16 Phuong Phap Giai Hoa

8/3/2019 16 Phuong Phap Giai Hoa

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Sch dnh tng hc sinh ph thng

16 Phng php v kthut gii nhanh ha hc

Cc cng thc gii nhanh trc nghim ha hc


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MC LCPHN I: 16 PHNG PHP V K THUT GII NHANH BI TP TRCNGHIM HA HC 3

Phng php 1:Phng php bo ton khi lng 4

Phng php 2:Phng php Bo ton nguyn t 16

Phng php 3:Phng php tng gim khi lng 24

Phng php 4:Phng php Bo ton in tch 40

Phng php 5:Phng php Bo ton electron 46

Phng php 6:Phng php trung bnh 62

Phng php 7:Phng php quy i 77

Phng php 8:Phng php ng cho 89

Phng php 9:Phng php h s 105

Phng php 10:Phng php s dng phng trnh ion thu gn 114

Phng php 11:Kho st th 125

Phng php 12:Phng php kho st t l s molCO2 v H2O133

Phng php 13:Phng php chia hn hp thnh hai phn khng

u nhau 145

Phng php 14:Phng php mi quan h gia cc i lng 150

Phng php 15:Phng php chn i lng thch hp 160

Phng php 16:Phng php chn i lng thch hp 170

Phng php 16+:Phng php s dng cng thc kinh nghim 178

PHN II: CC CNG THC GII NHANH TRC NGHIM HA HC 185CHNG I: CC CNG THC GII NHANH TRONG HA HC 186CHNG II: MT S BI TP THAM KHO 218CHNG III: HNG DN GII BI TP 228


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PHN I: 16 PHNG PHP V KTHUT GII NHANH

BI TP TRC NGHIM HA HC


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Phng php 1

Phng php bo ton khi lngPhng php bo ton khi lngPhng php bo ton khi lngPhng php bo ton khi lng

1. Ni dung phng php

- p dng nh lut bo ton khi lng (BTKL): Tng khi lng cc cht tham gia phnng bng tng khi lng cc cht sn phm

iu ny gip ta gii bi ton ha hc mt cch n gin, nhanh chng

Xt phn ng: A + B C + D

Ta lun c: mA + mB = mC + mD (1)

* Lu :iu quan trng nht khi p dng phng php ny l vic phi xc nh ng

lng cht (khi lng) tham gia phn ng v to thnh (c ch n cc cht kt ta, bay hi,

c bit l khi lng dung dch).

2. Cc dng bi ton thng gp

H qu 1: Bit tng khi lng cht ban u khi lng cht sn phm

Phng php gii: m(u) = m(sau) (khng ph thuc hiu sut phn ng)

H qu 2: Trong phn ng c n cht tham gia, nu bit khi lng ca (n 1) cht th ta d

dng tnh khi lng ca cht cn li.

H qu 3: Bi ton: Kim loi + axit mui + kh

m = m + m

- Bit khi lng kim loi, khi lng anion to mui (tnh qua sn phm kh) khi

lng mui

- Bit khi lng mui v khi lng anion to mui khi lng kim loi

- Khi lng anion to mui thng c tnh theo s mol kh thot ra:

Vi axit HCl v H2SO4 long

+ 2HCl H2 nn 2Cl H2

+ H2SO4 H2 nn SO42 H2

Vi axit H2SO4c, nng v HNO3: S dng phng php ion electron (xem thm

phng php bo ton electron hoc phng php bo ton nguyn t)

H qu 3: Bi ton kh hn hp oxit kim loi bi cc cht kh (H2, CO)

S: Oxit kim loi + (CO, H2) rn + hn hp kh (CO2, H2O, H2, CO)

Bn cht l cc phn ng: CO + [O] CO2

H2 + [O] H2O

n[O] = n(CO2) = n(H2O) m = m - m[O]

mui kim loi anion to mui

rn oxit


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3. nh gi phng php bo ton khi lng.

Phng php bo ton khi lng cho php gii nhanh c nhiu bi ton khi bit quan

h v khi lng ca cc cht trc v sau phn ng.

c bit, khi cha bit r phn ng xy ra hon ton hay khng hon ton th vic s

dng phng php ny cng gip n gin ha bi ton hn.Phng php bo ton khi lng thng c s dng trong cc bi ton nhiu cht.

4. Cc bc gii.

- lp s bin i cc cht trc v sau phn ng.

- T gi thit ca bi ton tmm

=m

(khng cn bit phn ng l hon ton hay

khng hon ton)

- Vn dng nh lut bo ton khi lng lp phng trnh ton hc, kt hp d kin khc

lp h phng trnh ton.

- Gii h phng trnh.

TH D MINH HA

V d 1: Ho tan hon ton 3,9 gam kali vo 36,2 gam nc thu c dung dch c nng

A. 15,47%. B. 13,97%. C. 14,0% D. 4,04%.

Gii:

2K + 2H2O 2KOH + H2

0,1 0,10 0,05(mol)

mdung dch = mK + OH2m - 2Hm = 3,9 + 36,2 - 0,05 2 = 40 gam

C%KOH =40

560,1100 % = 14% p n C

V d 2:in phn dung dch cha hn hp CuSO4 v KCl vi in cc trn khi thy kh bt

u thot ra c hai in cc th dng li thy c 448 ml kh (ktc) thot ra anot. Dung dch

sau in phn c th ho tan ti a 0,8 gam MgO. Khi lng dung dch sau in phn gim

bao nhiu gam (coi lng H2O bay hi l khng ng k) ?

A. 2,7 B. 1,03 C. 2,95. D. 2,89.

Gii:CuSO4 + 2KCl Cu + Cl2 + K2SO4 (1)

0,01 0,01

Dung dch sau in phn ho tan c MgO L dung dch axit, chng t sau phn

ng (1) CuSO4 d

2CuSO4 + 2H2O 2Cu + O2 + H2SO4 (2)

trc sau

0,02 0,01 0,02 (mol)


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n +2O

n =22400

480= 0,02 (mol)

H2SO4 + MgO MgSO4 + H2O (3)

0,02 0,02 (mol)

mdung dch gim = mCu + 2Clm + 2Om = 0,03 64 + 0,01x71 + 0,01x32 = 2,95 gam

p n C

V d 3: Cho 50 gam dung dch BaCl2 20,8 % vo 100 gam dung dch Na2CO3, lc b kt ta

c dung dch X. Tip tc cho 50 gam dung dch H2SO4 9,8% vo dung dch X thy ra 0,448 lt

kh (ktc). Bit cc phn ng xy ra hon ton. Nng % ca dung dch Na2CO3 v khi lng

dung dch thu c sau cng l:

A. 8,15% v 198,27 gam. B. 7,42% v 189,27 gam.

C. 6,65% v 212,5 gam. D. 7,42% v 286,72 gam.

Gii:

n = 0,05 mol; n = 0,05 mol

BaCl2 + Na2CO3 BaCO3 + 2NaCl

0,05 0,05 0,05 0,1

Dung dch B + H2SO4 kh dung dch B c Na2CO3 d

Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O

0,02 0,02

n ban u = 0,05 + 0,02 = 0,07 mol

C% =100

10607,0 %100 = 7,42%

LBTKL: mdd sau cng = 50 + 100 + 50 - m - m= 50 + 100 + 50 - 0,05.197 - 0,02.44 = 189,27 gam

p n B

V d 4: X l mt - aminoaxit, phn t cha mt nhm -NH2 v mt nhm -COOH. Cho 0,89

gam X phn ng va vi HCl thu c 1,255 gam mui. Cng thc to ra ca X l:

A. CH2 =C(NH2)-COOH. B. H2N-CH=CH-COOH.C. CH3-CH(NH2)-COOH. D. H2N-CH2-CH2-COOH.

Gii:

HOOC - R - NH2 + HCl HOOC -R-NH3Cl

mHCl = m mui - maminoaxit = 0,365 gam mHCl = 0,01 (mol)

Cl2

H2SO4BaCl2

Na2CO3

Na2CO3

CO2


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Maminoxit =01,0

89,0= 89

Mt khc X l -aminoaxit p n C

V d 5: Cho 15,6 gam hn hp hai ancol n chc, k tip nhau trong dy ng ng tc dng

ht vi 9,2 gam Na, thu c 24,5 gam cht rn. Hai ancol l:

A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.

C. C3H5OH v C4H7OH. D. C3H7OH v C4H9OH.

Gii:

2 OHR + 2Na 2 ONaR + H2

Theo bi hn hp ru tc dng vi ht Na Hc sinh thng nhm l: Na va , do

thng gii sai theo hai tnh hung sau:

Tnh hung sai 1: nNa= 23

2,9

= 0,4 nru = 0,4 ru = 4,0

6,15

= 39

p n A Sai.

Tnh hung sai 2: p dng phng php tng gim khi lng:

nru =22

6,155,24 = 0,405 ru =

405,0

6,15= 38,52 p n A Sai

p dng phng php bo ton khi lng ta c:

m = mru + mNa - mrn = 15,6 + 9,2 - 24,5 = 0,3 gam

nru= 2n = 0,3 (mol) ru = 3,06,15

= 52 p n B

V d 6: Trng hp 1,680 lt propilen (ktc) vi hiu sut 70%, khi lng polime thu c l:

A. 3,150 gam. B. 2,205 gam. C. 4,550 gam. D.1,850 gam.

Gii:

LBTKL: mpropilen = mpolime =4,22

680,1.42.

%100

%70= 2,205 gam p n B

V d 7: X phng ho hon ton 17,24 gam cht bo cn va 0,06 mol NaOH, c cn dung

dch sau phn ng thu c khi lng x phng l:

A. 17,80 gam. B.18,24 gam. C. 16,68 gam. D.13,38 gam.

(Trch thi tuyn sinh vo cc trngi hc, Cao ng khi B, 2008)

Gii:

(RCOO)3C3H5 + 3NaOH 3RCOONa + C3H5(OH)3

H2

H2

M

M

M

0,06 0,02 (mol)


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Theo nh lut bo ton khi lng:

17,24 + 0,06.40= mx phng + 0,02.92 mx phng =17,80 gam

p n: A

V d 8: Cho 3,60 gam axit cacboxylic no, n chc X tc dng hon ton vi 500ml dung dch

gm KOH 0,12M v NaOH 0,12M. C cn dung dch thu c 8,28 gam hn hp cht rn khan.Cng thc phn t ca X l:

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

C3H7COOH.

(Trch thi tuyn sinh vo cc trngi hc, Cao ng khi B, 2008)

Gii:

RCOOH + KOH RCOOK + H2O

RCOOH + NaOH RCOONa + H2O

nNaOH = nKOH = 0,5.0,12 = 0,06 mol

LBTKL: mX + mNaOH + mKOH = mrn + m

m = 1,08 gam n = 0,06 mol

nRCOOH = n = 0,06 mol MX = R + 45 =06,0

60,3= 60 R = 15

X: CH3COOH p n B

V d9: Nung 14,2 gam hn hp 2 mui cacbonat ca 2 kim loi ho tr 2 c 7,6 gam cht rn

v kh X. Dn ton b lng kh X vo 100ml dung dch KOH 1M th khi lng mui thu c

sau phn ng l:

A. 15 gam B. 10 gam C. 6,9 gam D. 5 gam

Gii:

X l CO2

LBTKL: 14,2 = 7,6 + mX mX = 6,6 gam nX = 0,15 mol

V:2CO

KOH

n

m=

15,0

1,0< 1 mui thu c l KHCO3

CO2 + KOH KHCO3

0,1 0,1 0,1 m = 0,1.100 = 10 gam p n B

V d 10: Nhit phn hon ton M gam hn hp X gm CaCO3 v Na2CO3 thu c 11,6 gam

cht rn v 2,24 lt khiu kin tiu chun. Hm lng % ca CaCO3 trong X l:

A. 6,25% B. 8,62% C. 50,2% D. 62,5%

H2O

H2O H2O

H2O

KHCO3


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Gii:

CaCO3 ot CaO + CO2

nCaCO 3 = nCO2= 0,1 (mol) mCaCO 3 = 10 gam

Theo LBTKL: mX = mcht rn = mkh= 11,6 + 0,144=16 gam %CaCO3=

16

10 100% = 62,5% p n: D

V d 11: un 27,6 gam hn hp 3 ancol n chc vi H2SO4c 140oC (H=100%) c 22,2

gam hn hp cc ete c s mol bng nhau. S mol mi ete trong hn hp l:

A. 0,3. B. 0,1 C. 0,2 D.0,05

Gii:

S ete thu c l:2

)13(3 += 6

LBTKL: 27,6= 22,2 + OH2m OH2m = 5,4 gam OH2n = 0,3 mol

OH2n = eten = 6nete nmi ete = 0,3: 6 = 0,5 mol p n: DV d 12: t chy hon ton 0,025 mol cht hu cX cn 1,12 lt O2 (ktc), dn ton b sn

phm thu c qua bnh 1 ng P2O5 khan v bnh 2 ng Ca(OH)2 d thy khi lng bnh 1

tng 0,9 gam, bnh 2 tng 2,2 gam. Cng thc phn t ca X l:

A. C2H4O. B. C3H6O. C. C3H6O2. D. C2H4O2.

Gii

mbnh 2 tng =2CO

m , mbnh 1 tng = OH2m

LBTKL: mx +2O

m =2CO

m + OH2m mx + 32.0,05 = 0,9 + 2,2

mx = 1,5 gam

Mx = 1,5:0,025=60 p n: D

V d 13: Cho 20,2 gam hn hp 2 ancol tc dng va vi K thy thot ra 5,6 lt H2(ktc) v

khi lng mui thu c l:

A. 3,92 gam B. 29,4 gam C. 32,9 gam D. 31,6 gam

Gii:

R (OH)a + aK R (OK)a +a

2H2

x xa 0,5 ax 2Hn = 0,5 ax = 0,25 ax = 0,5 mol

LBTKL: 20,2 + 39.0,5 = mmui + 2.0,25 mmui = 39,2 gam p n A


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V d 14: X phng ho cht hu cX n chc c 1 mui Y v ancol Z. t chy hon ton

4,8 gam Z cn 5,04 lt O2 (ktc) thu c lng CO2 sinh ra nhiu hn lng nc l 1,2 gam.

