10 PHUONG PHAP GIAI HOA

8/7/2019 10 PHUONG PHAP GIAI HOA

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10 PHNG PHP GII NHANH BI TP

TRC NGHIM HA HC

Phng php 1P DNG NH LUT BO TON KHI LNG

Nguyn tc ca phng php ny kh n gin, da vo nh lut bo ton khilng: Tng khi lng cc cht tham gia phn ng bng tng khi lng cc cht tothnh trong phn ng. Cn lu l: khng tnh khi lng ca phn khng tham giaphn ng cng nh phn cht c sn, v d nc c sn trong dung dch.

Khi c cn dung dch th khi lng mui thu c bng tng khi lng cc cationkim loi v anion gc axit.

V d 1: Hn hp X gm Fe, FeO v Fe2O3. Cho mt lung CO i qua ng s ng mgam hn hp X nung nng. Sau khi kt thc th nghim thu c 64 gam chtrn A trong ng s v 11,2 lt kh B (ktc) c t khi so vi H2 l 20,4. Tnh gitr m.

A. 105,6 gam. B. 35,2 gam. C. 70,4 gam. D. 140,8 gam.

Hng dn gii

Cc phn ng kh st oxit c th c:

3Fe2O3 + COot 2Fe3O4 + CO2 (1)

Fe3O4 + COot 3FeO + CO2 (2)

FeO + COo

t Fe + CO2 (3)

Nh vy cht rn A c th gm 3 cht Fe, FeO, Fe3O4 hoc t hn, iu khngquan trng v vic cn bng cc phng trnh trn cng khng cn thit, quan trng l smol CO phn ng bao gi cng bng s mol CO2 to thnh.

B

11,2n 0,5

22,5= = mol.

Gi x l s mol ca CO2 ta c phng trnh v khi lng ca B:

44x + 28(0,5 x) = 0,5 20,4 2 = 20,4

nhn c x = 0,4 mol v cng chnh l s mol CO tham gia phn ng.

Theo LBTKL ta c:

mX + mCO = mA + 2COm

m = 64 + 0,4 44 0,4 28 = 70,4 gam. (p n C)


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V d 2: un 132,8 gam hn hp 3 ru no, n chc vi H2SO4 c 140oC thu chn hp cc ete c s mol bng nhau v c khi lng l 111,2 gam. S molca mi ete trong hn hp l bao nhiu?

A. 0,1 mol. B. 0,15 mol. C. 0,4 mol. D. 0,2 mol.

Hng dn giiTa bit rng c 3 loi ru tch nc iu kin H2SO4 c, 140oC th to thnh 6

loi ete v tch ra 6 phn t H2O.

Theo LBTKL ta c

2H O etem m m 132,8 11,2 21,6= = =r u gam

2H O

21,6n 1,2

18= = mol.

Mt khc c hai phn t ru th to ra mt phn t ete v mt phn t H2O do s

mol H2O lun bng s mol ete, suy ra s mol mi ete l 1,2 0,26

= mol. (p n D)

Nhn xt: Chng ta khng cn vit 6 phng trnh phn ng t ru tch nc tothnh 6 ete, cng khng cn tm CTPT ca cc ru v cc ete trn. Nu cc bn xa vo vic vit phng trnh phn ng v t n s mol cc ete tnh ton th khngnhng khng gii c m cn tn qu nhiu thi gian.

V d 3: Cho 12 gam hn hp hai kim loi Fe, Cu tc dng va vi dung dch HNO363%. Sau phn ng thu c dung dch A v 11,2 lt kh NO 2 duy nht (ktc).Tnh nng % cc cht c trong dung dch A.

A. 36,66% v 28,48%. B. 27,19% v 21,12%.

C. 27,19% v 72,81%. D. 78,88% v 21,12%.

Hng dn gii

Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O

Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O

2NOn 0,5= mol

3 2HNO NOn 2n 1= = mol.

p dng nh lut bo ton khi lng ta c:

223

NOd HNOm m m m

1 63 10012 46 0,5 89 gam.

63

= +

= + =

2 2d mui h k.loi

t nFe = x mol, nCu = y mol ta c:

56x 64y 12

3x 2y 0,5

+ = + =

x 0,1

y 0,1

= =


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3 3Fe(NO )

0,1 242 100%m 27,19%

89

= =

3 2Cu(NO )

0,1 188 100%m 21,12%.

89

= = (p n B)

V d 4: Ho tan hon ton 23,8 gam hn hp mt mui cacbonat ca cc kim loi hotr (I) v mui cacbonat ca kim loi ho tr (II) trong dung dch HCl. Sau phnng thu c 4,48 lt kh (ktc). em c cn dung dch thu c bao nhiu gammui khan?

A. 13 gam. B. 15 gam. C. 26 gam. D. 30 gam.

Hng dn gii

M2CO3 + 2HCl 2MCl + CO2 + H2O

R2CO3 + 2HCl 2MCl2 + CO2 + H2O

2CO4,88n 0,222,4

= = mol

Tng nHCl = 0,4 mol v 2H On 0, 2 mol.=


23,8 + 0,4 36,5 = mmui + 0,2 44 + 0,2 18

mmui = 26 gam. (p n C)

V d 5: Hn hp A gm KClO3, Ca(ClO2)2, Ca(ClO3)2, CaCl2 v KCl nng 83,68 gam.Nhit phn hon ton A ta thu c cht rn B gm CaCl 2, KCl v 17,472 ltkh ( ktc). Cho cht rn B tc dng vi 360 ml dung dch K2CO3 0,5M (va) thu c kt ta C v dung dch D. Lng KCl trong dung dch D nhiu gp22/3 ln lng KCl c trong A. % khi lng KClO3 c trong A l

A. 47,83%. B. 56,72%. C. 54,67%. D. 58,55%.

Hng dn gii

o

o

o

2

t

3 2

t

3 2 2 2

t2 2 2 2

2 2

(A) (A)

h B

3KClO KCl O (1)

2

Ca(ClO ) CaCl 3O (2)

83,68 gam A Ca(ClO ) CaCl 2O (3)CaCl CaCl

KCl KCl

+

+

+

1 2 3

2On 0,78 mol.=


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mA = mB + 2Om

mB = 83,68 32 0,78 = 58,72 gam.

Cho cht rn B tc dng vi 0,18 mol K2CO3

Hn hp B

2 2 3 3

(B) (B)

CaCl K CO CaCO 2KCl (4)

0,18 0,18 0,36 mol

KCl KCl

+ +

hn hp D

( B) 2KCl B CaCl (B)

m m m

58,72 0,18 111 38,74 gam

=

= =

( D)KCl KCl (B) KCl (pt 4)m m m

38,74 0,36 74,5 65,56 gam

= +

= + =

( A ) ( D )KCl KCl

3 3m m 65,56 8,94 gam

22 22= = =

(B) (A)KCl pt (1) KCl KClm = m m 38,74 8,94 29,8 gam. = =

Theo phn ng (1):

3KClO

29,8m 122,5 49 gam.

74,5= =

3KClO (A)

49 100%m 58,55%.

