht) khng mu, ha nu trong khng kh, v c mt kim loi d. Sau cho thm dung dch H2SO4 2M, thy cht kh trn tip tc thot ra, ho tan ht kim loi cn 33,33ml. Khi lng kim loi Fe trong hn hp l

A. 8,4 gam B. 5,6 gam C. 2,8 gam D. 1,4 gam Cu 54: Thy phn hon ton mt lng mantoz, sau cho ton b lng glucoz thu c ln men thnh ancol etylic th thu c 100 ml ru 460. Khi lng ring ca ancol l 0,8gam/ml. Hp th ton b kh CO2 vo dung dch NaOH d thu c mui c khi lng l:

A. 84,8 gam. B. 42,4 gam. C. 212 gam. D. 169,6 gam. Cu 55: Cho cc cht sau y: 1)CH3COOH, 2)C2H5OH, 3)C2H2, 4)CH3COONa, 5)HCOOCH=CH2, 6)CH3COONH4. Dy gm cc cht no sau y u c to ra t CH3CHO bng mt phng trnh phn ng l:

A. 1, 2, 3, 4, 5, 6. B. 1, 2, 6. C. 1, 2. D. 1, 2, 4, 6. Cu 56: Cho bit th in cc chun ca cc cp oxi ho kh: 2H+/H2; Zn2+/Zn; Cu2+/Cu; Ag+/Ag ln lt l 0,00V; -0,76V; +0,34V; +0,80V. Sut in ng ca pin in ho no sau y ln nht?

A. Cu + 2Ag+ Cu2+ + 2Ag. B. 2Ag + 2H+ 2Ag+ + H2. C. Zn + Cu2+ Zn2+ + Cu. D. Zn + 2H+ Zn2+ + H2.

Cu 57: Cc kh thi cng nghip v ca cc ng c t, xe my...l nguyn nhn ch yu gy ra ma axit. Nhng thnh phn ha hc ch yu trong cc kh thi trc tip gy ra ma axit l:

A. SO2, CO, NO. B. SO2, CO, NO2. C. NO, NO2, SO2. D. NO2, CO2, CO. Cu 58: Cho isopren tc dng Br2 theo t l mol 1:1 thu c ti a x dn xut mono brom. un nng ancol bc 2 C5H12O vi H2SO4 c 1800C thu c ti a y sn phm hu c. Mi lin h gia x, y l :

A. x - y = 1 B. x = y C. y - x = 1 D. y - x = 2 Cu 59: Cho cc cht sau: axit glutamic; valin, lysin, alanin, trimetylamin, anilin. S cht lm qu tm chuyn mu hng, mu xanh, khng i mu ln lt l

A. 2, 1,3. B. 1, 2, 3. C. 3, 1, 2. D. 1, 1, 4. Cu 60: Cho s phn ng sau :

Cng thc cu to ca Y l A. CH2=C(CH3)-COOC6H5. B. CH2=CH-COOC6H5. C. C6H5COOCH=CH2. D. CH2=C(CH3)-COOCH2-C6H5.

Khng c s dng bng HTTH




P N THI TH I HC LP 12 - LN 3 - MN HA HC -NM 2010 THPT CHUYN L QU N QUNG TR

Cu 135 213 358 486 Cu 135 213 358 486

1 A C C B 31 B B B D

2 B A C D 32 D C C A

3 B A D A 33 B D D C

4 D D C C 34 A A A B

5 D D B D 35 B D A A

6 C C B C 36 A B C D

7 A D B D 37 A B B D

8 C A B A 38 A B D D

9 D C A C 39 B A D D

10 C D A C 40 C C D C

11 B B B D 41 C B C C

12 D C C A 42 A D B D

13 C A C A 43 A A D B

14 D C D B 44 D C C D

15 D A B B 45 D A B A

16 D A C A 46 B D A A

17 C B D C 47 C B A C

18 A C C C 48 B B A D

19 C D A A 49 C C A B

20 A A A B 50 C C D C

21 C B C C 51 A D C B

22 C A A B 52 C C D A

23 C C D B 53 B C B D

24 A D B A 54 A D B B

25 B D A C 55 D A B C

26 B B B A 56 C C A C

27 D D D B 57 C B C D

28 A B A D 58 D B C A

29 B B A B 59 B D B D

30 D C D B 60 B A D C Thc s. Nguyn ThoDownload ti liu, thi, bi ging ti : http://diendan.shpt.info ----T: 0905434602


TI LIU THAM KHO

1. thi i hc cc nm

2. thi GVG cp tnh Qung Tr

3.Phng php gii ton ha hc

4.Cc thi trn mng

5.Sch Bi tp Ha hc v Sch Gio khoa lp 10, 11, 12 nng cao




MC LC LI M U ................................................................................................................ 1 DNG I : VIT NG PHN ..................................................................................... 2

I.MT S LU ..................................................................................................... 2 II.BI TP ................................................................................................................. 2

DNG II : BI TON T CHY ............................................................................ 5 I.MT S LU ..................................................................................................... 5 II.BI TP ................................................................................................................. 5

DNG III : BI TON POLIME ................................................................................ 7 I.MT S LU ..................................................................................................... 7 II.BI TP ................................................................................................................. 7

TRONG GII BI TP................................................................................................ 9 I.MT S LU ..................................................................................................... 9 II.BI TP ................................................................................................................. 9

DNG V : BI TON V PHN NG GIA HIRO ...........................................11 I.MT S LU ....................................................................................................11 II.BI TP ................................................................................................................11

DNG VI : DNG KT QU TH .....................................................................15 I.MT S LU ....................................................................................................15 II.BI TP ................................................................................................................15

DNG VII : DNG BI TP THEO..........................................................................17 I.MT S LU ....................................................................................................17 II.BI TP ................................................................................................................17

DNG VIII : BI TON LIN QUAN N DY IN HA CA KIM LOI ..18 I.MT S LU ....................................................................................................18 II.BI TP ................................................................................................................18

DNG IX : DNG TON T L ...............................................................................23 I.MT S LU ....................................................................................................23 II.BI TP ................................................................................................................23

DNG X : DNG TON QUI V CHT, CHN CHT .........................................25 I.MT S LU ....................................................................................................25 II.BI TP ................................................................................................................25

DNG XI : BI TON NG DNG .......................................................................28 I.MT S LU ....................................................................................................28 II.BI TP ................................................................................................................28

DNG XII : DNG BI TP S DNG NGUYN L NGC...........................30 I.MT S LU ....................................................................................................30 II.BI TP ................................................................................................................30

THI TH I HC LN III NM 2010 .............................................................33 I. PHN CHUNG CHO TT C TH SINH (40 cu t cu 1 n cu 40) ...........33 II. PHN RING [10 cu] ........................................................................................36 P N THI TH I HC LP 12 - LN 3 - MN HA HC -NM 2010 THPT CHUYN L QU N QUNG TR ...............................................................39

TI LIU THAM KHO ............................................................................................40 MC LC .....................................................................................................................41



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