Nung mui Y vi vi ti xt thu c kh T c t khi hi i vi H2 l 8. Cng thc cu to ca

X l:

A. C2H5COOCH3 B. CH3COOCH3

C. HCOOCH3. D. CH3COOC2H5

Gii:

X + NaOH mui Y + ancol Z X: este n chc

RCOOR + NaOH ot RCOONa + ROH

RCOONa + NaOH RH + Na2CO3

MRH = 8.2 =16 RH: CH4 RCOONa : CH3COONa

CxHyO(Z) + O2 CO2 + H2O

LBTKL: 4,8 + 0,225.32 =2CO

m + OH2m = 12

2COm = OH2m + 1,2 2COm = 6,6 gam, OH2m = 5,4 gam

mC = 12.2CO

n =1,8 gam; mH = 2.2H O

n = 0,6 gam; mO = 2,4 gam

x: y: z =12

8,1:

1

6,0:

16

4,2= 0,15: 0,6: 0,15 = 1: 4: 1

Z: CH3OH X : CH3COOCH3 p n B

V d 15: t chy hon ton 4,3 gam mt axit cacboxylic X n chc thu c 4,48lt CO2

(ktc) v 2,7 gam H2O. S mol ca X l:

A. 0,01mol B. 0,02 mol C. 0,04 mol D. 0,05 mol

Gii:

Theo LBTKL: mX +2O

m =2CO

m + O2Hm

2O

m = 2,7 + 0,2 44 4,3 = 10,3 gam 2O

n = 0,225 (mol)

p dng bo ton nguyn ti vi oxi:

nX +2O

n =2CO

n +2

n OH2 nX =2CO

n +2

n OH2 -2O

n = 0,05(mol) p n D

V d 16: t chy hon ton x gam hn hp X gm propan, buten-2, axetilen thu c 47,96

gam CO2 v 21,42 gam H2O. Gi tr X l:

A. 15,46. B. 12,46. C. 11,52. D. 20,15.

CaO/t0


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Gii:

2COn = 1,09 mol ;

2H On = 1,19 mol

x = mC + mH = 12. 2COn + oH22.n = 15,46 gam p n A

V d 17: un nng 5,14 gam hn hp kh X gm metan, hiro v mt ankin vi xc tc Ni, thuc hn hp kh Y. Cho hn hp Y tc dng vi dung dch brom d thu c 6,048 lt hn hp

kh Z (ktc) c t khi i vi hiro bng 8. tng khi lng dung dch brom l:

A. 0,82 gam. B. 1,62 gam C. 4,6 gam D. 2,98 gam.

Gii:

X otNi, Y + 2Br Z

Nhn thy: mkh tc dng vi dung dch brom = mkhi lng bnh brom tng

mX = mY = mZ + mkhi lng bnh brom tng

mkhi lng bnh brom tng = mX - mZ = 5,14 -4,22

048,6 28 = 0,82 gam p n A

V d 18: Ho tan hon ton 8,9 gam hn hp 2 kim loi bng dung dch HCl dc 4,48 lt

(ktc). C cn dung dch thu c sau phn ng th lng mui khan thu c l:


Gii:

Cch 1: Gi cng thc chung ca hai kim loi M, ha tr n

2M + 2nHCl 2MCln + nH2

0,4 0,2 (mol)

Theo LBTKL: mkim loi + mHCl = mmui +2H

m

mmui = 8,9 + 0,4 36,5 0,2 2 =23,1 gam p n A

Cch 2: mCl-mui = nH+ =

2H2.n = 0,4 (mol)

mmui = mkim loi + mCl-(mui) = 8,9 + 0,435,5 = 23,1 gam p n A

V d 19. Ho tan hon ton 15,9 gam hn hp gm 3 kim loi Al, Mg v Cu bng dung dch

HNO3 thu c 6,72 lt kh NO (sn phm kh duy nht) v dung dch X. C cn cn thn dung

dch X th lng mui khan thu c l bao nhiu?A. 77,1 gam B. 71,7 gam C. 17,7 gam D. 53,1 gam

Gii:

5+

N + 3e 2+

N(NO)

0,9 0,3(mol)


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V sn phm kh duy nht l NO 3ON

n (trong mui) = n e nhng (hoc nhn) = 0,9 mol

(Xem thm phng php bo ton e)

mmui = mcation kim loi + mNO 3 (trong mui) 15,9 + 0,9 62 = 71,7 gam

p n B

BI TP TLUYN

Cu 1 : Trn 5,4 gam Al vi 6,0 gam Fe2O3 ri nung nng thc hin phn ng nhit nhm.

Sau phn ng ta thu c hn hp rn c khi lng l

A.11,40 gam. B. 9,40 gam. C. 22,40 gam. D. 9,45 gam.

Cu 2 : Trong bnh kn cha 0,5 mol CO v m gam Fe3O4. un nng bnh cho ti khi phn ng

xy ra hon ton, th kh trong bnh c t khi so vi kh CO ban u l 1,457. Gi tr ca m l.A. 16,8 B. 21,5 C. 22,8 D. 23,2

Cu 3:in phn 100 ml dung dch CuSO4 vi n cc, sau mt thi gian my khi lng dung

dch gim 12 gam. Dung dch sau in phn tc dng va vi 100ml dung dch H2S 1M. Nng

mi ca dung dch CuSO4 trc khi in phn l

A. 1M. B. 1,5 M. C. 2M. D. 2,5M.

Cu 4 : Cho mt lung CO i qua ng sng 0,04 mol hn hp A gm FeO v Fe2O3t nng

sau khi kt thc th nghim thu c cht rn B gm 4 cht nng 4,784 gam. Khi ra khi ng

s hp th vo dung dch Ca(OH)2 d, th thu c 4,6 gam kt ta. Phn trm khi lng FeOtrong hn hp A l

A. 13,03%. B. 31,03%. C. 68,03%. D. 68,97%.

Cu 5 : Dn kh CO t t qua ng sng 14 gam CuO, Fe2O3, FeO nung nng mt thi gian

thu c m gam cht rn X. Ton b kh thu c sau phn ng c dn chm qua dung dch

Ca(OH)2 d, kt ta thu c cho tc dng vi dung dch HCl dc 2,8 lt kh (ktc). Gi tr

ca m l

A. 6 gam. B. 12 gam. C. 8 gam. D. 10 gam.

Cu 6 : Nung hon ton 10,0 gam hn hp X gm CaCO3 v NaCl. Kt thc th nghim thu c7,8 gam cht rn khan. Khi lng CaCO3 c trong X l

A. 5,0 gam. B. 6,0 gam. C. 7,0 gam. D. 8,0 gam.

Cu 7 : Nung nng 34,8 gam hn hp X gm MCO3 v NCO3c m gam cht rn Y v 4,48 lt

CO2 (ktc). Nung Y cho n khi lng khng i c hn hp rn Z v kh CO2 dn ton b

CO2 thu c qua dung dch KOH d, tip tc cho thm CaCl2 d th c 10 gam kt ta. Ho


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tan hon ton Z trong V lt dung dch HCl 0,4M va c dung dch T. Gi tr m gam v V lt

ln lt l :

A. 26 v 1,5. B. 21,6 v 1,5. C. 26 v 0,6. D. 21,6 v 0,6.

Cu 8 : Ho tan 9,14 gam hp kim Cu, Mg, Al bng mt lng va dung dch HCl thu c

7,84 lt kh X (ktc), 2,54 gam cht rn Y v dung dch Z. Lc b cht rn Y, c cn cn thndung dch Z thu c lng mui khan l


Cu 9 : Cho 11,0 gam hn hp X gm Al v Fe vo dung dch HNO3 long d. thu c dung

dch Y (khng cha mui amoni), hn hp kh Y gm 0,2 mol NO v 0,3 mol NO2. C cn dung

dch Y th lng mui khan thu c l:


Cu 10 : Ho tan ht 7,8 gam hn hp Mg, Al trong dung dch HCl d. Sau phn ng thy khi

lng dung dch tng 7,0 gam so vi ban u. S mol axit phn ng lA. 0,08 mol B. 0,04 mol C. 0,4 mol D. 0,8 mol

Cu 11 : Cho x gam Fe ho tan trong dung dch HCl, sau khi c cn dung dch thu c 2,465

gam cht rn. Nu cho x gam Fe v y gam Zn vo lng dung dch HCl nh trn thu c 8,965

gam cht rn v 0,336 lt H2 (ktc). Gi tr ca x, y ln lt l:

A. 5,6 v 3,25 B. 0,56 v 6,5 C. 1,4 v 6,5. D. 7,06 v 0,84

Cu 12 : Ho tan hon ton 11,4 gam hn hp X gm kim loi M (ho tr I) v kim loi N (ho

tr II) vo dung dch cha ng thi H2SO4 v HNO3c nng thu c 4,48 lt (ktc) hn hp Y

gm NO2 v SO2 c t khi hi so vi hiro l 28,625 v mui khan c khi lng l:A. 44,7 gam B. 35,4 gam C. 16,05 gam D. 28,05 gam.

Cu 13: Ly 35,1 gam NaCl ho tan vo 244,9 gam H2O. Sau in phn dung dnh vi in

cc trc mng ngn cho ti khi catot thot ra 1,5 gam kh th dng li. Nng cht tan c

trong dung dch sau in phn l:

A. 9,2% B. 9,6% C. 10% D. 10,2%.

Cu 14:un a gam 1 ancol X vi H2SO4c 1700C c 1 olefin. Cho a gam X qua bnh ng

CuO d, nung nng (H = l00%) thy khi lng cht rn gim 0,4 gam v hn hp hi thu c

c t khi hi i vi H2 l l5,5. Gi tr a gam l:A. 23 B. 12,5 C. 1,15 D. 16,5.


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Cu 15 : Dn V lt (ktc) hn hp X gm axetilen v H2i qua ng sng Ni nung nng thu

c khi Y. Dn Y vo lng d dung dch AgNO3/NH3c 12 gam kt ta. Kh ra khi dung

dch phn ng va vi dung dch cha 16 gam Br2 v cn li kh Z. t chy hon ton Z thu

c 0,1 mol CO2 v 0,25 mol nc.

A. 11,2 B. 13,44 C. 5,6 D. 8,96.Cu 16 :un nng 7,6 gam hn hp X gm C2H2, C2H4 v H2 trong bnh kn vi xc tc Ni thu

c hn hp kh B. t chy hon ton hn hp Y, dn sn phm chy thu c ln lt qua

bnh 1 ng H2SO4c, bnh 2 ng Ca(OH)2 d thy khi lng bnh 1 tng 14,4 gam. Khi

lng tng ln bnh 2 l


Cu 17:t chy ht m gam hn hp X gm etan, etilen, axetilen v butaien-1,3 ri cho sn

phm chy hp th vo dung nh nc vi d, thu c 100 gam kt ta. Khi lng dung dch

nc vi sau phn ng gim 39,8 gam. Tr s ca m l:A. 58,75 gam B. 13,8 gam C. 37,4 gam D. 60,2 gam.

Cu 18 :t chy hon ton m gam hn hp X gm C2H2, CH4, C3H6 v C4H10 thu c 4,4 gam

CO2 v 2,52 gam H2O. m c gi tri l:

A. 1,48 gam B. 2,48 gam C. 14,8 gam D. 24,8 gam.

Cu 19: Thc hin phn ng ete ho hon ton 11,8 gam hn hp hai ru no n chc, mch

h, ng ng k tip thu c hn hp gm ba ete v l,98 gam nc. Cng thc hai ru l:

A. CH3OH, C2H5OH B. C4H9OH, C5H11OH.

C. C2H5OH, C3H7OH D. C3H7OH, C4H9OH.Cu 20 : Cho 10,1 gam hn hp 2 ancol n chc, k tip nhau trong dy ng ng tc dng ht

vi 5,75 gam Na c 15,6 gam cht rn. Hai ancol cn tm l

A. C2H5OH v C3H7OH. B. CH3OH v C2H5OH.

C. C3H7OH v C4H9OH. D. C3H5OH v C4H9OH .

Cu 21: Ho tan 25,2 gam tinh th R(COOH)n.2H2O vo 17,25ml etanol (D = 0,8g/ml) c

dung dch X. Ly 7,8 gam dung dnh X cho tc ng ht vi Na va thu c cht rn Y v

2,464 lt kh H2 (ktc). Khi lng ca Y l:

A. 12,64 gam B. 10,11 gam C. 12,86 gam D. 10,22 gam.Cu 22 : t chy hon ton a gam 1 este n chc ca ru metylic cn 1,68 lt kh O2 (ktc)

thu c 2,64 gam CO2; 1,26 gam H2O v 0,224 lt N2 (ktc). Cng thc cu to thu gn ca este

l:

A. CH3COOCH2NH2 B. CH3CH(NH2)COOCH3

C. H2NCH2CH2COOCH3 D. H2NCH2COOCH3


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15

15

Cu 23 : Cho 14,8 gam hn hp bn axit hu cn chc tc dng vi lng va Na2CO3 to

thnh 2,24 lt kh CO2 (ktc). Khi lng mui thu c l:

A. 15,9 gam B. 17,0 gam C. 19,3 gam D. 19,2 gam.

Cu 24 :t hon ton 34 gam este X cn 50,4 lt O2 (ktc) thu c OHCO 22 n:n = 2 . un

nng 1 mol X cn 2 mol NaOH. Cng thc cu to ca X l

A. CH3COOC6H5 B. C6H5COOCH3 C. C2H5COOC6H5 D. C6H5COOC2H5

Cu 25 : X phng ho hon ton m gam lipit X bng 200 gam dung dch NaOH 8%. Sau phn

ng c 9,2 gam glixerol v 94,6 gam cht rn khan. Cng thc cu to ca X l

A. (C17H35COO)3C3H5 B. (C15H31COO)3C3H5

C. (C17H33COO)3C3H5 D. (C17H31COO)3C3H5

Cu 26 :un nng 15 gam cht bo trung tnh vi 150ml dung dch NaOH 1M. Phi dnh 50ml

dung dch H2SO4 1M trung ho NaOH d. Khi lng x phng (cha 70% khi lng mui

nm ca axit bo) thu c t 2 tn cht bo trn l

A. 2062 kg B. 3238 kg. C. 2946 kg. D. 2266 kg.

Cu 27 : x phng ho hon ton 1 kg cht bo (c ln 1 lng nh axit bo t do) c ch s

axit bng 8,4 phi dng 450ml dung dch NaOH 1M. Khi lng x phng thu c l

A. 1001,6 kg. B. 978,7 gam. C. 987,7 kg D. 1006,1 gam.

Cu 28 : Cho 15 gam hn hp 3 amin n chc bc mt tc dng va vi dung dch HCl

1,2M th thu c 18,504 gam mui. Th tch ung dch HCl phi dng l

A. 0,8 lt. B. 0,08 lt. C. 0,4 lt. D. 0,04 lt

Cu 29 : Cho 0,01 mol amino axit X phn ng va vi 100ml dung dch HCl 0,1M thu c

1,695 gam mui. Mt khc 19,95 gam X tc dng vi 350ml dung dch NaOH 1M. C cn dung

dch thu c 28,55 gam cht rn. Cng thc cu to ca X l

A. HOOCCH(NH2)CH2NH2 B. NH2(CH2)3COOH.

C. HOOCCH2CH(NH2)COOH. D. HOOC(CH2)2CH(NH2)COOH.

P N1A 2D 3D 4A 5B 6A 7A 8A 9B 10D

11C 12D 13B 14C 15A 16C 17B 18A 19C 20B21A 22D 23D 24A 25D 26C 27D 28B 29C


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16

16

Phng php 2

Phng php Bo ton nguyn tI. PH

NG PHP GI

I

- Nguyn tc chung ca phng php l da vo nh lut bo ton nguyn t(BTNT); Trong

cc phn ng ha hc thng thng, cc nguyn tlun c bo ton

iu ny c ngha l: Tng smol nguyn tca mt nguyn tX bt k trc v sau phn ng

l lun bng nhau

- im mu cht ca phng php l phi xc nh c ng cc hp phn c cha nguyn t X

trc v sau phn ng, p dng LBT nguyn t vi X rt ra mi quan h gia cc hp

phn t a ra kt lun chnh.