83,68

= = (p n D)

V d 6: t chy hon ton 1,88 gam cht hu c A (cha C, H, O) cn 1,904 lt O 2(ktc) thu c CO2 v hi nc theo t l th tch 4:3. Hy xc nh cng thcphn t ca A. Bit t khi ca A so vi khng kh nh hn 7.

A. C8H12O5. B. C4H8O2. C. C8H12O3. D. C6H12O6.

Hng dn gii

1,88 gam A + 0,085 mol O2 4a mol CO2 + 3a mol H2O.


2 2CO H Om m 1,88 0,085 32 46 gam+ = + =Ta c: 44 4a + 18 3a = 46 a = 0,02 mol.

Trong cht A c:

nC = 4a = 0,08 mol

nH = 3a 2 = 0,12 mol


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nO = 4a 2 + 3a 0,085 2 = 0,05 mol

nC : nH : no = 0,08 : 0,12 : 0,05 = 8 : 12 : 5

Vy cng thc ca cht hu c A l C8H12O5 c MA < 203. (p n A)

V d 7: Cho 0,1 mol este to bi 2 ln axit v ru mt ln ru tc dng hon ton viNaOH thu c 6,4 gam ru v mt lng mi c khi lng nhiu hnlng este l 13,56% (so vi lng este). Xc nh cng thc cu to ca este.

A. CH3COOCH3.

B. CH3OCOCOOCH3.

C. CH3COOCOOCH3.

D. CH3COOCH2COOCH3.

Hng dn gii

R(COOR )2 + 2NaOH R(COONa)2 + 2R OH0,1 0,2 0,1 0,2 mol

R OH

6,4M 32

0,2 = = Ru CH3OH.


meste + mNaOH = mmui + mru

mmuimeste = 0,2 40 64 = 1,6 gam.

m mmui meste =13,56

100meste

meste =1,6 100

11,8 gam13,56

= Meste = 118 vC

R + (44 + 15) 2 = 118 R = 0.

Vy cng thc cu to ca este l CH3OCOCOOCH3. (p n B)

V d 8: Thu phn hon ton 11,44 gam hn hp 2 este n chc l ng phn ca nhaubng dung dch NaOH thu c 11,08 gam hn hp mui v 5,56 gam hn hpru. Xc nh cng thc cu to ca 2 este.

A. HCOOCH3 v C2H5COOCH3,

B. C2H5COOCH3 vCH3COOC2H5.

C. HCOOC3H7 v C2H5COOCH3.

D. C B, C u ng.

Hng dn gii

t cng thc trung bnh tng qut ca hai este n chc ng phn l RCOOR .


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RCOOR + NaOH RCOONa + R OH

11,44 11,08 5,56 gam


MNaOH = 11,08 + 5,56 11,44 = 5,2 gam

NaOH5,2n 0,13 mol40

= =

RCOONa11,08

M 85,230,13

= = R 18,23=

R OH5,56

M 42,770,13

= = R 25,77 =

RCOOR11,44

M 880,13

= =

CTPT ca este l C4H8O2Vy cng thc cu to 2 este ng phn l:

HCOOC3H7 v C2H5COOCH3

hoc C2H5COOCH3 vCH3COOC2H5. (p n D)

V d 9: Chia hn hp gm hai anehit no n chc lm hai phn bng nhau:

- Phn 1: em t chy hon ton thu c 1,08 gam H2O.

- Phn 2: Tc dng vi H2 d (Ni, to) th thu c hn hp A. em t chyhon ton th th tch kh CO2 (ktc) thu c l

A. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 0,672 lt.

Hng dn gii

Phn 1: V anehit no n chc nn2 2CO H O

n n= = 0,06 mol.

2CO C

n n 0,06(phn2) (phn2)= = mol.

Theo bo ton nguyn t v bo ton khi lng ta c:

C C (A)n n 0,06(phn2) = = mol.

2CO (A)

n = 0,06 mol

2CO

V = 22,4 0,06 = 1,344 lt. (p n C)

V d 10: Cho mt lung CO i qua ng s ng 0,04 mol hn hp A gm FeO v Fe 2O3t nng. Sau khi kt thc th nghim thu c B gm 4 cht nng 4,784 gam.Kh i ra khi ng s cho hp th vo dung dch Ba(OH) 2 d th thu c9,062 gam kt ta. Phn trm khi lng Fe2O3 trong hn hp A l


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A. 86,96%. B. 16,04%. C. 13,04%. D.6,01%.

Hng dn gii

0,04 mol hn hp A (FeO v Fe2O3) + CO 4,784 gam hn hp B + CO2.

CO2 + Ba(OH)2 d BaCO3 + H2O

2 3CO BaCOn n 0,046 mol= =

v2CO( ) CO

n n 0,046 molp. = =


mA + mCO = mB + 2COm

mA = 4,784 + 0,046 44 0,046 28 = 5,52 gam.

t nFeO = x mol, 2Fe O 3n y mol= trong hn hp B ta c:

x y 0,0472x 160y 5,52

+ = + = x 0,01 mol

y 0,03 mol= =

%mFeO =0,01 72 101

13,04%5,52

=

%Fe2O3 = 86,96%. (p n A)

MT S BI TP VN DNG GII THEO PHNG PHP S

DNG NH LUT BO TON KHI LNG

01. Ha tan 9,14 gam hp kim Cu, Mg, Al bng mt lng va dung dch HCl thuc 7,84 lt kh X (ktc) v 2,54 gam cht rn Y v dung dch Z. Lc b cht rn Y,c cn cn thn dung dch Z thu c lng mui khan l


02. 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 dung dch HCl phi dng l

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

03. Trn 8,1 gam bt Al vi 48 gam bt Fe2O3 ri cho tin hnh phn ng nhit nhm

trong iu kin khng c khng kh, kt thc th nghim lng cht rn thu c lA. 61,5 gam. B. 56,1 gam. C. 65,1 gam. D. 51,6 gam.

04. Ha tan hon ton 10,0 gam hn hp X gm hai kim loi (ng trc H trong dyin ha) bng dung dch HCl d thu c 2,24 lt kh H 2 (ktc). C cn dung dchsau phn ng thu c lng mui khan l



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05. Nhit phn hon ton m gam hn hp X gm CaCO3 v Na2CO3 thu c 11,6 gamcht rn v 2,24 lt kh (ktc). Hm lng % CaCO3 trong X l

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

06. Cho 4,4 gam hn hp hai kim loi nhm I A hai chu k lin tip tc dng vi dungdch HCl d thu c 4,48 lt H2 (ktc) v dung dch cha m gam mui tan. Tn haikim loi v khi lng m l

A. 11 gam; Li v Na. B. 18,6 gam; Li v Na.

C. 18,6 gam; Na v K. D. 12,7 gam; Na v K.

07. t chy hon ton 18 gam FeS2 v cho ton b lng SO2 vo 2 lt dung dchBa(OH)2 0,125M. Khi lng mui to thnh l

A. 57,40 gam. B. 56,35 gam. C. 59,17 gam. D.58,35 gam.

08. Ha tan 33,75 gam mt kim loi M trong dung dch HNO 3 long, d thu c 16,8 ltkh X (ktc) gm hai kh khng mu ha nu trong khng kh c t khi hi so vihiro bng 17,8.

a) Kim loi l

A. Cu. B. Zn. C. Fe. D. Al.

b) Nu dng dung dch HNO3 2M v ly d 25% th th tch dung dch cn ly l

A. 3,15 lt. B. 3,00 lt. C. 3,35 lt. D. 3,45 lt.

09. Ho tan hon ton 15,9 gam hn hp gm 3 kim loi Al, Mg v Cu bng dung dchHNO3 thu c 6,72 lt kh NO v dung dch X. em c cn dung dch X thu cbao nhiu gam mui khan?