II. CC DNG BI TP THNG GPPhng php bo ton nguyn t c th p dng cho hu ht cc dng bi tp, c bit l cc

dng bi hn hp nhiu cht, xy ra nhiu bin i phc tp. Di y l mt s dng bi tp in

hnh.

Dng 1. Tnhiu cht ban u to thnh mt sn phm.

T d kin bi s mol ca nguyn t X trong cc cht u tng s mol trong sn phm

to thnh s mol sn phm.

- Hn hp kim loi v oxit kim loi hyroxit kim loi oxit

- Al v Al2O3 + cc oxit st hn hp rn hyroxit Al2O3 + Fe2O3

2 3Al O

n (cui) = Aln

2+

2 3Al On (u) ;

2 3Fe On (cui) =

Fen

2

Dng 2. Tmt cht ban u to thnh hn hp nhiu sn phm

T d kin bi tng s mol ban u, s mol ca cc hp phn cho s mol ca cht

cn xc nh.

- Axit c tnh oxi ha (HNO3, H2SO4c, nng) Mui + kh

nX (axit) = nX (mui) + nX (kh) (X: N hoc S)

- Kh CO2 (hoc SO2) hp th vo dung dch kim:

CO2 CO32 + HCO3 SO2 SO32 + HSO3

2CO

n = 23CO

n +3HCO

n 2SO

n = 23SO

n +3HSO

n

t0

(u)

Kim loi


17/236

17

17

- Tnh lng tnh ca Al(OH)3

Trng hp 1 Trng hp 2

Al3+ OH

Al(OH)3 + [Al(OH)4] [Al(OH)4] H+

Al(OH)3 + Al3+

3Al

n + =3Al(OH)

n[ ]

+3

Al(OH)n 4Al(OH)

n[ ]

= 3Al

n + +3

Al(OH)n

- Hn hp cc oxit kim loi + CO (H2)0t hn hp cht rn + CO2 (H2O)

Theo nh lut bo ton nguyn t vi O:

* Khi H = 100%: nO (oxit) = nO (rn) + nhn hp kh sau = nO (rn) + nhn hp kh trc

* Khi H < 100%:

nO (oxit) = nO (rn) +

- Bi ton cracking ankan:

Ankan X hn hp Y

Mc d c nhng bin i ha hc xy ra trong qu trnh cracking, v Y thng l hn hp phctp (c th c H2), do phn ng cracking xy ra theo nhiu hng, vi hiu sut H < 100%.

Nhng ta ch quan tm n s bo ton nguyn ti vi C, H t d dng xc nh c tng

lng ca 2 nguyn t ny.

Thng thng bi cho s mol ankan X C(Y) C(X)

H(Y) H(X)

n n

n n

=

=

Dng 3. Tnhiu cht ban u to thnh hn hp nhiu sn phm

Trong trng hp ny khng cn thit phi tm chnh xc s mol ca tng cht, m ch quan tm

n h thc: X(n = X(n

Tc l ch quan tm n tng s mol ca nguyn t trc v sau phn ng. Nu bit X(n

X(n v ngc li.

Vi dng ny, bi thng yu cu thit lp mt h thc di dng tng qut v s mol cc cht.

Dng 4. Bi ton it chy trong ha hu c

Xt bi t chy tng qut: CxHyOzNt + O2 CO2 + H2O + N2nC =

2COn

Theo LBT nguyn t: nH = 2.2H O

n x y z tO(C H O N )

n = 2.2CO

n +2H O

n - 2.2O

n

nN = 2.2N

n

mhn hp kh sau - mhn hpkh16

cracking

u) cui)

u)

cui)

t0


18/236

18

18

Phng php bo ton khi lng nguyn t vi O c s dng rt ph bin trong cc bi ton

ha hu c.

* Ch :i vi trng hp t chy hp cht hu ccha Nitbng khng kh, lng nitthu

c sau phn ng l:2N

n (sau phn ng) =2N

n (t phn ng t chy) +2N

n (t khng kh)

p dng tt phng php BTNT, cn ch mt sim sau:

* Hn ch vit phng trnh phn ng m thay vo nn vit s phn ng (s hp

thc, c ch h s) biu din cc bin i cbn ca cc nguyn t quan tm.

* bi thng cho (hoc qua d kin bi ton s tnh c) s mol ca nguyn t quan tm,

t xc nh c lng (mol, khi lng) ca cc cht.

III. CC V D

V d 1: Ho tan hn hp X gm 0,2 mol Fe v 0,1 mol Fe2O3 vo dung dch HCl dc dung

dch D. Cho dung dch D tc dng vi NaOH d thu c kt ta. Lc kt ta, ra sch em nung

trong khng khn khi lng khng i thu c m gam cht rn Y. Gi tri ca m l

A. 16,0. B. 30,4. C. 32,0. D. 48,0.

Gii:

S : }{ 32t

3

2NaOH

3

2HCl

32

OFeYFe(OH)

Fe(OH)

FeCl

FeCl

OFe

FeX

0

Theo BTNT vi Fe: nFe2O3(Y) = mol0,20,12

0,2n

2n

(X)OFeFe

32=+=+

m = 0,2.160 = 32,0 p n CV d 2:un nng hn hp bt X gm 0,06 mol Al, 0,01 mol Fe3O4, 0,015 mol Fe2O3 v 0,02

mol FeO mt thi gian. Hn hp Y thu c sau phn ng c ho tan hon ton vo dung dch

HCl d, thu c dung dch Z. Thm NH3 vo Z cho n d, lc kt ta T, em nung ngoi

khng khn khi lng khng i thu c m gam cht rn. Gi tr ca m l

A. 6,16. B. 6,40. C. 7,78. D. 9.46

Gii:

Theo BTNT vi Al:32OAl

n =2

nAl= 0,03 mol


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19

19

Theo BTNT vi Fe: 32OFen = mol0,04n23n

2n

(X)OFe(X)OFeFe

32

43 =++

m = =+=+ 9,460,04.1600,06.102nn3232 OFeOAl

p n D

V d 3:t chy 9,8 gam bt Fe trong khng kh thu c hn hp rn X gm FeO, Fe3O4 v

Fe2O3. ho tan X cn dng va ht 500ml dung dch HNO3 1,6M, thu c V lt kh NO (sn

phm kh duy nht, do ktc). Gi tr ca V l

A. 6,16. B. 10,08. C. 11,76. D. 14,0.

Gii:

S phn ng : Fe + ++ NO)Fe(NOX 33HNOtO 3

02,

Theo BNTN vi Fe:33 )Fe(NO

n = nFe = 0,175mol

Theo BNTN vi N: nNO =3HNO

n 333 )Fe(NO

n = 0,5.1,6 3.0,175 = 0,275 mol

V = 0,275. 22,4 = 6,16 p n A

V d 4: Ly a mol NaOH hp th hon ton 2,64 gam kh CO2, thu c ng 200ml dung dch

X. Trong dung dch X khng cn NaOH v nng ca ion 23CO l 0,2M. a c gi tr l :

A. 0,06. B. 0,08. C. 0,10. D. 0,12.

Gii:

S phn ng :

CO2 + NaOH Na2CO3 + NaHCO3

Theo BNTN vi C : 0,02mol0,2.0,244

2,64nnn3223 CONaCONaHCO

===

Theo BNTN vi Na: a = 232CONa

n +3NaHCO

n = 2. 0,04 + 0,02 = 0,1 p n C

V d 5: Ho tan hon ton hn hp gm x mol FeS2 v y mol Cu2S vo axit HNO3 (va ), thu

c dung dch X (ch cha hai mui sunfat) v kh duy nht NO. T s x/y l

A. 6/5. B. 2/1. C. 1/2. D. 5/6.

Gii:

X ch cha 2 mui sunfat, kh NO l duy nht S chuyn ht thnh 24SO

S bin i:

2yy0,5xx

2CuSOSCu;)(SOFe2FeS 423422

Theo BTNT vi S: 2x + y = 3.0,5x + 2y 0,5x = y x/y = 2/1 p n B


20/236

20

20

V d 6:t chy hon ton m gam hn hp X gm C3H8, C4H6, C5H10 v C6H6 thu c 7,92

gam CO2 v 2,7 gam H2O, m c gi tr l

A. 2,82. B. 2,67. C. 2,46. D. 2,31.

Gii:

S phn ng: X {C3H8 , C4H6 , C5H10 , C6H6} +0

2 ,tO

OH

CO

2

2

Theo BTNT vi C v H: m = mc + mH = =+ 2,469

2,7x12

447,92

p n C

V d 7: Tin hnh cracking nhit cao 5,8 gam butan. Sau mt thi gian thu c hn hp

kh X gm CH4 , C2H6, C2H4, C3H6 v C4H10. t chy hon ton X trong kh oxi d, ri dn ton

b sn phm sinh ra qua bnh ng H2SO4c. tng khi lng ca bnh H2SO4c l


Gii:

S phn ng : C4H100

2,O tcracking X+ H2O

Khi lng bnh H2SO4c tng ln l khi lng ca H2Ob hp th

Theo BTNT vi H: mol0,5585,8

5.2

10n

2n

n 1042

HCHOH ====

OH2n = 0,5.18 = 9,0 gam p n A

V d 8: t chy hon ton 0,1 mol anehit n chc X cn dng va 12,32 lt kh O2 (ktc),

thu c 17,6 gam CO2, X l anehit no di y?

A. CH=C-CH2-CHO. B. CH3-CH2-CH2-CHO.

C. CH2=CH-CH2-CHO. D. CH2=C=CH-CHO.

Gii:

2On = 0,55 mol;

2COn = 0,4 mol

Nhn xt: X l anehit n chc nO(X) = nX = 0,1 mol

Theo LBT nguyn t vi O :

OH2n =

O)O(H2n = nX + 2

2On - 2

2COn = 0,1+2.0,55-2.0,4 = 0,4 mol

Nhn thy:

=

==

XCO

COOH

4nn

0,4molnn

2

22 X l CH3 CH2 CH2 CHO p n B


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21

21

V d 9: X l mt ancol no, mch h. t chy hon ton 0,05 mol X cn 5,6 gam oxi, thu c

hi nc v 6,6 gam CO2. Cng thc ca X l

A. C2H4(OH)2 B. C3H7OH. C. C3H6(OH)2 D. C3H5(OH)3

Gii:

2On = 0,175mol; 2COn = 0,15mol

S chy: X + O2 CO2 + H2O

V X l ancol no, mnh h 22 COXOH

nnn += = 0,05+0,15 = 0,2 mol

Theo LBT nguyn t vi O :

nO(X) =222 OOHCO

n2nn2 + = 2.0,15 + 0,2 2.0,175 = 0,15mol

Nhn thy

=

=

XO(X)

XCO

3nn

3nn2 X l C3H5(OH)3 p n D

V d 10:t chy hon ton m gam mt amin n chc X bng lng khng kh va thu

c 1,76 gam CO2; 1,26 gam H2O v V lt N2 (ktc). Gi thit khng kh ch gm N2 V O2

trong oxi chim 20% v th tch. Cng thc phn t ca X v th tch V ln lt l

A. X l C2H5NH2 ; V = 6,72 1t. B. X l C3H7NH2 ; V = 6,944 1t.

C. X l C3H7NH2 ; V = 6,72 1t. D. X l C2H5NH2 ; V = 6,944 1t.

Gii:

2COn = 0,04 mol; OH2n = 0,07 mol

Nhn thy: == 270,040,07.2nnCH X l C2H5NH2

S chy: 2C2H5NH2 + O2 4CO2 + 7H2O + N2

Theo LBT nguyn t vi N:2N

n (t phn ng t chy) = 0,01mol2

0,022

nX ==

Theo LBT nguyn t vi O: CO2n + 0,075mol20,07

0,042

n OH2 =+=

2N

n (t khng kh) =2O

4n = 4. 0,075 = 0,3 mol

2Nn (thu c) = 2Nn (t phn ng t chy) + 2Nn (t khng kh)= 0,01 + 0,3 = 0,31 mol

V= 22,4.0,31 = 6,944 lt p n D


22/236

22

22

IV. BI TP TLUYN

Cu 1 : Hn hp cht rn X gm 0,1 mol Fe2O3 v 0,1 mol Fe3O4. Ho tan hon ton X bng

dung dch HCl d, thu c dung dch Y. Cho NaOH d vo Y, thu c kt ta Z. Lc ly kt

ta, ra sch ri em nung trong khng khn khi lng khng i th thu c cht rn ckhi lng l


Cu 2 : Cho 4,48 lt kh CO (ktc) t ti qua ng s nung nng ng 8 gam mt oxit st n

khi phn ng xy ra hon ton. Kh thu c sau phn ng c t khi so vi hiro bng 20. Cng

thc ca oxit st v phn trm th tch ca kh CO2 trong hn hp kh sau phn ng ln lt l:

A. FeO; 75%. B. Fe2O3; 75%. C. Fe2O3; 65%. D. Fe3O4; 75%.

Cu 3 : Hn hp A gm etan, etilen, axetilen v butaien-1,3. t chy ht m gam hn hp A.

Cho sn phm chy hp th vo dung dch nc vi d, thu c 100 gam kt ta v khi lngdung dch nc vi sau phn ng gim 39,8 gam. Tr s ca m l


Cu 4 : Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo axit HNO3 (va ),

thu c dung dch X (ch cha hai mui sunfat) v kh duy nht NO. Gi tr ca m l

A. 0,06. B. 0,04. C. 0,12. D. 0,075.

Cu 5 :t chy hon ton mt th tch kh thin nhin gm metan, etan, propan bng oxi khng

kh (trong khng kh, oxi chim 20% th tch), thu c 7,84 lt kh CO2 (ktc) v 9,9 gam

nc. Th tch khng kh (ktc) nh nht cn dng t chy hon ton lng kh thin nhintrn l

A. 70,0 lt B. 78,4 lt. C. 84,0 lt. D. 56,0 lt.

Cu 6 : Dn V lt (ktc) hn hp X gm axetilen v hiro i qua ng sng bt niken nung

nng, thu c kh Y. Dn Y vo lng d AgNO3 (hoc Ag2O) trong dung dch NH3 thu c

12 gam kt ta. Khi 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. 5,6. B. 13,44. C. 11,2. D. 8,96.

Cu 7: Ho tan hon ton 0,3 mol hn hp gm Al v Al4C3 vo dung dch KOH (d), thu cx mol hn hp kh v dung dch X. Sc kh CO2 (d) vo dung dch X, lng kt ta thu c l

46,8 gam. Gi tr ca x l

A. 0,55. B. 0,60. C. 0,40. D. 0,45.


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23

23

Cu 8 : Ho tan hon ton m gam oxit FexOy bng dung dch H2SO4c nng va , c cha

0,075 mol H2SO4, thu c z gam mui v thot ra 168ml kh SO2 (sn phm kh duy nht, o

ktc). Oxit FexOy l

A. FeO. B. Fe2O3 C. Fe3O4 D. FeO hoc Fe3O4

Cu 9: Ho tan hon ton hn hp gm 0,27 gam bt nhm v 2,04 gam bt Al2O3 trong dungdch NaOH d thu c dung dch X. Cho CO2 d tc dng vi dung dch X thu c kt ta Y,

nung Y nhit cao n khi lng khng i thu c cht rn Z. Bit hiu sut cc phn ng

u t 100%. Khi lng ca Z l

A. 2,04 gam B. 2,31 gam. C. 3,06 gam. D. 2,55 gam.

Cu 10 :un nng 7,6 gam hn hp A gm C2H2, C2H4 v H2 trong bnh kn vi xc tc Ni thu

c hn hp kh B. t chy hon ton hn hp B, dn sn phm chy thu c ln lt qua

bnh 1 ng H2SO4c, bnh 2 ng Ca(OH)2 d thy khi lng bnh 1 tng 14,4 gam. Khi

lng tng ln bnh 2 lA. 6,0 gam B. 9,6 gam. C. 35,2 gam. D. 22,0 gam.

Cu 11 :t chy hon ton m gam hn hp hai ancol n chc cng dy ng ng dng va

V lt kh O2 (ktc), thu c 10,08 lt CO2 (ktc) v 12,6 gam H2O. Gi tr ca V l

A. 17,92 lt. B. 4,48 lt. C. 15,12 lt. D. 25,76 lt.

Cu 12 :t chy mt hn hp hidrocacbon X thu c 2,24 lt CO2 (ktc) v 2,7 gam H2O.

Th tch O2 tham gia phn ng chy (ktc) l

A. 2,80 lt B. 3,92 lt. C. 4,48 lt. D. 5,60 lt.

Cu 13 : Dung dch X gm Na2CO3, K2CO3, NaHCO3. Chia X thnh hai phn bng nhau :- Phn 1: tc dng vi nc vi trong dc 20 gam kt ta.

- Phn 2: tc dng vi dung dch HCl dc V lt kh CO2 (ktc). Gi tr ca V l:

A. 2,24. B. 4,48. C. 6,72. D. 3,36.

Cu 14 : Chia hn hp gm : C3H6, C2H4, C2H2 thnh 2 phn bng nhau:

- t chy phn 1 thu c 2,24 lt kh CO2 (ktc).

- Hiro ho phn 2 ri t chy ht sn phm th th tch CO2 (ktc) thu c l:

A. 2,24 lt. B. 1,12 lt. C. 3,36 lt. D. 4,48 lt.

P N

1D 2B 3A 4A 5A 6C 7B

8C 9D 10D 11C 12B 13B 14A


24/236

24

24

Phng php 3

Phng php tng gim khi lngI. PH

NG PHP GI

I

1. Ni dung phng php

- Mi s bin i ha hc (c m t bng phng trnh phnng) u c lin quan n s

tng hoc gim khi lng ca cc cht.

+ Da vo s tng hoc gim khi lng khi chuyn 1 mol cht X thnh 1 hoc nhiu mol

cht Y (c th qua cc giai on trung gian) ta d dng tnh c s mol ca cc cht v ngc

li, t s mol hoc quan h v s mol ca 1 cc cht m tas bit c s tng hay gim khi

lng ca cc cht X, Y.

+ Mu cht ca phng php l: * Xc nh ng mi lin h t l mi gia cc cht bit

(cht X) vi cht cn xc nh (cht Y) (c th khng cn thit phi vit phng trnh phn ng,

m ch cn lp s chuyn ha gia 2 cht ny, nhng phi da vo LBT nguyn t xc

nh t l mi gia chng).

*Xem xt khi chuyn t cht X thnh Y (hoc ngc li)

th khi lng tng ln hay gim i theo t l phn ng v theo cho.

* Sau cng, da vo quy tc tam sut, lp phng trnh

ton hc gii.

2. Cc dng bi ton thng gp

Bi ton 1: Bi ton kim loi + axit (hoc hp cht c nhm OH linh ng) mui + H2

2M + 2nHX 2MXn+ nH2 (l)

2M + nH2SO4 M2(SO4)n+ nH2 (2)

2R(OH)n+ 2nNa 2R(ONa)n+ nH2 (3)

T (l), (2) ta thy: khi lng kim loi gim v tan vo dung dch di dng ion, nhng nu

c cn dung dch sau phn ng th khi lng cht rn thu c s tng ln so vi khi lng

kim loi ban u, nguyn nhn l do c anion gc axit thm vo.

T (3) ta thy: khi chuyn 1 mt Na vo trong mui s gii phng 0,5 mol H2 tng ng vi s

tng khi lng l m = MRO.Do , khi bit s mol H2 v m => R.

Th d: Cho m gam ancol n chc X vo bnh ng Na d, sau phn ng c 0,1 mol H2 v

khi lng bnh tng 6,2gam. Xc nh CTPT ca X.


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RO = 31 R = 15 (CH3) X l CH3OH

Hng dn gii

Theo (3), vi n = 1 : 1 mol Na 1 mol R- ONa

0,5 mol H2: m = MRO

0,1 mol H2: m = 6,2gam

Bi ton 2: Bi ton nhit luyn

Oxit (X) + CO (hoc H2) rn (Y) + CO2 (hoc H2O)

Ta thy: d khng xc nh c Y gm nhng cht g nhng ta lun c v oxi b tch ra khi

oxit v thm vo CO (hoc H2) to CO2 hoc H2O

m = mX - mY = mO nO =16

m = nCO = n 2CO (hoc = 2Hn = n 2H )

Bi ton 3: Bi ton kim loi + dung dch mui: nA + mBn+ nAm+ + mB

Ta thy: tng (gim) khi lng ca kim loi chnh l gim (tng) khi lng camui (v manion = const) .

* Ch : Coi nh ton b kim loi thot ra l bm ht ln thanh kim loi nhng vo dung dch

mui.

Bi ton 4: Bi ton chuyn ha mui ny thnh mui khc.

Khi lng mui thu c c th tng hoc gim, do s thay th anion gc axit ny bng

anion gc axit khc, s thay th ny lun tun theo quy tc ha tr (nu ha tr ca nguyn t kim

loi khng thay i).

* T 1 mol CaCO3 CaCl2: m = 71 - 60 = 11

( c 1 mol CO32ha tr 2 phi c thay th bng 2 mol Cl ha tr 1)

* T 1 mol CaBr2 2 mol AgBr: m = 2. 108 - 40 = 176

( c 1 mol Ca2+ ha tr 2 phi c thay th bng 2 mol Ag+ ha tr 1)

Bi ton 5: Bi ton chuyn oxit thnh mui:

MxOy MxCl2y (c 1 mol O-2c thay th bng 2 mol Cl)

MxOy Mx(SO4)y (c 1 mol O-2c thay th bng 1 mol SO42)

* Ch : Cc iu ny chng khi kim loi khng thay i ha tr.

Bi ton 6: Bi ton phn ng este ha:

RCOOH + HO R RCOOR + H2O


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- meste < m : m tng = m - meste

- meste > m : m gim = meste m

Bi ton 7: Bi ton phn ng trung ha: - OHaxit, phenol + kim

- OH(axit, phenol) + NaOH - ONa + H2O(c 1 mol axit (phenol) mui: m = 23 1 = 22)

3.nh gi phng php tng gim khi lng

- Phng php tng gim khi lng cho php gii nhanh c nhiu bi ton khi bit quan

h v khi lng v t l mi ca cc cht trc v sau phn ng.

- c bit, khi cha bit r phn ng xy ra l hon ton hay khng hon ton th vic s

dng phng php ny cng gip n gin ha bi ton hn.

- Cc bi ton gii bng phng php tng gim khi lng u c th gii c theo

phng php bo ton khi lng, v vy c th ni phng php tng gim khi lng v boton khi lng l 2 anh em sinh i. Tuy nhin, ty tng bi tp m phng php ny hay

phng php kia s l u vit hn.

- Phng php tng gim khi lng thng c s dng trong cc bi ton hn hp nhiu

cht.

4. Cc bc gii.

- Xc nh ng mt quan h t l mi gia cht cn tm v cht bit (nh vn dng

LBTNL).

- Lp s chuyn ho ca 2 cht ny.- Xem xt s tng hoc gim ca M v m theo phng trnh phn ng v theo d kin bi

ton

- Lp phng trnh ton hc gii.

II. TH D MINH HA

V d 1: Khi oxi ho hon ton 2,2 gam mt anehit n chc thu c 3 gam axit tng ng.

Cng thc anehit l

A. HCHO. B. C2H3CHO. C. C2H5CHO. D. CH3CHO.

Gii:

RCHO [O] RCOOH

x mol x mol

m tng= 16x = 3 2,2 x = 0,05

mui mui

mui mui


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Manehit = (R+29) = == CHOCH15R440,05

2,23 p n D

V d 2 : Oxi ho m gam X gm CH3CHO, C2H3CHO, C2H5CHO bng oxi c xc tc, sn phm

thu c sau phn ng gm 3 axit c khi lng (m + 3,2) gam. Cho m gam X tc dng vi

lng d dung dch AgNO3/NH3 th thu c x gam kt ta. Gi tr ca x l


Gii

2 +0txt,

2OCHOR 2 COOOHR

Khi lng tng 3,2 gam l khi lng ca oxi tham gia phn ng

nx = 22O

n = 2 x 0,2(mol)323,2 =

V cc anehit l n chc (khng c HCHO) nAg= 2nx= 2.0,2 = 0,4 (mol)

mAg = x = 0,4.108 = 43,2 gam p n D

V d 3 : Cho 3,74 gam hn hp 4 axit, n chc tc dng vi dung dch Na2CO3 thu c V lt kh

CO2 (ktc) v dung dch mui. C cn dung dch th thu c 5,06 gam mui. Gi tr ca V lt l:

A. 0,224 B. 0,448. C. 1,344. D. 0,672

Gii:

OHCOCOONaR2NaCOCOOHR 223 +++

a mol a mol 0,5a mol

m tng = (23 - 1)a = 5,06 3,74 a = 0,06 mol

2COV = 0,06. 0,5. 22,4 = 0,672 lt p n D

V d 4: Cho 2,02 gam hn hp hai ancol n chc, ng ng k tip tc dng va vi Na

c 3,12 gam mui khan. Cng thc phn t ca hai ancol l :

A. CH3OH, C2H5OH. B. C2H5OH, C3H7OH.

C. C3H7OH, C4H9OH. D. C4H9OH, C5H11OH.

Gii:

2

1

ROH Na RONa H+ + a mol a mol

mtng = 22a = 3,12 2,02 a = 0,05 mol

M 2 ru = M R +17 =


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V d 5: Trung ho 5,48 gam hn hp X gm axit axetic, phenol v axit benzoic cn dng 600ml

dung dch NaOH 0,10M. C cn dung dch sau phn ng thu c hn hp cht rn khan c khi

lng l:


Gii:nNaOH = 0,06mol

Hn hp X + NaOH Mui + H2, trong nguyn t H trong nhm OH hoc COOH c

thay th bi nguyn t Na

tng khi lng = 22. 0,06 = 1,32 gam

Khi lng mui = 5,48 + 1,32 = 6,80gam p n D

V d 6 :t chy hon ton m gam hn hp cc este no, n nhc, mch h. Dn ton b sn

phm chy vo bnh ng dung dch Ba(OH)2 d thy khi lng bnh tng 1,55 gam. Khi

lng kt ta thu c l:

A. 2,5 gam. B. 4,925 gam. C. 6,94 gam. D. 3.52 gam.

Gii:

OHnCOnOOHC 22t

22n2n

0

++

a mol n a n a

OHBaCOBa(OH)CO 2322 ++

n a n a

= bnhm 2COm + OH2m = 0,025an1,55an18an44 ==+ mkt ta = 0,025.197 = 4,925 gam p n B

V d 7: Cho m gam hn hp bt Zn v Fe vo lng d dung dch CuSO4. Sau khi kt thc phn

ng lc b phn dung dch thu c m gam bt rn. Thnh phn % theo khi lng ca Zn trong

hn hp ban u l:

A. 90,28% B. 85,30% C. 82,20% D. 12,67%

Gii:

Zn + CuSO4 ZnSO4 +Cu (1)

x x

mgim = (65 - 64)x = x

Fe + CuSO4 FeSO4 + Cu (2)

y y

m tng = (64 - 56)y = 8y


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V khi lng hn hp rn trc v sau phn ng i mgim = mtng x = 8y

%Zn = =+

90,28%100%x56y65x

65xp n A

V d 8: Cho 4,48 lt CO (ktc) tc dng vi FeO nhit cao mt thi gian, sau phn ng thu

c cht rn X c khi lng b hn 1,6gam so vi khi lng FeO ban u. Khi lng Fe thu

c v % th tch CO2 trong hn hp kh sau phn ng ln lt l:

A. 5,6gam; 40% B. 2,8gam; 25%

C. 5,6gam; 50% C. 11,2gam; 60%

Gii:

FeO + CO 2t COFe0

+

mgim = mO(oxit phn ng )= 0,1(mol)16

1,6=

Fen = 2COn = 0,1 (mol) mFe = 0,1.56 = 5,6gam (*)

Theo bo ton nguyn t: n hn hp kh sau phn ng = nCO(ban u) = 0,2 (mol)

% th tch kh CO2 = 50%(**)x100%0,2

0,1=

T (*) v (**) p n C

V d 9 : Tin hnh 2 th nghim :

- TN 1 : Cho m gam bt Fe d vo V1 (lt) dung dch Cu(NO3)2 1M.

- TN2 : Cho m gam bt Fe d vo V2 (lt) dung dch AgNO3 0,1M.Sau khi cc phim ng xy ra hon ton, khi lng cht rn thu c 2 th nghim u bng

nhau. Gi tr ca Vl so vi V2 l

A. V1 = V2 B. Vl = l0V2 C. Vl = 5V2 D. Vl = 2V2

Gii:

Fe + Cu2+ Fe2+ + Cu

V1 mol V1 mol

m tng = 64V1 56V1 = 8V1 gam

Fe + 2Ag+

Fe2+

+ 2Ag0,05V2 mol 0,1V2 mol

mtng = 108.0,1V2 56.0,05V2 = 8V2 gam

Theo mrn(TN1) = mrn(TN2) 8V1= 8V2V1 = V2 p n A


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V d 10 : Nung 1 hn hp rn gm a mol FeCO3 v b mol FeS2 trong bnh kn cha khng kh

d. Sau khi cc phn ng xy ra hon ton, a bnh v nhit ban u thu c cht rn duy

nht l Fe2O3 v hn hp kh. Bit p sut kh trong bnh trc v sau phn ng bng nhau v sau

cc phn ng lu hunh mc oxi ho +4, th tch cc cht rn l khng ng k. Mi lin h

gia a v b lA. a = 0,5b. B. a = b. C. a = 4b.