10. Ha tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500 ml axit H2SO40,1M (va ). Sau phn ng, hn hp mui sunfat khan thu c khi c cn dungdch c khi lng l


p n cc bi tp vn dng:

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

6. B 7. D 8. a-D, b-B 9. B 10. A

Phng php 2

BO TON MOL NGUYN T

C rt nhiu phng php gii ton ha hc khc nhau nhng phng php boton nguyn t v phng php bo ton s mol electron cho php chng ta gp nhiuphng trnh phn ng li lm mt, qui gn vic tnh ton v nhm nhanh p s. Rt phhp vi vic gii cc dng bi ton ha hc trc nghim. Cch thc gp nhng phng


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trnh lm mt v cch lp phng trnh theo phng php bo ton nguyn t s c giithiu trong mt s v d sau y.

V d 1: kh hon ton 3,04 gam hn hp X gm FeO, Fe 3O4, Fe2O3 cn 0,05 mol H2.Mt khc ha tan hon ton 3,04 gam hn hp X trong dung dch H 2SO4 c thu

c th tch kh SO2 (sn phm kh duy nht) iu kin tiu chun lA. 448 ml. B. 224 ml. C. 336 ml. D. 112 ml.

Hng dn gii

Thc cht phn ng kh cc oxit trn l

H2 + O H2O

0,05 0,05 mol

t s mol hn hp X gm FeO, Fe3O4, Fe2O3 ln lt l x, y, z. Ta c:

nO = x + 4y + 3z = 0,05 mol (1)

Fe 3,04 0,05 16n 0,04 mol56 = =

x + 3y + 2z = 0,04 mol (2)

Nhn hai v ca (2) vi 3 ri tr (1) ta c:

x + y = 0,02 mol.

Mt khc:

2FeO + 4H2SO4 Fe2(SO4)3 + SO2 + 4H2O

x x/2

2Fe3O4 + 10H2SO4 3Fe2(SO4)3 + SO2 + 10H2Oy y/2

tng: SO2x y 0,2

n 0,01 mol2 2

+= = =

Vy:2SO

V 224 ml.= (p n B)

V d 2: Thi t t V lt hn hp kh (ktc) gm CO v H 2 i qua mt ng ng 16,8gam hn hp 3 oxit: CuO, Fe3O4, Al2O3 nung nng, phn ng hon ton. Sauphn ng thu c m gam cht rn v mt hn hp kh v hi nng hn khi

lng ca hn hp V l 0,32 gam. Tnh V v m.A. 0,224 lt v 14,48 gam. B. 0,448 lt v 18,46 gam.

C. 0,112 lt v 12,28 gam. D. 0,448 lt v 16,48 gam.

Hng dn gii

Thc cht phn ng kh cc oxit trn l

CO + O CO2


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H2 + O H2O.

Khi lng hn hp kh to thnh nng hn hn hp kh ban u chnh l khi lngca nguyn t Oxi trong cc oxit tham gia phn ng. Do vy:

mO = 0,32 gam.

O 0,32n 0,02 mol16= =

( )2CO H

n n 0,02 mol+ = .


moxit = mcht rn + 0,32

16,8 = m + 0,32

m = 16,48 gam.

2

hh(CO H )V 0,02 22,4 0,448+ = =lt. (p n D)

V d 3: Thi rt chm 2,24 lt (ktc) mt hn hp kh gm CO v H 2 qua mt ng sng hn hp Al2O3, CuO, Fe3O4, Fe2O3 c khi lng l 24 gam d ang cun nng. Sau khi kt thc phn ng khi lng cht rn cn li trong ng s l


Hng dn gii

2hh(CO H )

2,24n 0,1 mol

22,4+= =

Thc cht phn ng kh cc oxit l:CO + O CO2

H2 + O H2O.

Vy:2O CO H

n n n 0,1 mol= + = .

mO = 1,6 gam.

Khi lng cht rn cn li trong ng s l: 24 1,6 = 22,4 gam. (p n A)

V d 4: Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung

nng. Sau khi phn ng hon ton, khi lng cht rn trong bnh gim 0,32gam. Hn hp hi thu c c t khi i vi hiro l 15,5. Gi tr ca m l


Hng dn gii

CnH2n+1CH2OH + CuOot CnH2n+1CHO + Cu + H2O


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Khi lng cht rn trong bnh gim chnh l s gam nguyn t O trong CuO phnng. Do nhn c:

mO = 0,32 gam O0,32

n 0,02 mol16

= =

Hn hp hi gm:n 2n 1

2

C H CHO : 0,02 molH O : 0,02 mol.

+

Vy hn hp hi c tng s mol l 0,04 mol.

C M = 31

mhh hi = 31 0,04 = 1,24 gam.

mancol + 0,32 = mhh hi

mancol = 1,24 0,32 = 0,92 gam. (p n A)

Ch : Vi ru bc (I) hoc ru bc (II) u tha mn u bi.

V d 5: t chy hon ton 4,04 gam mt hn hp bt kim loi gm Al, Fe, Cu trongkhng kh thu c 5,96 gam hn hp 3 oxit. Ha tan ht hn hp 3 oxit bngdung dch HCl 2M. Tnh th tch dung dch HCl cn dng.

A. 0,5 lt. B. 0,7 lt. C. 0,12 lt. D. 1 lt.

Hng dn gii

mO = moxitmkl = 5,96 4,04 = 1,92 gam.

O

1,92n 0,12 mol

16= = .

Ha tan ht hn hp ba oxit bng dung dch HCl to thnh H2O nh sau:

2H+ + O2 H2O

0,24 0,12 mol

HCl0,24

V 0,122

= = lt. (p n C)

V d 6: t chy hon ton 0,1 mol mt axit cacbonxylic n chc cn va V lt O2( ktc), thu c 0,3 mol CO2 v 0,2 mol H2O. Gi tr ca V l

A. 8,96 lt. B. 11,2 lt. C. 6,72 lt. D. 4,48 lt.

Hng dn gii

Axit cacbonxylic n chc c 2 nguyn t Oxi nn c th t l RO2. Vy:

2 2 2 2O (RO ) O (CO ) O (CO ) O (H O)n n n n+ = +

0,1 2 + nO (p.) = 0,3 2 + 0,2 1

nO (p.) = 0,6 mol


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2On 0,3mol=

2O

V 6,72= lt. (p n C)

V d 7: (Cu 46 - M 231 - TSC Khi A 2007)

Cho 4,48 lt CO ( ktc) t t i qua ng s nung nng ng 8 gam mt oxit stn khi phn ng xy ra hon ton. Kh thu c sau phn ng c t khi so vihiro bng 20. Cng thc ca oxit st v phn trm th tch ca kh CO2 tronghn hp kh sau phn ng l

A. FeO; 75%. B. Fe2O3; 75%.

C. Fe2O3; 65%. D. Fe3O4; 65%.

Hng dn gii

FexOy + yCO xFe + yCO2

Kh thu c c M 40= gm 2 kh CO2 v CO d

2CO

CO

n 3

n 1=

2CO%V 75%= .