D. a = 2b.

Gii:

2FeCO3+ 32t

2 OFeO2

1 0 +2CO2

a4

aa

Phn ng lm tng 1 lng kh l (a -4a )= mol

43a

2FeS2 + 232t

2 4SOOFeO2

11 0+

b4

11b2b

Phn ng lm gim mt lng kh l: mol4

3b2b

4

11b=

V ptrc = psau == ba4

3b4

3a p n B

V d 11: Cho 5,90 gam amin n chc X tc dng va vi dung dch HCl sau khi phn ng

xy ra hon ton thu c dung dch Y. Lm bay hi dung dch Y c 9,55 gam mui khan. S

cng thc cu to ng vi cng thc phn t ca X l:

A. 5. B. 4. C. 2. D. 3.

Gii:

RNH2 + HCl RNH3Cl

x mol x mol x molm tng = 36,5x = 9,55 5,9 x = 0,1

Mamin = MR +16 =1,0

9,5=59 MR = 43 X: C3H7NH2

CH3 CH2 CH2 NH2 ; (CH3)2CHNH2; CH3NHCH3CH2; (CH3)3N p n B


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V d 12: Trong phn t amino axit X c 1 nhm amino v 1 nhm cacboxyl. Cho 15,0 gam X

tc dng va vi dung dch NaOH. C cn dung dch sau phn ng thu c 19,4 gam mui

khan. Cng thc ca X l

A. H2NC3H6COOH. B. H2NCH2COOH.

C. H2NC2H4COOH. D. H2NC4H8COOH.Gii:

H2NRCOOH + NaOH H2NRCOONa + H2O

x mol x mol

mtng = 22x = 19,4 15,0 x = 0,2 mol

Mx = MR +61 = 75 MR = 14 X: H2NCH2COOH p n B

V d 13:t chy hon ton 4,40 gam cht hu cX n chc thu c sn phm chy gm

4,48 lt CO2 (ktc) v 3,60 gam H2O. Nu cho 4,40 gam X tc dng vi dung dch NaOH va

n khi phn ng hon ton c 4,80 gam mui ca axit hu cY v cht hu cZ. Tn ca X l

A. etyl propionat. B. metyl propionat

C. isopropyl axetat. D. etyl axetat.

Gii :

2COn = OH2n = 0,2mol X l este no n

CnH2nO2 + ( )2

13n O2

0t nCO2 + nH2O

moln

0,20,2 mol

mX = (14n + 32) n

0,2= 4,4 n = 4 X: C4H8O2 v nX = 4

0,2= 0,05 mol

RCOOR + NaOH RCOONa + ROH

0,05 mol 0,05 mol

mX < mmui mtng = (23-R) 0,05 = 4,8 4,4 = 0,4 R= 15

Cng thc cu to ca X l: C2H5OHCOOCH3 p n B

V d 14: Hn hp X gm HCOOH v CH3COOH (t l mol 1:1). Ly 5,30 gam hn hp X tc

dng vi 5,75 gam C2H5OH (xc tc H2SO4c) thu c m gam este (hiu sut ca cc phn

ng este ho u bng 80%). Gi tr ca m l:

A. 10,12 gam. B. 6,48 gam.

C. 16,20 gam. D. 8,10 gam.


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Gii:

x mol x mol x mol

532x

60x46xMX =+=

nX = 5,3 : 53 = 0,1 mol < OHHC 52n = 0,125 mol khi lng este tnh theo s mol ca axit

mtng = (29-1)x = m - 5,3 m = 8,1 gam

Khi lng este thc t thu c l 6,48gam100%

8,1.80%=

p n B

V d 15: Dn t t hn hp kh CO v H2 qua ng sng 55,4 gam hn hp bt CuO, MgO,

ZnO, Fe3O4un nng. Sau khi phn ng xy ra hon ton thu c 10,08 lt (ktc) hn hp kh

v hi ch cha CO2 v H2O, trong ng s cn li mt lng cht rn c khi lng l

A. 48,2 gam. B. 36,5 gam. C. 27,9 gam D. 40,2 gam

Gii:

Bn cht ca cc phn ng CO, H2 + [O] CO2 , H2O

nO =2CO

n + OH2n = nCO + 2Hn = 0,45mol

m rn = moxit mO = 55,4 0,45.16 = 48,2 gam p n A

V d 16: Nung 47,40 gam kali pemanganat mt thi gian thy cn li 44,04 gam cht rn. %khi lng kali pemanganat b nhit phn l

A. 50%. B. 70%. C. 80%. D. 65%.

Gii:

2KMnO4 0t

K2MnO4 + MnO2 + O2

gim khi lng ca cht rn =2O

m = 47,4 44,04 = 3,36gam

2O

n = 3,36: 32 = 0,105 mol 4KMnO

m tham gia = 0,105.2 = 0,21 mol

% 4KMnOm phn ng = 4,47 158.21,0 .100%= 70% p n B

V d 17 : Nhit phn a gam Zn(NO3)2 sau 1 thi gian dng li lm ngui v em cn thy khi

lng gim i 2,700 gam (hiu sut phn ng l 60%). Gi tr a l

A. 4,725 gam. B. 2,835 gam. C. 7,785 gam. D. 7.875 gam.


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Gii:

Zn(NO)2 0

t ZnO + 2NO2 +2

1O2

xmol 2xmol 0,5xmol

m rn gim = 2NOm + 2Om = 92x + 16x = 2,7 x = 0,025mol

H = 7,875gama60%.100%a

189x== p n C

V d 18 : Cho 3,06 gam hn hp K2CO3 v MgCO3 tc dng vi dung dch HCl thu c V lt

kh (ktc) v dung dch X. C cn dung dch X c 3,39 gam mui khan. Gi tr V (lt) l:

A. 0,224 B. 0,448 C. 0,336 D. 0,672.

Gii:

mtng = 112CO

n = 3,39 3,06 2CO

n = 0,03 mol 2CO

V = 0,672 lt

p n D

V d 19 : Ho tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500ml dung dch

H2SO4 0,1M va . Sau phn ng hn hp mui sunfat khan thu c khi c cn dung dch c

khi lng l


Gii:

O2-(trong oxit)-2

4SO

Khi lng tng: 0,05 (96 -16) = 4,0 gam mmui = moxit + mmui = 2,81 + 4 = 6,81 gam

p n D

III. BI TP TLUYN

Cu 1: Dn 130 cm3 hn hp X gm 2 hirocacbon mch hqua dung dch Br2 d kh thot ra

khi bnh c th tch l 100cm3, bit dx/He = 5,5 v phn ng xy ra hon ton. Hai hirocacbon

cn tm l

A. metan, propen. B. metan, axetilen.C. etan, propen. D. metan, xiclopropan.

Cu 2 :un nng 1,77 gam X vi 1 lng va 1,68 gam KOH c 2,49 gam mui ca axit

hu cY v 1 ancol Z vi s mol Z gp 2 ln s mol Y (bit phn ng xy ra hon ton). X l

A. CH2(COOCH3)2 B. (COOCH3)2

C. HCOOC2H5 D. C2H4(COOCH3)2


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Cu 3: Trung ho 5,48 gam hn hp axit axetic, phenol v axit benzoic cn dng 600ml dung

dch NaOH 0,1M. C cn dung dch sau phn ng c hn hp cht rn khan c khi lng l


Cu 4: Cho 5,76 gam axit hu cX n chc mch h tc dng ht vi CaCO3c 7,28 gam

mui ca axit hu c. Cng thc cu to thu gn ca X l:A. CH2=CH-COOH B. CH3COOH

C. CH C-COOH D. CH3-CH2-COOH

Cu 5: Ho tan hon ton 2,1 gam mui cacbonat ca kim loi ho tr II trong dung dch H2SO4

long c 3 gam cht rn khan. Cng thc mui cacbonat ca kim loi ho tri II l:

A. CaCO3 B. Na2CO3 C. FeCO3 D. MgCO3

Cu 6: Cho ancol X tc dng vi Na d thy s mol khi bay ra bng s mol X phn ng. Mt

khc, X tc dng vi lng d CuO nung nng n phn ng hon ton thy lng rn gim 1,2

gam v c 2,7 gam cht hu ca chc Y. Cng thc cu to thu gn ca Y l:A. OHC-CH2-CH2-CHO B. OHC-CH2-CHO

C. CH3-CO-CO-CH3 D. OHC-CO-CH3

Cu 7: Cho 26,80 gam hn hp KHCO3 v NaHCO3 tc dng ht vi dung dch HCl dc

6,72 lt kh (ktc). Sau phn ng c cn c a gam mui khan. Gi tr ca a gam l:

A. 34,45. B. 20,15. C. 19,15. D. 19,45.

Cu 8: Dn V lt (ktc) hn hp gm CO v H2 qua ng s nung nng cha hn hp FeO, Al2O3

(cc phn ng xy ra hon ton) c hn hp kh v hi nng hn hn hp kh ban u 2 gam.

Gi tr ca V lt lA. 2,80. B. 5,60. C. 0,28. D. 0,56

Cu 9: Nung hn hp rn gm FeCO3 v FeS2 (t l mol 1 : 1) trong 1 bnh kn cha khng kh

d vi p sut l p1 atm. Sau khi cc phn ng xy ra hon ton a bnh v nhit ban u thu

c cht rn duy nht l Fe2O3 v p sut kh trong bnh lc ny l p2 atm (th tch cc cht rn

khng ng k v sau cc phn ng lu hunh mc oxi ho + 4). Mi lin h gia pl v p2 l:

A. pl = p2 B. pl = 2p2 C. 2pl = p2 D. pl = 3p2

Cu 10: Dn kh CO i qua ng s nung nng cha 0,02 mol hn hp X gm FeO v Fe2O3

phn ng xy ra hon ton thu c 1,96 gam cht rn Y, khi ra khi ng s hp th hon tonvo dung dch Ca(OH)2 d th thy khi lng bnh tng 2,20 gam. Hn hp X c:

A. 50%FeO v 50% Fe2O3 B. 13,04%FeO v 86,96% Fe2O3

C. 20%FeO v 80% Fe2O3 D. 82%FeO v 18%Fe2O3


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Cu 11: Ho tan ht 1,625 gam kim loi M vo dung dch Ca(OH)2 thy khi lng dung dch

sau phn ng tng 1,575 gam. M l

A. Al. B. Be. C. Zn. D. Cr.

Cu 12: Dn V lt kh CO2 (ktc) hp th hon ton vo 750ml dung dch Ba(OH)2 0,1M, sau

phn ng khi lng dung dch gim 5,45 gam v c hn hp 2 mui. Gi tr V lt lA. l,68. B. 2,24. C. 1,12. D. 3,36.

Cu 13: Cho 1,825 gam amin X tc dng va vi dung dch HCl, sau khi phn ng xy ra

hon ton thu c dung dch Y. Lm bay hi dung dch Y c 2,7375 gam mui RNH3Cl. X c

tng sng phn cu to amin bc 1 l:

A. 4. B. 6. C. 7. D. 8.

Cu 14: Cho a gam hn hp gm metanol v propan-2-ol qua bnh ng CuO d, nung nng.

Sau khi phn ng xy ra hon ton a hn hp kh v hi c khi lng l (a + 0,56) gam. Khi

lng CuO tham gia phn ng lA. 0,56 gam. B. 2,80 gam C. 0,28 gam. D. 5,60 gam.

Cu 15: Cho a gam hn hp cc ankanol qua bnh ng CuO d, nung nng. Sau khi phn ng

xy ra hon ton c hn hp kh v hi c khi lng l (a + 1,20) gam v c t khi hi i

vi H2 l 15. Gi tr ca a gam l


Cu 16: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phn

ng. Ly a gam X tc dng vi dung dch HCl dc (a + 0,9125) gam Y. un ton b lng

Y thu c vi 200ml dung dch NaOH thu c dung dch Z. Bit X lm qu tm ho. Nng mol ca dung dch NaOH phn ng l

A. 0,2500M. B. 0,1250M. C. 0,3750M. D. 0,4750M.

Cu 17: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phn

ng. Ly a gam X tc dng vi dung dch HCl dc (a + 0,9125) gam Y. em ton b lng

Y tc dng va vi dung dch NaOH un nng c dung dch Z. C cn Z c 5,8875 gam

mui khan. Bit X lm qu tm ho . Gi tr a gam l

A. 3,325. B. 6,325. C. 3,875. D. 5,875.

Cu 18: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phnng. Ly a gam X tc dng vi dung dch HCl dc (a + 0,9125) gam Y. em ton b lng

Y tc dng va vi dung dch NaOH un nng c dung dch Z. C cn Z c 5,8875 gam

mui khan. Bit X lm qu tm ho . Cng thc cu to ca X l

A.HOOC-CH(NH2)-COOH

B. HOOC-CH2CH(NH2)CH2-COOH


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C. HOOC-CH2CH2CH2NH2

D. HOOC-CH2CH(NH2)-COOH

Cu 19: Cho amino axit x tc dng va vi Na thy s mol kh to ra bng s mol X phn

ng. Ly a gam X tc dng vi dung dch HCl dc (a + 0,9125) gam Y. em ton b lng

Y tc dng va vi dung dch NaOH un nng c dung dch Z. C cn Z c 1 lngmui khan. Bit X lm qu tm ho . Khi lng mui khan thu c so vi khi lng ca Y

s

A. tng 1,65 gam. B. gim 1,65 gam.

C. tng 1,10 gam. D. gim 1,10 gam.

Cu 20:t chy hon ton 3,72 gam hp cht hu cX (bit2X/H

d < 70), dn ton b sn phm

chy thu c qua bnh ng dung dch Ba(OH)2 d thy to ra 41,37 gam kt ta ng thi khi

lng dung dch gim 29,97 gam. Bit s mol NaOH cn dng phn ng ht vi X bng s

mol kh hiro sinh ra khi cho X tc dng vi Na d. Cng thc cu to thu gn ca X l:

A. CH3-C6H4(OH)2 B. C6H7COOH.

C. C5H6(COOH)2 D. HO-C6H4-CH2OH.

Cu 21: Th tch oxi phn ng l bao nhiu nu chuyn 1 th tch oxi thnh ozon thy th tch

gim i 7,0 cm3 (th tch cc kho cng iu kin)

A. 21,0 dm3 B. 7,0 cm3 C. 21,0 cm3 D. 4,7 cm3

Cu 22: Trong 1 bnh kn dung tch khng i cha 0,2 mo1 CO v 1 lng hn hp X gm

Fe3O4 v FeCO3 (t l mol 1 : l). Nung bnh nhit cao cc phn ng xy ra hon ton v

a bnh v nhit ban u (th tch cc cht rn khng ng k) thy p sut trong bnh tng 2

ln so vi ban u. Tng s mol ca Fe3O4 v FeCO3 l:

A 0,4 B. 0,3. C. 0,2. D. 0,1.

Cu 23:t chy hon ton 16,8 gam mui sunfua ca kim loi ho tri II khng i thu c

cht rn X v kh B. Ho tan ht X bng 1 lng va dung dch H2SO4 35% c dung dch

mui c nng 44,44%. Ly dung dch mui ny lm lnh xung nhit thp thy tch ra 25

gam tinh th ngm nc Y v dung dch bo ho c nng 31,58%. Y c cng thc l

A. CuSO4.3H2O. B. MgSO4.2H2O.

C. CuSO4.5H2O. D. CuSO4.2H2O.

Cu 24: Thu phn hon ton 1,76 gam X n chc bng 1 lng va dung dch NaOH un

nng c 1,64 gam mui Y v m gam ancol Z. Ly m gam Z tc dng vi lng d CuO nung

nng n phn ng hon ton thy lng cht rn gim 0,32 gam. Tn gi ca X l

A. etyl fomat. B. etyl propionat.

C. etyl axetat. D. metyl axetat.


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Cu 25: Cho hn hp X gm 2 axit ng ng k tip nhau tc dng vi Na d thy s mol H2

bay ra bng2

1mol X. un 20,75 gam X vi 1 lng d C2H5OH (xc tc H2SO4c) c

18,75 gam hn hp este (hiu sut ca cc phn ng este ho u bng 60%). % theo khi lng

cc cht c trong hn hp X l:A. 27,71% HCOOH v 72,29% CH3COOH.

B. 27,71 % CH3COOH v 72,29% C2H5COOH.

C. 40% C2H5COOH v 60% C3H7COOH.

D. 50% HCOOH v 50% CH3COOH.

Cu 26: Ho tan 5,4 gam Al vo 0.5 lt dung d ch X gm AgNO3 v Cu(NO3)2c 42 gam rn

Y khng tc dng vi dung dch H2SO4 long v dung dch Z. Ly ton b dung dch Z cho tc

dng vi dung dch NaOH d th c 14,7 gam kt ta (cho phn ng xy ra hon ton). Nng

mi ca AgNO3 v Cu(NO3)2 trong dung dch X ln lt l:A. 0,6M v 0,3M. B. 0,6M v 0,6M.

C. 0,3M v 0,6M. D. 0,3M v 0,3M.

Cu 27: Nhng m gam kim loi M ho tr II vo dung dch CuSO4 sau 1 thi gian ly thanh kim

loi thy khi lng gim 0,075%. Mt khc, khi nhng m gam thanh kim loi trn vo dung dch

Pb(NO3)2 sau 1 thi gian ly thanh kim loi thy khi lng thanh kim loi tng 10,65% (bit s

mol ca CuSO4 v Pb(NO3)2 tham gia 2 trng hp l nh nhau). M l

A. Mg. B. Zn. C. Mn. D. Ag.

Cu 28: Nhng 1 thanh Al v 1 thanh Fe vo dung dch Cu(NO3)2 sau 1 thi gian ly 2 thanh kimloi ra thy dung dch cn li cha Al(NO3)3 v Fe(NO3)2 vi t l mol 3 : 2 v khi lng dung

dch gim 2,23 gam (cc phn ng xy ra hon ton). Khi lng Cu bm vo thanh Al v Fe

l:


Cu 29 : Cho 32,50 gam Zn vo 1 dung dch cha 5,64 gam Cu(NO3)2 v 3,40 gam AgNO3 (cc

phn ng xy ra hon ton v tt c kim loi thot ra u bm vo thanh kim loi). Khi lng

sau cng ca thanh kim loi l


Cu 30:in phn l00ml dung dch M(NO3)n. Vi in cc trcho n khi b mt catot xut

hin bt kh th ngng in phn. Phi dng 25ml dung dch KOH 2M trung ho dung dch sau

khi in phn. Mt khc, nu ngm 20 gam Mg vo 100ml dung dch M(NO3)n. Sau mt thi gian


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ly thanh Mg ra, sy kh v cn li thy khi lng tng thm 24% so vi lng ban u. Bit

cc phn ng xy ra hon ton. Cng thc ho hc ca M(NO3)n l

A. Cu(NO3)2 B. Ni(NO3)2 C. Pb(NO3)2 D. AgNO3

Cu 31: Nung 46,7 gam hn hp Na2CO3 v NaNO3n khi lng khng i thu c 41,9

gam cht rn. Khi lng Na2CO3 trong hn hp u lA. 21,2 gam. B. 25,5 gam. C. 21,5 gam. D. 19,2 gam.

Cu 32: Nung 104,1 gam hn hp K2CO3 v NaHCO3 cho n khi khi lng khng i thu

c 88,6 gam cht rn % khi lng ca cc cht trong hn hp u l

A. 20% v 80%. B. 45,5% v 54,5%.

C. 40,35% v 59,65%. D. 35% v 65%.

Cu 33: Dn kh CO qua ng s cha 7,6 gam hn hp gm FeO v CuO nung nng, sau 1 thi

gian c hn hp kh X v 6,8 gam rn Y. Cho hn hp kh X hp th hon ton vo dung dch

Ca(OH)2 d thy c kt ta. Khi lng kt taA. 5 gam. B. 10 gam. C. 15 gam. D. 20 gam.

Cu 34:t chy hon ton m gam hai kim loi Mg, Fe trong khng kh, thu c (m + 0,8) gam

hai oxit. hon tan ht lng oxit trn th khi lng dung dch H2SO4 20% ti thiu phi dng

l

A. 32,6 gam. B. 32 gam. C. 28,5 gam. D. 24,5 gam.

Cu 35: Ly 2,98 gam hn hp X gm Zn v Fe cho vo 200ml dung dch HCl 1M, sau khi phn

ng hon ton ta c cn (trong iu kin khng c oxi) th c 6,53 gam cht rn. Th tch kh

H2 bay ra (ktc) lA. 0,56 lt. B. 1,12 lt. C. 2,24 lt. D. 4,48 lt.

Cu 36:em nung nng m gam Cu(NO3)2 mt thi gian ri dng li, lm ngui v em cn thy

khi lng gim 0,54 gam so vi ban u. Khi lng mui Cu(NO3)2 b nhit phn l


Cu 37: trung ho 7,4 gam hn hp 2 axit hu cn chc cn 200ml dung dch NaOH

0,5M. Khi lng mui thu c khi c cn dung dch l


Cu 38: Cho 5,615 gam hn hp gm ZnO, Fe2O3, MgO tc dng va vi 100ml dung dchH2SO4 1M th khi lng mui sunfat thu c l



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Cu 39: t chy hon ton 33,4 gam hn hp X gm Al, Fe, Cu ngoi khng kh thu c 41,4

gam hn hp Y gm ba oxit. Th tch ti thiu dung dch H2SO4 20% (D =1,14 g/ml) cn dng

ho tan ht hn hp Y l:

A. 215ml. B. 8,6ml. C. 245ml. D. 430ml.

Cu 40: X l mt -aminoaxit ch cha 1 nhm -NH2 v 1 nhm -COOH. Cho 0,445 gam Xphn ng va vi NaOH to ra 0,555 gam mui. Cng thc cu to ca X c th l

A. H2N-CH2-COOH. B. CH3-CH(NH2)-COOH.

C. H2N-CH2-CH2-COOH. D. H2N-CH=CH-COOH.

Cu 41: Cho hn hp X gm NaCl v NaBr tc dng vi dung dch AgNO3 d th lng kt ta

thu c sau phn ng bng khi lng AgNO3 tham gia phn ng. Thnh phn % khi lng

NaCl trong X l

A. 27,88%. B. 13,44%. C. 15,20%. D. 24,50%.

Cu 42: Cho 1,52 gam hn hp hai ancol n chc l ng ng k tip nhau tc dg vi Na va, sau phn ng thu c 2,18 gam cht rn. Cng thc phn t ca hai ancol v th tch kh thu

c sau phn ng ktc ln lt l:

A. CH3OH; C2H5OH v 0,336 lt. B. C2H5OH; C3H7OH v 0,336 lt

C. C3H5OH; C4H7OH v 0,168 lt. D. C2H5OH; C3H7OH v 0,672 lt.

Cu 43: Hn hp X c khi lng 25,1 gam gm ba cht l axit axetic, axit acrylic v phenol.

Lng hn hp X trn c trung ho va bng 100ml dung dch NaOH 3,5M. Tnh khi

lng ba mui thu c sau phn ng trung ho l

A. 32,80 gam. B. 33,15 gam. C. 34,47 gam. D. 31,52 gam.Cu 44: Ngm mt inh st sch trong 200ml dung dch CuSO4n khi dung dch ht mu xanh,

ly inh st ra khi dung dch, ra sch, sy kh, cn thy inh st tng 0,8 gam. Nng mi

ca dung dch CuSO4 l

A. 0,5M. B. 5M. C. 0,05M. D. 0,1M

Cu 45: Nung l00 gam hn hp gm Na2CO3 v NaHCO3 cho n khi khi lng hn hp khng

i c 69 gam cht rn. Xc nh phn trm khi lng ca mi cht trong hn hp ln lt l:

A. 16% v 84%. B. 84% v 16%.

C. 26% v 74%. D. 74% v 26%.Cu 46: Ly 2,98 gam hn hp X gm Zn v Fe cho vo 200ml dung dch HCl 1M, sau khi phn

ng hon ton ta c cn (trong iu kin khng c oxi) th c 6,53 gam cht rn. Th tch kh

H2 bay ra (ktc) l

A. 0,56 lt. B. 1,12 lt. C. 2,24 lt. D. 4,48 lt.


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Cu 47: Cho mt anken X tc dng ht vi H2O (H+, t0) c cht hu cY, ng thi khi

lng bnh ng nc ban u tng 4,2 gam. Cng cho mt lng X nh trn tc dng vi HBr

va , thu c cht Z, thy khi lng Y, Z thu c khc nhau 9,45 gam (gi s cc phn ng

xy ra hon ton). Cng thc phn t ca X l:

A. C2H4 B. C3H6 C. C4H8 D. C5H10

P N

1A 2B 3D 4A 5D 6B 7C 8A 9A 10B

11C 12B 13A 14B 15B 16C 17A 18D 19A 20D

21C 22A 23C 24C 25A 26B 27B 28A 29B 30D

31A 32C 33A 34D 35B 36D 37A 38B 39A 40B

41A 42B 43A 44A 45A 46B 47A

Phng php 4

Phng php Bo ton in tch

I. CSCA PHNG PHP

1. Cs: Nguyn t, phn t, dung dch lun lun trung ha vin

- Trong nguyn t: s proton = s electron

- Trong dung dch:

s mol in tch ion dng = s mol in tch ion m

2. p dng v mt s ch

a, Khi lng dung dch mui (trong dung dch) = khi lng cc ion to mui

b, Qu trnh p dng nh lut bo ton in tch thng kt hp:

- Cc phng php bo ton khc: Bo ton khi lng, bo ton nguyn t

- Vit phng trnh ha hc dng ion thu gn


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II. CC DNG BI TON THNG GP

Dng 1: p dng n thun nh lut bo ton in tch

V d 1 : Mt dung dch c cha 4 ion vi thnh phn : 0,01 mol Na+, 0,02 mol Mg2+, 0,015 mol

24SO , x mol

Cl . Gi tr ca x l

A. 0,015. B. 0,035. C. 0,02. D. 0,01.

Gii:

p dng nh lut bo ton in tch ta c:

0,01.1 + 0,02.2 = 0.015.2 + x.1 x = 0,02 p n C

Dng 2: Kt hp vi nh lut bo ton khi lng

V d 2 : Dung dch A cha hai cation l Fe2+: 0,1 mol v Al3+: 0,2 mol v hai anion l Cl : x

mol v 24SO : y mol. em c cn dung dch A thu c 46,9 gam hn hp mui khan. Gi tr ca

x v y ln lt l:A. 0,6 v 0,1 B. 0,3 v 0,2 C. 0,5 v 0,15 D. 0,2 v 0,3

Gii:


0,01.2 + 0,2.3 = x.1 +y.2 x + 2y = 0,8 (*)

Khi c cn dung dch khi lng mui = khi lng cc ion to mui

0,1.56 + 0,2.27 + x.35,5 + y.96 = 46,9 35,5x + 96y = 35,9 (**)

T (*) v (**) x = 0,2; y = 0,3 p n D.

V d 3 : Chia hn hp X gm hai kim loi c ho tr khng i thnh 2 phn bng nhau.Phn 1: Ho tan hon ton bng dung dch HCl d thu c 1,792 lt H2 (ktc).

Phn 2 : Nung trong khng kh d thu c 2,84 gam hn hp rn ch gm cc oxit. Khi lng

hn hp X l


Gii:

Nhn xt: Tng s mol in tch ion dng (ca hai kim loi) trong hai phn l bng nhau

Tng s mol in tch ion m trong hai phn cng bng nhau

O2- 2Cl

Mt khc: -Cln = +Hn = 2 2Hn = 0,08mol22,4

1,792=

nO(trong oxit) = 0,04(mol)

Trong mt phn: mkim loi= moxit moxi = 2,84 0,08.16 = 1,56 gam


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khi lng hn hp X = 2.1,56 = 3,12gam p n D

Dng 3: Kt hp vi bo ton nguyn t

V d 4 : Cho hn hp X gm x mol FeS2 v 0,045 mol Cu2S tc dng va vi HNO3 long,

un nng thu c dung dch ch cha mui sunfat ca cc kim loi v gii phng kh NO duy

cht. Gi tr ca x l:A. 0,045 B. 0,09. C. 0,135. D. 0,18.