Mt khc:2CO ( ) CO

75n n 0,2 0,15

100p. = = = mol nCO d = 0,05 mol.

Thc cht phn ng kh oxit st l do

CO + O (trong oxit st) CO2

nCO = nO = 0,15 mol mO = 0,15 16 = 2,4 gam

mFe = 8 2,4 = 5,6 gam nFe = 0,1 mol.

Theo phng trnh phn ng ta c:

2

Fe

CO

n x 0,1 2

n y 0,15 3= = = Fe2O3. (p n B)

V d 8: Cho hn hp A gm Al, Zn, Mg. em oxi ho hon ton 28,6 gam A bng oxi

d thu c 44,6 gam hn hp oxit B. Ho tan ht B trong dung dch HCl thuc dung dch D. C cn dung dch D c hn hp mui khan l

A. 99,6 gam. B. 49,8 gam.

C. 74,7 gam. D. 100,8 gam.

Hng dn gii

Gi M l kim loi i din cho ba kim loi trn vi ho tr l n.

2CO

CO

n 44 12

40

n 28 4


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M +n

2O2 M2On (1)

M2On + 2nHCl 2MCln + nH2O (2)

Theo phng trnh (1) (2) 2HCl O

n 4.n= .

p dng nh lut bo ton khi lng 2O

m 44,6 28,6 16= = gam

2O

n 0,5= mol nHCl = 4 0,5 = 2 mol

Cln 2 mol =

mmui = mhhkl + Clm = 28,6 + 2 35,5 = 99,6 gam. (p n A)

V d 9: Cho mt lung kh CO i qua ng ng 0,01 mol FeO v 0,03 mol Fe2O3 (hnhp A) t nng. Sau khi kt thc th nghim thu c 4,784 gam cht rn Bgm 4 cht. Ho tan cht rn B bng dung dch HCl d thy thot ra 0,6272 lt

H2 ( ktc). Tnh s mol oxit st t trong hn hp B. Bit rng trong B s moloxit st t bng 1/3 tng s mol st (II) oxit v st (III) oxit.

A. 0,006. B. 0,008. C. 0,01. D. 0,012.

Hng dn gii

Hn hp A2 3

FeO : 0,01 mol

Fe O : 0,03 mol

+ CO 4,784 gam B (Fe, Fe2O3, FeO, Fe3O4)

tng ng vi s mol l: a, b, c, d (mol).

Ho tan B bng dung dch HCl d thu c2H

n 0,028= mol.

Fe + 2HCl FeCl2 + H2

a = 0,028 mol. (1)

Theo u bi: ( )3 4 2 3Fe O FeO Fe O

1n n n

3= + ( )

1d b c

3= + (2)

Tng mB l: (56.a + 160.b + 72.c + 232.d) = 4,78 gam. (3)

S mol nguyn t Fe trong hn hp A bng s mol nguyn t Fe trong hn hp B.Ta c:

nFe (A) = 0,01 + 0,03 2 = 0,07 mol

nFe (B) = a + 2b + c + 3d

a + 2b + c + 3d = 0,07 (4)

T (1, 2, 3, 4) b = 0,006 mol

c = 0,012 mol

d = 0,006 mol. (p n A)


14/91

V d 10: Kh hon ton 24 gam hn hp CuO v FexOy bng H2 d nhit cao thuc 17,6 gam hn hp 2 kim loi. Khi lng H2O to thnh l


Hng dn gii

mO (trong oxit) = moxitmkloi = 24 17,6 = 6,4 gam.

( )2O H Om 6,4= gam ; 2H O6,4

n 0,416

= = mol.

2H O

m 0,4 18 7,2= = gam. (p n C)

V d 11: Kh ht m gam Fe3O4 bng CO thu c hn hp A gm FeO v Fe. A tan va trong 0,3 lt dung dch H2SO4 1M cho ra 4,48 lt kh (ktc). Tnh m?


Hng dn gii

Fe3O4 (FeO, Fe) 3Fe2+

n mol

( ) 24 4Fe trong FeSO SOn n 0,3= = mol

p dng nh lut bo ton nguyn t Fe:

( ) ( )43 4 Fe FeSOFe Fe On n=

3n = 0,3 n = 0,1

3 4Fe O

m 23,2= gam (p n A)

V d 12: un hai ru n chc vi H2SO4 c, 140oC c hn hp ba ete. Ly 0,72gam mt trong ba ete em t chy hon ton thu c 1,76 gam CO2 v 0,72gam H2O. Hai ru l

A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. C2H5OH v C4H9OH. D. CH3OH v C3H5OH.

Hng dn gii

t cng thc tng qut ca mt trong ba ete l CxHyO, ta c:

C

0,72

m 12 0,4844= = gam ; H0,72

m 2 0,0818= = gam

mO = 0,72 0,48 0,08 = 0,16 gam.

0,48 0,08 0,16x : y :1 : :

12 1 16= = 4 : 8 : 1.

Cng thc phn t ca mt trong ba ete l C4H8O.


15/91

Cng thc cu to l CH3OCH2CH=CH2.

Vy hai ancol l CH3OH v CH2=CHCH2OH. (p n D)

MT S BI TP VN DNG GII THEO PHNG PHP BO

TON MOL NGUYN T

01. Ha tan hon ton hn hp X gm 0,4 mol FeO v 0,1mol Fe 2O3 vo dung dchHNO3 long, d thu c dung dch A v kh B khng mu, ha nu trong khngkh. Dung dch A cho tc dng vi dung dch NaOH d thu c kt ta. Ly ton bkt ta nung trong khng kh n khi lng khng i thu c cht rn c khilng l


02. Cho kh CO i qua ng s cha 16 gam Fe2O3 un nng, sau phn ng thu c hn

hp rn X gm Fe, FeO, Fe3O4, Fe2O3. Ha tan hon ton X bng H2SO4 c, nngthu c dung dch Y. C cn dung dch Y, lng mui khan thu c l


03. Kh hon ton 17,6 gam hn hp X gm Fe, FeO, Fe2O3 cn 2,24 lt CO ( ktc).Khi lng st thu c l


04. t chy hn hp hirocacbon X thu c 2,24 lt CO2 (ktc) v 2,7 gam H2O. Thtch O2 tham gia phn ng chy (ktc) l

A. 5,6 lt. B. 2,8 lt. C. 4,48 lt. D. 3,92 lt.

05. Ho tan hon ton a gam hn hp X gm Fe v Fe2O3 trong dung dch HCl thu c2,24 lt kh H2 ktc v dung dch B. Cho dung dch B tc dng dung dch NaOH d,lc ly kt ta, nung trong khng kh n khi lng khng i thu c 24 gam chtrn. Gi tr ca a l


06. Hn hp X gm Mg v Al2O3. Cho 3 gam X tc dng vi dung dch HCl d giiphng V lt kh (ktc). Dung dch thu c cho tc dng vi dung dch NH 3 d, lcv nung kt ta c 4,12 gam bt oxit. V c gi tr l:

A. 1,12 lt. B. 1,344 lt. C. 1,568 lt. D. 2,016 lt.

07. Hn hp A gm Mg, Al, Fe, Zn. Cho 2 gam A tc dng vi dung dch HCl d giiphng 0,1 gam kh. Cho 2 gam A tc dng vi kh clo d thu c 5,763 gam hnhp mui. Phn trm khi lng ca Fe trong A l

A. 8,4%. B. 16,8%. C. 19,2%. D. 22,4%.

08. (Cu 2 - M 231 - TSC - Khi A 2007)


16/91

t chy hon ton mt th tch kh thin nhin gm metan, etan, propan bng oxikhng kh (trong khng kh Oxi chim 20% th tch), thu c 7,84 lt kh CO2 (ktc)v 9,9 gam H2O. Th tch khng kh (ktc) nh nht cn dng t chy honton lng kh thin nhin trn l

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

09. Ho tan hon ton 5 gam hn hp 2 kim loi X v Y bng dung dch HCl thu c dungdch A v kh H2. C cn dung dch A thu c 5,71 gam mui khan. Hy tnh th tchkh H2 thu c ktc.

A. 0,56 lt. B. 0,112 lt. C. 0,224 lt D. 0,448 lt

10. t chy hon ton m gam hn hp Y gm C2H6, C3H4 v C4H8 th thu c 12,98gam CO2 v 5,76 gam H2O. Vy m c gi tr l



1. D 2. C 3. C 4. D 5. C

6. C 7. B 8. A 9. C 10. C

Phng php 3

BO TON MOL ELECTRON

Trc ht cn nhn mnh y khng phi l phng php cn bng phn ng oxiha - kh, mc d phng php thng bng electron dng cn bng phn ng oxi ha -kh cng da trn s bo ton electron.

Nguyn tc ca phng php nh sau: khi c nhiu cht oxi ha, cht kh trong mthn hp phn ng (nhiu phn ng hoc phn ng qua nhiu giai on) th tng selectron ca cc cht kh cho phi bng tng s electron m cc cht oxi ha nhn. Tach cn nhn nh ng trng thi u v trng thi cui ca cc cht oxi ha hoc chtkh, thm ch khng cn quan tm n vic cn bng cc phng trnh phn ng. Phngphp ny c bit l th i vi cc bi ton cn phi bin lun nhiu trng hp c thxy ra.


17/91

Sau y l mt s v d in hnh.

V d 1: Oxi ha hon ton 0,728 gam bt Fe ta thu c 1,016 gam hn hp hai oxit st(hn hp A).

1. Ha tan hn hp A bng dung dch axit nitric long d. Tnh th tch kh NO duy

nht bay ra ( ktc).A. 2,24 ml. B. 22,4 ml. C. 33,6 ml. D. 44,8 ml.

2. Cng hn hp A trn trn vi 5,4 gam bt Al ri tin hnh phn ng nhit nhm(hiu sut 100%). Ha tan hn hp thu c sau phn ng bng dung dch HCl d.Tnh th tch bay ra ( ktc).

A. 6,608 lt. B. 0,6608 lt. C. 3,304 lt. D. 33,04. lt

Hng dn gii

1. Cc phn ng c th c:

2Fe + O2o

t 2FeO (1)2Fe + 1,5O2

ot Fe2O3 (2)

3Fe + 2O2ot Fe3O4 (3)

Cc phn ng ha tan c th c:

3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O (4)

Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O (5)

3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O (6)

Ta nhn thy tt c Fe t Fe0

b oxi ha thnh Fe+3

, cn N+5

b kh thnh N+2

, O20

bkh thnh 2O2 nn phng trnh bo ton electron l:

0,7283n 0,009 4 3 0,039

56+ = = mol.

trong , n l s mol NO thot ra. Ta d dng rt ra

n = 0,001 mol;

VNO = 0,001 22,4 = 0,0224 lt = 22,4 ml. (p n B)

2. Cc phn ng c th c:

2Al + 3FeOot 3Fe + Al2O3 (7)

2Al + Fe2O3ot 2Fe + Al2O3 (8)

8Al + 3Fe3O4o

t 9Fe + 4Al2O3 (9)

Fe + 2HCl FeCl2 + H2 (10)

2Al + 6HCl 2AlCl3 + 3H2 (11)


18/91

Xt cc phn ng (1, 2, 3, 7, 8, 9, 10, 11) ta thy Fe0 cui cng thnh Fe+2, Al0 thnh

Al+3, O20 thnh 2O2 v 2H+ thnh H2 nn ta c phng trnh bo ton electron nh sau:

5,4 30,013 2 0,009 4 n 2

27

+ = +

Fe0Fe+2 Al0Al+3 O202O2 2H+H2

n = 0,295 mol

2H

V 0,295 22,4 6,608= = lt. (p n A)

Nhn xt: Trong bi ton trn cc bn khng cn phi bn khon l to thnh haioxit st (hn hp A) gm nhng oxit no v cng khng cn phi cn bng 11 phngtrnh nh trn m ch cn quan tm ti trng thi u v trng thi cui ca cc cht oxiha v cht kh ri p dng lut bo ton electron tnh lc bt c cc giai ontrung gian ta s tnh nhm nhanh c bi ton.

V d 2:Trn 0,81 gam bt nhm vi bt Fe2O3 v CuO ri t nng tin hnh phnng nhit nhm thu c hn hp A. Ho tan hon ton A trong dung dchHNO3 un nng thu c V lt kh NO (sn phm kh duy nht) ktc. Gi trca V l

A. 0,224 lt. B. 0,672 lt. C. 2,24 lt. D. 6,72 lt.

Hng dn gii

Tm tt theo s :

o2 3 t

NO

Fe O

0,81 gam Al V ?CuO 3ha tan hon ton

dungdch HNOhn h p A

+ =Thc cht trong bi ton ny ch c qu trnh cho v nhn electron ca nguyn t Al

v N.

Al Al+3 + 3e

0,81

27 0,09 mol

v N+5 + 3e N+2

0,09 mol 0,03 mol

VNO = 0,03 22,4 = 0,672 lt. (p n D)

Nhn xt: Phn ng nhit nhm cha bit l hon ton hay khng hon ton do hn hp A khng xc nh c chnh xc gm nhng cht no nn vic vit phngtrnh ha hc v cn bng phng trnh phc tp. Khi ha tan hon ton hn hp A trongaxit HNO3 th Al0 to thnh Al+3, nguyn t Fe v Cu c bo ton ha tr.


19/91

C bn s thc mc lng kh NO cn c to bi kim loi Fe v Cu trong hn hpA. Thc cht lng Al phn ng b li lng Fe v Cu to thnh.