Gii:

- p dng bo ton nguyn t

Fe3+: x mol; Cu2+: 0,09 mol; 24SO : (x + 0,045) mol

- p dng nh lut bo ton in tch (trong dung dch ch cha cc mui sunfat) ta c:

3x + 2.0,09 = 2(x + 0,045) x = 0,09p n B

V d 5 : Dung dch X c cha 5 ion: Mg2+, Ba2+, Ca2+, 0,1 mol Cl v 0,2 mol 3NO . Thm dn

V lt dung dch K2CO3 1M vo X n khi c lng kt ta ln nht th gi tr V ti thiu cn

dng l

A. 150ml B. 300ml C. 200ml D. 250ml

Gii:

C th quy i cc ion Mg2+, Ba2+, Ca2+ thnh M2+ (xem thm phng php quy i)

M2+ + 23CO 3MCO

Khi phn ng kt thc, phn dung dch cha K+, Cl v 3NO


+Kn = Cln + -3NOn = 0,15 (lt) = 150ml p n A

Dng 4: Kt hp vi vic vit phng trnh dng ion thu gn

V d 6 : Cho tan hon ton 15,6 gam hn hp gm Al v Al2O3 trong 500ml dung dch NaOH

1M thu c 6,72 lt H2 (ktc) v dung dch X. Th tch HCl 2M ti thiu cn cho vo X thu

c lng kt ta ln nht l

A. 0,175 lt. B. 0,25 lt. C. 0,125 lt. D. 0,52 lt.

Gii:Dung dch X cha cc ion Na+; 2AlO ;

OH d (c th).

p dng nh lut bo ton in tch: 2AlO

n + OHn = +Nan = 0,5

Khi cho HCl vo dung dch X:

H+ + OH H2O (1)


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H+ + 2AlO+ H2O Al(OH)3 (2)

3H+ + Al(OH)3 Al3+ + 3H2O (3)

kt ta l ln nht khng xy ra (3) v nH+ = 2AlO

n + nOH-= 0,5

VHCl = 25,025,0 = (lt) p n B

Dng 5: Bi ton tng hp

V d 7: Hon ton 10 gam hn hp X gm Mg v Fe bng dung dch HCl 2M. Kt thc th

nghim thu c dung dch Y v 5,6 lt H2 (ktc). kt ta hon ton cc cation c trong Y cn

va 300ml dung dch NaOH 2M. Th tch dung dch HCl dng l

A. 0,2 lt. B. 0,24 lt. C. 0,3 lt. D. 0,4 lt

Gii:

== + OHNa nn nNaOH = 0,6 (mol)Khi cho NaOH vo dung dch Y (cha cc ion: Mg2+; Fe2+; H+ d; Cl ) cc ion dng s tc dng

vi OH to thnh kt ta. Nh vy dung dch thu c sau phn ng ch cha Na+ v Cl

+ = NaCl nn = 0,6 +Hn = 0,6 VHCl= = 0,3lt20,6

p n C

V d 8 : ho tan hon ton 20 gam hn hp X gm Fe, FeO, Fe3O4, Fe2O3 cn va 700ml

dung dch HCl 1M thu c dung dch X v 3,36 lt H2 (ktc). Cho NaOH d vo dung dch X

ri ly ton b kt ta thu c em nung trong khng khn khi lng khng i th lng

cht rn thu c l

A. 8 gam B. 16 gam C. 24 gam D. 32 gam

Gii:

Vi cch gii thng thng, ta vit 7 phng trnh ho hc, sau t n s, thit lp h phng

trnh v gii

Nu p dng nh lut bo ton in tch ta c:

Fe + 2HCl FeCl2 + H2

S mol HCl ho tan l Fe l: nHCl =2H

2n = 0,3(mol)

S mol HCl ho tan cc oxit = 0,7 0,3 = 0,4 mol

Theo nh lut bo ton in tch ta c:

nO2-

(oxit) = 0,3(mol)56

0,2.1620

56

mmn0,2(mol)n

2

1 oxioxitFe(trongX)Cl

=

=

==


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C th coi: 2Fe (trong X) Fe2O3

32 OFe

n = 0,15mol 32 OFe

m = 24 gam p n C

BI TP TLUYN

Cu 1: Dung dch X c cha a mol Na+ ; b mol Mg2+ ; c mol Cl v d mol 24SO . Biu thc lin

h gia a, b, c, d l

A. a + 2b = c + 2d B. a+ 2b = c + d.

C. a + b = c + d D. 2a + b = 2c + d

Cu 2: C hai dung dch, mi dung dch u cha hai cation v hai anion khng trng nhau trong

cc ion sau : K+: 0,15 mol, Mg2+: 0,1 mol, NH4+ : 0,25 mol, H+ : 0,2 mol. Cl : 0,1 mol, 24SO :

0,075 mol, 3NO : 0,25 mol v 23CO : 0,15 mol. Mt trong hai dung dch trn cha:

A. K+, Mg2+, 24SO vCl B. K+, NH4

+, 23CO vCl

C. NH4+, H+,

3NO v

24SO D. Mg

2+, H+, 24SO vCl

Cu 3 : Dung dch Y cha Ca2+ 0,1 mol, Mg2+ 0,3 mol, Cl 0,4 mol, 3HCO y mol. Khi c cn

dung dch Y th lng mui khan thu c l

A. 37,4 gam B. 49,8 gam. C. 25,4 gam. D. 30,5 gam.

Cu 4 : Mt dung dch cha 0,02 mol Cu2+, 0,03 mol K+, x mol Cl v y mol 24SO . Tng khi

lng cc mui tan c trong dung dch l 5,435 gam. Gi tr ca x v y ln lt l :

A. 0,03 v 0,02. B. 0,05 v 0,01 C. 0,01 v 0,03 D. 0,02 v 0,05

Cu 5 : Ho tan hon ton hn hp gm 0,12 mol FeS2 v x mol Cu2S vo dung dch HNO3 va

, thu c dung dch X ch cha 2 mui sunfat ca cc kim loi v gii phng kh NO duy

nht. Gi tr X l

A. 0,03 B. 0,045 C. 0,06. D. 0,09.

Cu 6 : Cho m gam hn hp Cu, Zn, Mg tc dng hon ton vi dung dch HNO3 long, d. C

cn cn thn dung dch thu c sau phn ng 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

A. (m + 4) gam. B. (m + 8) gam. C. (m + 16) gam. D. (m + 32) gam.


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Cu 7 : Cho 24,4 gam hn hp Na2CO3, K2CO3 tc dng va vi dung dch BaCl2 sau phn

ng thu c 39,4 gam kt ta. Lc tch kt ta, c cn dung dch th thu dc bao nhiu gam

mui clorua khan


Cu 8 : Trn dung dch cha Ba2+; OH 0,06 mol v Na+ 0,02 mol vi dung dch cha 3HCO

0,04 mol; 23CO 0,03 mol v Na+. Khi lng kt ta thu c sau khi trn l

A. 3,94 gam. B. 5,91 gam. C. 7,88 gam. D. 1,71 gam

Cu 9 : Ho tan hon ton 5,94 gam hn hp hai mui clorua ca 2 kim loi nhm IIA vo nc

c 100ml dung dch X. lm kt ta ht ion Cl c trong dung dch X trn ta cho ton b

lng dung dch X trn tc dng va vi dung dch AgNO3. Kt thc th nghim, thu c

dung dch Y v 17,22 gam kt ta. Khi lng mui khan thu c khi c cn dung dch Y l

A. 4,86 gam. B. 5,4 gam. C. 7,53 gam. D. 9,12 gam.Cu 10 : Dung dch X cha 0,025 mol 23CO ; 0,1 mol Na

+ ; 0,25 mol NH4+ v 0,3 mol Cl .Cho

270ml dung dch Ba(OH)2 0,2M vo v un nng nh (gi s H2O bay hi khng ng k). Tng

khi lng dung dch X v dung dch Ba(OH)2 sau qu trnh phn ng gim i l.


Cu 11 : Trn 100ml dung dch AlCl3 1M vi 200ml dung dch NaOH l,8M n phn ng hon

ton th lng kt ta thu c l


Cu 12 : Dung dch B cha ba ion K+ ; Na+ ; 34PO . 1 lt dung dch B tc dng vi CaCl2 d thu

c 31 gam kt ta. Mt khc, nu c cn mt lt dung dch B thu c 37,6 gam cht rn

khan. Nng ca hai ba ion K+ ; Na+ ;

34PO ln lt l .

A. 0,3M ; 0,3M v 0,6M B. 0,1M ; 0,1M v 0,2M

C. 0,3M ; 0,3M v 0,2M D. 0,3M ; 0,2M v 0,2M

Cu 13 : Cho dung dnh Ba(OH)2n d vo 100ml dung dch X gm cc ion : NH4+ , 24SO

,

3NO ri tin hnh un nng th thu c 23,3 gam kt ta v 6,72 lt (ktc) mt cht kh duy

nht. Nng kt ta (NH4)2SO4 v NH4NO3 trong dung dch X ln lt l:

A. 1M v 1M. B. 2M v 2M. C. 1M v 2M. D. 2M v 1M.

Cu 14 : Dung dch X cha cc ion : Fe3+, SO42 , NH4

+ ,Cl . Chia dung dch X thnh hai phn

bng nhau :


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- 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.

P N

1A 2B 3A 4A 5C 6B 7C

8A 9D 10C 11A 12C 13A 14C

Phng php 5

Phng php Bo ton electronI. CSCA PHNG PHP

1. Csca phng php

Trong phn ng oxi ha kh: s electron nhng = s electron nhn

s mol electron nhng = s mol electron nhn2. Mt s ch .

- Ch yu p dng cho bi ton oxi ha khcc cht v c

- C thp dng bo ton electron cho mt phng trnh, nhiu phng trnh hoc ton b qu trnh.

- Xc nh chnh xc cht nhng v nhn electron. Nu xt cho mt qu trnh, chcn xc nh

trng thi u v trng thi cui soxi ha ca nguyn t , thng khng quan tm n trng

thi trung gian soxi ha ca nguyn t.

- Khi p dng phng php bo ton electron thng sdng km cc phng php bo ton

khc (bo ton khi lng, bo ton nguyn t)

- Khi cho kim loi tc dng vi dung dch HNO3 v dung dch sau phn ng khng cha mui amoni:

3NOn = s mol electron nhng (hoc nhn)


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47

II. CC DNG BI TON THNG GP

V d 1 : Ho tan hon ton 19,2 gam Cu bng dung dch HNO3 ton b lng kh NO (sn phm

kh duy nht) thu c em oxit ho thnh NO2 ri chuyn ht thnh HNO3 Th tch kh oxi

(ktc) tham gia vo qu trnh trn l

A. 2,24 lt. B. 4,48 lt. C. 3,36 lt. D. 6,72 lt.

Gii :

Cch 1:

Gii thng thng: nCu = 0,3mol64

19,2=

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O (1)

0,3 0,2 mol

2NO + O2 2NO2 (2)

0,2 0,1 0,2

4NO2 + O2 + 2H2O 4HNO3 (3)

0,2 0,05

2On = 0,1 + 0,05 = 0,15 (mol) V = 0,15.22,4 = 3,36 lt p n C

Cch 2:

p dng phng php bo ton e.

Nhn xt:

Xt ton b qu trnh+ Nitcoi nh khng c s thay i s oxi ha (HNO3 ban u HNO3)

+ Nh vy ch c 2 nguyn t c s thay i s oxi ha l Cu v O2

Cu - 2e Cu2+

0,3 2.0,3

O2 + 4e 2O2-

0,15 0,6 V= 0,15.22,4 = 5,6 lt p n C

V d 2 : Oxi ho hon ton 0,728 gam bt Fe ta thu c 1,016 gam hn hp X gm hai oxit st.Ho tan hon ton X bng dung dch axit HNO3 long d. Th tch kh NO (sn phm kh duy

nht ktc) thu c sau phn ng l

A. 2,24ml. B. 22,4ml. C. 33,6ml. D. 44,8ml.


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48

Gii :

Cc phn ng c th c

2Fe +O2 2FeO0t 1)

2Fe + 1,5O2

0t Fe2O3 (2)

3Fe +2O2 0t Fe3O4 (3)

Cc phn ng ho tan c th c:

3FeO + 10HNO3 3Fe(NO3)3+NO+5H2O (4)

Fe2O3 +6HNO3 O3H)2Fe(NO 233 + (5)

3Fe3O4 +28HNO3 O14HNO)9Fe(NO 233 ++ (6)

Xt c qu trnh ta thy c 3 qu trnh thay i s oxi ho l:

+Fe t Fe0

b oxi ho thnh Fe+3

, cn N+5

b kh thnh N+2

,02O b kh thnh 2O

-2

.p dng bo ton khi lng:

2Om = mx mFe(ban u)= 1,016 0,728

2On = 0,009

Thc cht cc qu trnh oxi ho - kh trn l:

Fe - 3e Fe3+ O2 + 4e 2O2-

0,013 0,039 0,009 0,036

N+5 + 3e N+2(NO)

3nNO nNO

p dng bo ton eletron, ta c: 3nNO + 0,036 = 0,039

nNO = 0,001 mol VNO= 0,001.22,4 = 0,0224 lt = 22,4ml p n B.

V d 3 : Nung m gam bt st trong oxi, thu c 3 gam hn hp nht rn X. Ho tan ht hn

hp X bng dung dch HNO3 d thu c 0,56 lt NO (sn phm kh duy nht ktc). Gi tr ca

m l


Gii :

m gam

+

+

+++

33

3

2

HNOO0

)(NOFe

NOXFe

5

3

0

2

p dng nh lut bo ton khi lng ta c :


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49

2Om = mx mFe(ban u) = 3- m

2On =

32

m3

Thc cht cc qu trnh oxi ho - kh trn l :

Fe - 3e Fe3+ O2 + 4e 2O2-

56m

563m

32m-3

324(3-m)

N+5 + 3e N+2

0,075 0,025 (mol)

2,52gamm0,07532

m)4(3

56

3m=+

= p n A

V d 4 : Cho m gam bt Fe vo dng dch HNO3 ly d, ta c hn hp gm hai kh NO2 v

NO c VX = 8,96 lt (ktc) v t khi i vi O2 bng 1,3125. Thnh phn % NO v % NO2 theo

th tch trong hn hp X v khi lng m ca Fe dng ln lt lA. 25% v 75% ; 1,12 gam. B. 25% v 75% ; 11,2 gam.

C. 35% v 65% ; 11,2 gam. D. 45% v 55% ; 1,12 gam.

Gii :

Ta c : nX = 0,4 mol; Mx= 42

Sng cho :

NO2:46 42 30 =12

42

NO:30 46 30 =12

=+ mol0,4nn

3=4:12=n:n

NONO

NONO

2

2

==

= 75%%V25%%V

mol0,3n

0,1mol=n

22 NO

NO

NO

NO

Fe 3e Fe3+ N+5 +3e N+2

x 3x 0,3 0,1

N+5 +1e N+4

0,3 0,3Theo nh lut bo ton electron: 3x = 0,3 + 0,3 x = 0,2 mol

mFe= 0,2.56 =11,2 g p n B


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V d 5: m gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hp X c

khi lng l 75,2 gam gm Fe, FeO, Fe2O3 v Fe3O4. Cho hn hp X phn ng ht vi dung

dch H2SO4m c, nng thu c 6,72 lit kh SO2 (ktc). Gi tr ca m l:

A. 56 B. 11,2 C. 22,4 D. 25,3

Gii:

nFe(ban u) =56

mmol

p dng nh lut bo ton khi lng 2On (phn ng) =75,2 - m

(mol)32

Fe Fe3+ + 3e O2 + 4e 2O-2

56

m 56

3m 32

m-75,2 32

m-75,2.4

en nhng = 563m mol S+6 + 2e S+4(SO2)

0,6 0,3

ne nhn = 32

m-75,2.4 + 0,6

32

m-75,2.4 + 0,6 = 56

3m

m = 56 gam.

p n A.V d 6 : Ho 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 lt. B. 4,48 lt C. 5,6 lt. D. 3,36 lt.