V d 3: Cho 8,3 gam hn hp X gm Al, Fe (nAl = nFe) vo 100 ml dung dch Y gmCu(NO3)2 v AgNO3. Sau khi phn ng kt thc thu c cht rn A gm 3 kim

loi. Ha tan hon ton cht rn A vo dung dch HCl d thy c 1,12 lt khthot ra (ktc) v cn li 28 gam cht rn khng tan B. Nng C M caCu(NO3)2 v ca AgNO3 ln lt l

A. 2M v 1M. B. 1M v 2M.

C. 0,2M v 0,1M. D. kt qu khc.

Tm tt s :

Al Fe

8,3 gam hn h p X

(n =n )

Al

Fe

+ 100 ml dung dch Y 3

3 2

AgNO : x mol

Cu(NO ) :y mol

Cht rn A

(3 kim loi)

2

HCl d 1,12 lt H

2,8 gamcht rn khng tan B

+Z]

Hng dn gii

Ta c: nAl = nFe =8,3

0,1mol.83

=

t3AgNO

n x mol= v3 2Cu(NO )

n y mol=

X + Y Cht rn A gm 3 kim loi.

Al ht, Fe cha phn ng hoc cn d. Hn hp hai mui ht.

Qu trnh oxi ha:

Al Al3+ + 3e Fe Fe2+ + 2e

0,1 0,3 0,1 0,2

Tng s mol e nhng bng 0,5 mol.

Qu trnh kh:

Ag+ + 1e Ag Cu2+ + 2e Cu 2H+ + 2e H2

x x x y 2y y 0,1 0,05Tng s e mol nhn bng (x + 2y + 0,1).

Theo nh lut bo ton electron, ta c phng trnh:

x + 2y + 0,1 = 0,5 hay x + 2y = 0,4 (1)

Mt khc, cht rn B khng tan l: Ag: x mol ; Cu: y mol.

108x + 64y = 28 (2)


20/91

Gii h (1), (2) ta c:

x = 0,2 mol ; y = 0,1 mol.

3M AgNO

0,2C

0,1= = 2M; 3 2MCu(NO )

0,1C

0,1= = 1M. (p n B)

V d 4: Ha tan 15 gam hn hp X gm hai kim loi Mg v Al vo dung dch Y gmHNO3 v H2SO4 c thu c 0,1 mol mi kh SO2, NO, NO2, N2O. Phn trmkhi 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%.

Hng dn gii

t nMg = x mol ; nAl = y mol. Ta c:

24x + 27y = 15.(1)

Qu trnh oxi ha:

Mg Mg2+ + 2e Al Al3+ + 3e

x 2x y 3y

Tng s mol e nhng bng (2x + 3y).

Qu trnh kh:

N+5 + 3e N+2 2N+5 + 24e 2N+1

0,3 0,1 0,8 0,2

N+5 + 1e N+4 S+6 + 2e S+4

0,1 0,1 0,2 0,1

Tng s mol e nhn bng 1,4 mol.

Theo nh lut bo ton electron:

2x + 3y = 1,4 (2)

Gii h (1), (2) ta c: x = 0,4 mol ; y = 0,2 mol.

27 0,2

%Al 100% 36%.15

= =

%Mg = 100% 36% = 64%. (p n B)

V d 5: Trn 60 gam bt Fe vi 30 gam bt lu hunh ri un nng (khng c khngkh) thu c cht rn A. Ho tan A bng dung dch axit HCl d c dungdch B v kh C. t chy C cn V lt O2 (ktc). Bit cc phn ng xy ra honton. V c gi tr l


21/91

A. 11,2 lt. B. 21 lt. C. 33 lt. D. 49 lt.

Hng dn gii

V Fe S30

n n32

> = nn Fe d v S ht.

Kh C l hn hp H2S v H2. t C thu c SO2 v H2O. Kt qu cui cng ca qutrnh phn ng l Fe v S nhng e, cn O2 thu e.

Nhng e: Fe Fe2+ + 2e

60

mol56

60

256

mol

S S+4 + 4e

30mol

32

304

32 mol

Thu e: Gi s mol O2 l x mol.O2 + 4e 2O-2

x mol 4x

Ta c:60 30

4x 2 456 32

= + gii ra x = 1,4732 mol.

2O

V 22,4 1,4732 33= = lt. (p n C)

V d 6: Hn hp A gm 2 kim loi R1, R2 c ho tr x, y khng i (R1, R2 khng tcdng vi nc v ng trc Cu trong dy hot ng ha hc ca kim loi).Cho hn hp A phn ng hon ton vi dung dch HNO 3 d thu c 1,12 ltkh NO duy nht ktc.

Nu cho lng hn hp A trn phn ng hon ton vi dung dch HNO3 th thuc bao nhiu lt N2. Cc th tch kh o ktc.

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

Hng dn gii

Trong bi ton ny c 2 th nghim:

TN1: R1 v R2 nhng e cho Cu2+ chuyn thnh Cu sau Cu li nhng e cho

5N+ thnh 2N+ (NO). S mol e do R1 v R2 nhng ra l5

N+

+ 3e 2

N+

0,15 05,04,22

12,1=


22/91

TN2: R1 v R2 trc tip nhng e cho5

N+

to ra N2. Gi x l s mol N2, th s mol

e thu vo l

25

N+

+ 10e 02N

10x x molTa c: 10x = 0,15 x = 0,015

2NV = 22,4.0,015 = 0,336 lt. (p n B)

V d 7: Cho 1,35 gam hn hp gm Cu, Mg, Al tc dng ht vi dung dch HNO 3 thuc hn hp kh gm 0,01 mol NO v 0,04 mol NO 2. Tnh khi lng muito ra trong dung dch.


Hng dn gii

Cch 1: t x, y, z ln lt l s mol Cu, Mg, Al.

Nhng e: Cu =2

Cu+

+ 2e Mg =2

Mg+

+ 2e Al =3

Al+

+ 3e

x x 2x y y 2y z z 3z

Thu e:5

N+

+ 3e =2

N+

(NO)5

N+

+ 1e =4

N+

(NO2)

0,03 0,01 0,04 0,04

Ta c: 2x + 2y + 3z = 0,03 + 0,04 = 0,07

v 0,07 cng chnh l s mol NO3

Khi lng mui nitrat l:

1,35 + 62 0,07 = 5,69 gam. (p n C)

Cch 2:

Nhn nh mi: Khi cho kim loi hoc hn hp kim loi tc dng vi dung dch axitHNO3 to hn hp 2 kh NO v NO2 th

3 2HNO NO NOn 2n 4n= +

3HNOn 2 0,04 4 0,01 0,12= + = mol

2H On 0,06= molp dng nh lut bo ton khi lng:

3 2 2KL HNO mui NO NO H Om m m m m m+ = + + +

1,35 + 0,12 63 = mmui + 0,01 30 + 0,04 46 + 0,06 18

mmui = 5,69 gam.


23/91

V d 8: (Cu 19 - M 182 - Khi A - TSH - 2007)

Ha tan hon ton 12 gam hn hp Fe, Cu (t l mol 1:1) bng axit HNO3, thuc V lt ( ktc) hn hp kh X (gm NO v NO2) v dung dch Y (ch chahai 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. 5,60 lt. D. 3,36 lt.Hng dn gii

t nFe = nCu = a mol 56a + 64a = 12 a = 0,1 mol.