Gii :

t nFe = nCu = a mol 56a + 64a = 12 a = 0,1mol

Fe 3e Fe3+ N+5+ 3e N+2 (NO)

0,10,3mol 3x x

Cu 2e Cu2+ N+5 +1e N+4 (NO2)

0,10,2 mol y y

Theo phng php bo ton e: ne(nhng) = ne(nhn)

3x + y = 0,5 (*)


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Mt khc: 19,2yx

46y30x=

++

(**)

T (*) v (**) x = y = 0,125 mol

V hn hp kh (ktc) = (0,125 +0,125). 22,4 = 5,6 lt p n C

V d 7 : Ho tan 15 gam hn hp X gm hai kim loi Mg v Al vo dung dch Y gm HNO3 v

H2SO4c thu c 0,1 mol mi kh SO2, NO, NO2 , N2O. Thnh phn % khi lng ca Al v

Mg trong X ln lt l

A. 63% v 37%. B. 36% v 64%.

C. 50% v 50%. D. 46% v 54%.

Gii :

t nMg = x mol, nAl = y mol. Ta c : 24x +27y = 15 (1)

Mg 2e Mg2+ N+5 + 3e N+2(NO)

x 2x 0,3 0,1

Al 3e Al3+ N+5 + e N+4(NO)

ne nhng = 2x+3y 0,1 0,1

N+5 + 4e N+1(N2O)

0,80,1.2

S+6 + 2e S+4(SO2)

0,2 0,1

ne nhn

= 1,4

Theo nh lut bo ton eletron: 2x +3y = 1,4 (2)

Gii h (1), (2) ta c: x = 0,4 mol ; y = 0,2 mol

% Al = %36%100.15

2,0.27 =

%Mg = 100% - 36% = 64% p n B.

V d 8 : Hn hp X gm 2 kim loi R1, R2 c ho tr x,y khng i (R1, R2 khng tc dng vi

nc v ng trc Cu trong dy hot ng ho hc ca kim loi). Cho hn hp X tan ht trong

dung dch Cu(NO3)2 sau ly cht rn thu c phn ng hon ton vi dung dch HNO3 d thu

c 1,12 lt kh NO duy nht ktc. Nu cng lng hn hp X trn phn ng hon ton vi

dung dch HNO3 long d th thu c bao nhiu lt N2 (sn phm kh duy nht ktc) ?

A. 0,224 lt. B. 0,336 lt. C. 0,448 lt. D. 0,672 lt.


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52

Gii:

Trong bi ton ny c hai th nghim:

TN1: R1 v R2 nhng e cho Cu2+ chuyn thnh Cu sau Cu li nhng e cho

5

N+

thnh

(NO)N

2+

. S mol e do R1v R2 nhng ra l:

5

N+

+ 3e 5

N+

0,15 05,04,22

12,1=

TN2. R1; R2 trc tip nhng e cho5

N+

to ra N2. Gi x l s mol N2, th s mol e thu c vo

l:

25

N+

+10e 02N

10x x mol

Ta c: 10x = 0,15

2N

V = 22,4.0,015 = 0,336 lt p n B

V d 9 : Hn hp X gm hai kim loi ng trc H trong dy in ho v c ho tr khng i

trong cc hp cht. Chia m gam X thnh hai phn bng nhau

- Phn 1 : Ho tan hon ton trong dung dch cha axit HCl v H2SO4 long to ra 3,36 lt kh H2

- Phn 2 : Tc dng hon ton vi dung dch HNO3 thu c V lt kh NO (sn phm kh duy

nht).

Bit cc th tch kho iu kin tiu chun. Gi tr ca V l

A. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 6,72 lt.

Gii:

Nhn xt:

V tng s mol e nhng trong 2 phn l nh nhau, nn s e nhn trong 2 phn cng nh nhau

- Phn 1: 2H+ + 2e H20,03 0,015

- Phn 2: N+5 + 3e N+2(NO)0,03 0,01

VNO = 0,1.22,4 = 2,24 lt

p n A.


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V d 10: Cho 1,35 gam hn hp gm Cu, Mg, Al tc dng ht vi dung dch HNO3 thu c

hn hp kh gm 0,01 mol NO v 0,04 mol NO2. Bit phn ng khng to mui NH4NO3. Khi

lng mui to ra trong dung dch l:


Gii:

N+5 + 3e N+2(NO)

0,03 0,01

N+5 + 1e N+4(NO2)

0,04 0,04

-3NO

n(mui) = n electron nhng (hoc nhn) = 0,03 + 0,04 = 0,07 (mol)

mmui = mkim loi +-3NOm (mui) = 1,35 + 0,07.63 = 5,69 gam

p n C.

V d 11: Cho 3 kim loi Al, Fe, Cu vo 2 lt dung d ch HNO3 phn ng va thu c 1,792

lt kh X (ktc) gm N2 v NO2 c t khi hi so vi He bng 9,25. Nng mol ca HNO3 trong

dung dch u l:

A. 0,28 M B. 1,4 M C. 1,7 M D. 1,2 M

Gii:

Ta c2 2N NO

X

(M M )

M 9,25. 4 37

+

= = = l trung bnh cng khi lng phn t ca 2 kh N2

v NO2 nn:2N

n = 2

nn XNO2 = = 0,04 mol

2N+5 + 10e N2

0,4 0,04

N+5 + 1e N+4(NO2)

0,04 0,04

-3NO

n(mui) =

nelectron nhng (hoc nhn) = 0,4 + 0,04 = 0,44 mol

p dng bo ton nguyn t ta c:

3HNOn

(b kh) = -3NOn

(mui) + nN(trong kh) = 0,44 + 0,04.2 + 0,04 = 0,56 mol

[HNO3] = 0,28M2

0,56 = p n A


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V d 12 : Chia m gam hn hp X gm Al, Fe thnh hai phn bng nhau :

- Phn 1 : Ho tan hon ton trong dung dch HCl d thu c 7,28 lt H2

- Phn 2 : Ho tan hon ton trong dung dch HNO3 long d thu c 5,6 lt NO (sn phm kh

duy nht).

- Bit th tch cc kho ktc Khi lng Fe, Al c trong X ln lt l:A. 5,6 gam v 4,05 gam. B. 16,8 gam v 8,1 gam.

C. 5,6 gam v 5,4 gam. D. 11,2 gam v 4,05 gam.

Gii:

Tc dng vi HCl

Al - 3e Al3+ 2H+ + 2e H2

Fe - 2e Fe2+ 0,65 0,325

Tc dng vi HNO3

M - 3e M3+ N+5 + 3e N+2

0,25 0,75 0,75 0,25

Nhn xt:

S mol e hn hp Al; Fe nhng khi tc dng HCl : 0,65 mol

S mol e hn hp Al; Fe nhng khi tc dng HNO3: 0,75 mol

S mol e m Al nhng l nh nhau vi HCl v HNO3; 1 mol Fe nhng cho HNO3

nhiu hn cho HCl l 1 mol e;

nFe=0,75 - 0,65 = 0,1 mol mFe = 5,6 gam

nAl =0,25 - 0,1 = 0,15 mol mAl = 4,05 gam

p n A.

V d 13 : Ho tan hon ton 11,2 gam hn hp Cu - Ag bng 19,6 gam dung dch H2SO4c

un nng sau phn ung thu c kh X v dung dch Y. Ton b kh X c dn chm qua dung

dch nc clo d, dung dch thu c cho tc dng vi BaCl2 d thu c 18,64 gam kt ta.

Khi lng Cu, Ag v nng ca dung dch H2SO4 ban u ln lt l :

A. 2,56 ; 8,64 v 96%. B. 4,72 ; 6,48 v 80%.

C. 2,56 ; 8,64 v 80%. D. 2,56 ; 8,64 v 90%.

Gii:

t : nCu = x; nAg = y 64x + 108y = 11,2 (*)


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Cu 2e Cu2+ S+6 +2e S+4(SO2)

x 2x 0,16 0,08

Ag e Ag+

y y

Ta c s chuyn ho

SO 4BaCl2

4OHCl

224 BaSOSOSO 222

+++

0,08 mol0,08233

18,64 =

p dng bo ton eletron: 2x + y = 0,16 (**)

T (*) (**) x = 0,04, y = 0,08

mCu = 0,04. 64 = 2,56gam; mAg = 8,64gam

p dng bo ton nguyn t ca lu hunh

24SO

n(axit) = 24SO

n(mui) + 2SO

n = 16,008,0)2

08,004,0( =++

C%(H2SO4) = %80%100.6,19

98.16,0 = p n C

V d 14 : Cho hn hp X gm 0,1 mol Al v 0,1 mol Fe vo 100ml dung d ch Y gm Cu(NO3)2

v AgNO3 sau khi phn ng kt thc thu c cht rn Z gm 3 kim loi. Ho tan hon ton Z

bng dung dch HCl d thu c 0,05 mol H2 v cn li 28 gam cht rn khng tan. Nng mi

ca Cu(NO3)2 v ca AgNO3 trong Y ln lt l :

A. 2M v 1M. B. 1M v 2M.

C. 0,2M v 0,1M. D. 0,5M v 0,5M.

Gii:

Tm tt s:

8,3gam hn hp X +

Fe

Al 100ml dung dch Y

moly:)Cu(NO

molx:AgNO

23

3

1,12 lt H2

Cht rn A + HCl (3 kim loi) 2,8 gam cht rn khng tan B

t3AgNO

n = x mol v23)Cu(NO

n = y mol

Cht rn Z gm 3 kim loi 3 kim loi phi l: Ag, Cu, Fe

Al, Cu(NO3)2 v AgNO3 tham gia phn ng ht, Fe cha phn ng hoc d

(nAl = nFe)

d


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Xt cho ton b qu trnh, ta c:

Ag+ +1e Ag0 Al 3e Al3+

x x x 0,1 0,3

Cu2+ +2e Cu0 Fe 2e Fe2+

y 2y y 0,1 0,2

2H+ +2e H2

0,10,05

Theo nh lut bo ton eletron, ta c phng trnh:

x + 2y + 0,1 = 0,3 + 0,2 x + 2y = 0,4 (1)

Mt khc, cht rn khng tan l: Ag: x mol; Cu: y mol

108x + 64y = 28 (2)

Gii h (1), (2) ta c: x = 0,2 mol; y = 0,1 mol

[AgNO3] = 3 20,2 0,1

M; [Cu(NO ) ] 1M0,1 0,1

= = = p n B

V d 15 : Trn 0,54 gam bt Al vi hn hp bt Fe2O3 v CuO ri tin hnh phn ng nhit

nhm trong iu kin khng c khng kh mt thi gian. thu c hn hp cht rn X. Ho tan

hon ton X trong dung dch HNO3c, nng, d th th tch NO2 (sn phm kh duy nht

ktc) thu c l

A. 0,672 lt. B. 0,896 lt. C. 1,12 lt. D. 1,344 lt.

Gii:

Phn tch:

Nu gii theo cch thng thng s gp rt nhiu kh khn:

+ Phn ng nhit nhm l khng hon ton (tin hnh phn ng mt thi gian ), do c

nhiu sn phm v vy phi vit rt nhiu phng trnh

+ Sn s cn t ln, trong khi bi ton ch cho mt d kin

Xt cho ton b qu trnh, ch c Al v N (trong HNO3) c s thay i s oxi ho trng thi

u v cui, do ch cn vit hai qu trnh:

Al - 3e Al3+ N+5 +1e N+4 (NO2)

0,02 0,06 0,06 0,06

2NO

V = 0,06. 22,4 = 1,344 lt p n D


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V d 16 : Trn 60 gam bt Fe vi 30 gam bt lu hunh ri un nng (khng c khng kh) thu

c cht rn X. Ho tan X bng dung dch axit HCl dc dung dch Y v kh Z. t chy

hon ton Z cn ti thiu V lt O2 (ktc). Bit cc phn ng xy ra hon ton. Gi tr ca V l

A. 11,2. B. 21. C. 33. D. 49.

Gii:

V nFe > nS =30

32nn Fe d v S ht

Kh C l hn hp H2S v H2. t Z thu c SO2 v H2O. Kt qu cui cng ca qu trnh phn

ng l Fe v S nhng e, cn O2 thu e

Fe - 2e Fe2+ O2 + 4e 2O-2

mol56

602.

56

60 x 4x

S 4e S+4

mol32

304.

32

30

2.60 30

4. 4x x 1,4732 mol56 32

+ = =

2O

V = 22,4. 1,4732 = 33 lt p n C

V d 17 : Ho tan hon ton 1,08 gam Al bng dung dch HNO3 d, sn phm ng thu c

0,336 lt kh X (sn phm kh duy nht ktc). Cng thc phn t ca X l

A. NO2 B. N2O C. N2 D. NO

Gii:

nAl = 0,04 ; nX = 0,015

Al 3e Al3+ N+5 + ne X5-n

0,04 0,12 mol 0,12 moln

0,12

8n0,015n

0,12 == ng vi 2N+5 + 8e 2N+1(N2O) p n B

V d 18 : Khi cho 9,6 gam Mg tc dng ht vi dung dch H2SO4m c, thy c 49 gam

H2SO4 tham gia phn ng to mui MgSO4, H2O v sn phm kh X l :

A. SO2 B. S. C. H2S. D. H2


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Gii:

Dung dch H2SO4m c va l cht oxi ho va l mi trng

Gi a l s oxi ho ca S trong X

Mg Mg2+ + 2e S+6 + (6-a)e Sa

0,4 mol 0,8mol 0,1 mol 0,1(6-a)mol

Tng s mol H2SO4 dng l: 0,5(mol)98

49 =

S mol H2SO4 dng to mui bng s mol Mg = 9,6: 24 = 0,4mol

S mol H2SO4 dng oxi ho Mg = 0,5 0,4 = 0,1mol

Ta c: 0,1.(6 - a) = 0,8 x = - 2. Vy Z l H2S p n C

V d 19 : Cho 13,92 gam Fe3O4 tc dng hon ton vi dung dch HNO3 sau phn ng thu c

dung dch X v 0,448 lt kh NxOy (Sn phm kh duy nht (ktc). Khi lng HNO3 nguyn

cht tham ra phn ng l


Gii:

Cch 1: Vit v cn bng phng trnh ho hc:

(5x 2y )Fe3O4 + (46x-18y) HNO3 (15x -6y)Fe(NO3)3 +NxOy +(23x-9y)H2O

0,06 0,02(mol)

Cch 2:

3Fe+8

3 e 3Fe+3 xN+5 + (5x-2y)e xN+2y/x

0,06 0,06 0,02 (5x- 2y) 0,02x

iu kin : x 2; y 5 (x,y N)

0,02(5x-2y) = 0,06 x =1; y = 1 (hp l)

3HNOn

(phn ng) = 3NOn

(mui) + Nn (trong kh) = 3. 0,06. 3 + 0,02 = 0,56 mol

3HNO

m(phn ng)

= 0,56. 63 = 35,28 gam p n A

V d 20 : Cho 18,56 gam st oxit tc dng hon ton vi dung dch HNO3 sau phn ng thu

c dung dch X v 0,224 lt kh mt oxit ca nit(sn phm kh duy nht ktc). Cng thc

ca hai oxit ln lt l

A. FeO v NO. B. Fe3O4 v NO2

C. FeO v N2O. D. Fe3O4 v N2O.

Gii:

t cng thc tng qut ca 2 oxit l: Fe2On; N2Om(n


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59

2Fe+n - 2(3 - n)e 2Fe+3

2.18,65 2.(3-n).18,65

112 + 16n 112 + 16n

2N+5 + 2(5 - m)e 2N+m (2)

0,02.(5-m) 2

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16 Phuong Phap Giai Hoa