Cho e: Fe Fe3+ + 3e Cu Cu2+ + 2e

0,1 0,3 0,1 0,2

Nhn e: N+5 + 3e N+2 N+5 + 1e N+4

3x x y y

Tng ne cho bng tng ne nhn.

3x + y = 0,5

Mt khc: 30x + 46y = 19 2(x + y).

x = 0,125 ; y = 0,125.

Vhh kh (ktc) = 0,125 2 22,4 = 5,6 lt. (p n C)

V d 9: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan hthn hp X trong dung dch HNO3 (d), thot ra 0,56 lt ( ktc) NO (l snphm kh duy nht). Gi tr ca m l


Hng dn gii

m gam Fe + O2 3 gam hn hp cht rn X 3HNO d 0,56 lt NO.

Thc cht cc qu trnh oxi ha - kh trn l:

Cho e: Fe Fe3+ + 3e

m

56

3m

56mol e

Nhn e: O2 + 4e 2O2 N+5 + 3e N+2

3 m32 4(3 m)

32 mol e 0,075 mol 0,025 mol

3m

56=

4(3 m)

32

+ 0,075

m = 2,52 gam. (p n A)


24/91

V d 10: Hn hp X gm hai kim loi A v B ng trc H trong dy in ha v cha tr khng i trong cc hp cht. Chia m gam X thnh hai phn bng nhau:

- Phn 1: Ha tan hon ton trong dung dch cha axit HCl v H 2SO4 long to ra3,36 lt kh H2.

- Phn 2: Tc dng hon ton vi dung dch HNO3 thu c V lt kh NO (snphm kh duy nht).

Bit cc th tch kh o 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.

Hng dn gii

t hai kim loi A, B l M.

- Phn 1: M + nH+ Mn+ + 2n

H2

(1)

- Phn 2: 3M + 4nH+

+ nNO3

3Mn+

+ nNO + 2nH2O (2)Theo (1): S mol e ca M cho bng s mol e ca 2H+ nhn;

Theo (2): S mol e ca M cho bng s mol e ca N+5 nhn.

Vy s mol e nhn ca 2H+ bng s mol e nhn ca N+5.

2H+ + 2e H2 v N+5 + 3e N+2

0,3 0,15 mol 0,3 0,1 mol

VNO = 0,1 22,4 = 2,24 lt. (p n A)

V d 11: Cho m gam bt Fe vo dung 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 O 2 bng 1,3125. Xc nh%NO v %NO2 theo th tch trong hn hp X v khi lng m ca Fe dng?

A. 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.

Hng dn gii

Ta c: nX = 0,4 mol; MX = 42.

S ng cho:

2

2

NO NO

NO NO

n : n 12 : 4 3

n n 0,4 mol

= = + =

2NO : 46 42 30 12

42

NO : 30 46 42 4

=

=


25/91

2

NO

NO

n 0,1 mol

n 0,3 mol

= =

2

NO

NO

%V 25%

%V 75%

= =

v Fe 3e Fe3+ N+5 + 3e N+2 N+5 + 1e N+4

3x x 0,3 0,1 0,3 0,3

Theo nh lut bo ton electron:

3x = 0,6 mol x = 0,2 mol

mFe = 0,2 56 = 11,2 gam. (p p B).

V d 12: Cho 3 kim loi Al, Fe, Cu vo 2 lt dung dch HNO 3 phn ng va thu c1,792 lt kh X (ktc) gm N2 v NO2 c t khi hi so vi He bng 9,25. Nng mol/lt HNO3 trong dung dch u l

A. 0,28M. B. 1,4M. C. 1,7M. D. 1,2M.

Hng dn gii

Ta c: ( )2 2N NOX

M MM 9,25 4 37

2

+= = =

l trung bnh cng khi lng phn t ca hai kh N2 v NO2 nn:

2 2

XN NO

nn n 0,04 mol

2= = =

v NO3 + 10e N2 NO3 + 1e NO2

0,08 0,4 0,04 mol 0,04 0,04 0,04 mol

M Mn+ + n.e0,04 mol

3HNO (bkh)n 0,12 mol.=

Nhn nh mi: Kim loi nhng bao nhiu electron th cng nhn by nhiu gc

NO3 to mui.

3HNO ( ) ( ) ( )

n n.e n.e 0,04 0,4 0,44 mol.to mui nh ng nhn= = = + =

Do :3HNO ( )

n 0,44 0,12 0,56 molphn ng = + =

[ ]30,56HNO 0,28M.

2= = (p n A)

V d 13: Khi cho 9,6 gam Mg tc dng ht vi dung dch H 2SO4 m c, thy c 49gam H2SO4 tham gia phn ng, to mui MgSO4, H2O v sn phm kh X. Xl

A. SO2 B. S C. H2S D. SO2, H2S


26/91

Hng dn gii

Dung dch H2SO4 m c va l cht oxi ha va l mi trng.

Gi a l s oxi ha ca S trong X.

Mg Mg2+ + 2e S+6 + (6-a)e S a

0,4 mol 0,8 mol 0,1 mol 0,1(6-a) mol

Tng s mol H2SO4 dng l :49

0,598

= (mol)

S mol H2SO4 dng to mui bng s mol Mg = 9,6 : 24 = 0,4 mol.

S mol H2SO4 dng oxi ha Mg l:

0,5 0,4 = 0,1 mol.

Ta c: 0,1 (6 a) = 0,8 x = 2. Vy X l H2S. (p n C)

V d 14: a gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hpA c khi lng l 75,2 gam gm Fe, FeO, Fe 2O3 v Fe3O4. Cho hn hp Aphn ng ht vi dung dch H2SO4 m c, nng thu c 6,72 lt kh SO2(ktc). Khi lng a gam l:

A. 56 gam. B. 11,2 gam. C. 22,4 gam. D. 25,3 gam.

Hng dn gii

S mol Fe ban u trong a gam: Fea

n56

= mol.

S mol O2 tham gia phn ng: 2O75,2 a

n 32

= mol.

Qu trnh oxi ha:

3Fe Fe 3ea 3a

mol mol56 56

+ +(1)

S mol e nhng: e3a

n mol56

=

Qu trnh kh: O2 + 4e 2O2 (2)

SO42 + 4H+ + 2e SO2 + 2H2O (3)

T (2), (3) cho 2 2e O SOn 4n 2n= +

75,2 a 3a

4 2 0,332 56

= + =

a = 56 gam. (p n A)


27/91

V d 15:Cho 1,35 gam hn hp A gm Cu, Mg, Al tc dng vi HNO3 d c 1,12 ltNO v NO2 (ktc) c khi lng mol trung bnh l 42,8. Tng khi lng muinitrat sinh ra l:

A. 9,65 gam B. 7,28 gam C. 4,24 gam D. 5,69 gam

Hng dn giiDa vo s ng cho tnh c s mol NO v NO2 ln lt l 0,01 v 0,04

mol. Ta c cc bn phn ng:

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

NO3 + 2H+ + 1e NO2 + H2O

Nh vy, tng electron nhn l 0,07 mol.

Gi x, y, z ln lt l s mol Cu, Mg, Al c trong 1,35 gam hn hp kim loi. Ta ccc bn phn ng:

Cu Cu2+

+ 2e Mg Mg2+

+ 2e Al Al3+

+ 3e 2x + 2y + 3z = 0,07.

Khi lng mui nitrat sinh ra l:

m =3 2Cu(NO )

m +3 2Mg(NO )

m +3 3Al(NO )

m

= 1,35 + 62(2x + 2y + 3z)

= 1,35 + 62 0,07 = 5,69 gam.

MT S BI TP VN DNG GIAI THEO PHNG PHP BO

TOM MOL ELECTRON

01. Ho tan hon ton m gam Al vo dung dch HNO3 rt long th thu c hn hpgm 0,015 mol kh N2O v 0,01mol kh NO (phn ng khng to NH4NO3). Gi trca m l


02. Cho mt lung CO i qua ng s ng 0,04 mol hn hp A gm FeO v Fe 2O3 tnng. Sau khi kt thc th nghim thu c cht rn B gm 4 cht nng 4,784 gam.Kh i 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 FeO trong hn hp A lA. 68,03%. B. 13,03%. C. 31,03%. D. 68,97%.

03. Mt hn hp gm hai bt kim loi Mg v Al c chia thnh hai phn bng nhau:- Phn 1: cho tc dng vi HCl d thu c 3,36 lt H2.- Phn 2: ho tan ht trong HNO3 long d thu c V lt mt kh khng mu, honu trong khng kh (cc th tch kh u o ktc). Gi tr ca V l

A. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 5,6 lt.


28/91

04. Dung dch X gm AgNO3 v Cu(NO3)2 c cng nng . Ly mt lng hn hp gm0,03 mol Al; 0,05 mol Fe cho vo 100 ml dung dch X cho ti kh phn ng kt thcthu c cht rn Y cha 3 kim loi.Cho Y vo HCl d gii phng 0,07 gam kh. Nng ca hai mui l

A. 0,3M. B. 0,4M. C. 0,42M. D. 0,45M.

05. Cho 1,35 gam hn hp Cu, Mg, Al tc dng vi HNO3 d c 896 ml hn hp gm NOv NO2 c M 42= . Tnh tng khi lng mui nitrat sinh ra (kh ktc).


06. Ha tan ht 4,43 gam hn hp Al v Mg trong HNO 3 long thu c dung dch A v1,568 lt (ktc) hn hp hai kh (u khng mu) c khi lng 2,59 gam trong cmt kh b ha thnh mu nu trong khng kh. Tnh s mol HNO3 phn ng.

A. 0,51 mol. B. A. 0,45 mol. C. 0,55 mol. D. 0,49 mol.

07. Ha tan hon ton m gam hn hp gm ba kim loi bng dung dch HNO 3 thu c1,12 lt hn hp kh D (ktc) gm NO2 v NO. T khi hi ca D so vi hiro bng18,2. Tnh th tch ti thiu dung dch HNO3 37,8% (d = 1,242g/ml) cn dng.

A. 20,18 ml. B. 11,12 ml. C. 21,47 ml. D. 36,7 ml.

08. Ha tan 6,25 gam hn hp Zn v Al vo 275 ml dung dch HNO 3 thu c dung dchA, cht rn B gm cc kim loi cha tan ht cn nng 2,516 gam v 1,12 lt hn hp khD ( ktc) gm NO v NO2. T khi ca hn hp D so vi H2 l 16,75. Tnh nng mol/l ca HNO3 v tnh khi lng mui khan thu c khi c cn dung dch sau phnng.

A. 0,65M v 11,794 gam. B. 0,65M v 12,35 gam.

C. 0,75M v 11,794 gam. D. 0,55M v 12.35 gam.

09. t chy 5,6 gam bt Fe trong bnh ng O 2 thu c 7,36 gam hn hp A gmFe2O3, Fe3O4 v Fe. Ha tan hon ton lng hn hp A bng dung dch HNO3 thuc V lt hn hp kh B gm NO v NO 2. T khi ca B so vi H2 bng 19. Th tchV ktc l

A. 672 ml. B. 336 ml. C. 448 ml. D. 896 ml.

10. Cho a gam hn hp A gm oxit FeO, CuO, Fe2O3 c s mol bng nhau tc dng honton vi lng va l 250 ml dung dch HNO 3 khi un nng nh, thu c dung dchB v 3,136 lt (ktc) hn hp kh C gm NO2 v NO c t khi so vi hiro l 20,143.Tnh a.

A. 74,88 gam. B. 52,35 gam. C. 61,79 gam. D. 72,35 gam.p n cc bi tp vn dng

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

6. D 7. C 8. A 9. D 10. A

Phng php 4


29/91

S DNG PHNG TRNH ION - ELETRON

lm tt cc bi ton bng phng php ion iu u tin cc bn phi nm chcphng trnh phn ng di dng cc phn t t suy ra cc phng trnh ion, i khi

c mt s bi tp khng th gii theo cc phng trnh phn t c m phi gii datheo phng trnh ion. Vic gii bi ton ha hc bng phng php ion gip chng tahiu k hn v bn cht ca cc phng trnh ha hc. T mt phng trnh ion c thng vi rt nhiu phng trnh phn t. V d phn ng gia hn hp dung dch axit vidung dch baz u c chung mt phng trnh ion l

H+ + OH H2O

hoc phn ng ca Cu kim loi vi hn hp dung dch NaNO3 v dung dch H2SO4 l

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O...

Sau y l mt s v d:

V d 1: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tanht vo dung dch Y gm (HCl v H2SO4 long) d thu c dung dch Z. Nht t dung dch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot kh NO.Th tch dung dch Cu(NO3)2 cn dng v th tch kh thot ra ktc thucphng n no?

A. 25 ml; 1,12 lt. B. 0,5 lt; 22,4 lt.

C. 50 ml; 2,24 lt. D. 50 ml; 1,12 lt.

Hng dn gii

Quy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4.Hn hp X gm: (Fe3O4 0,2 mol; Fe 0,1 mol) tc dng vi dung dch Y

Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O

0,2 0,2 0,4 mol

Fe + 2H+ Fe2+ + H2

0,1 0,1 mol

Dung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2:

3Fe2+ + NO3 + 4H+ 3Fe3+ + NO + 2H2O

0,3 0,1 0,1 mol VNO = 0,1 22,4 = 2,24 lt.

3 2 3Cu(NO ) NO

1n n 0,05

2= = mol

3 2dd Cu(NO )

0,05V 0,05

1= = lt (hay 50 ml). (p n C)


30/91

V d 2: Ha tan 0,1 mol Cu kim loi trong 120 ml dung dch X gm HNO3 1M vH2SO4 0,5M. Sau khi phn ng kt thc thu c V lt kh NO duy nht (ktc).

Gi tr ca V l

A. 1,344 lt. B. 1,49 lt. C. 0,672 lt. D. 1,12 lt.

Hng dn gii

3HNOn 0,12= mol ;

2 4H SOn 0,06= mol

Tng: Hn 0,24+ = mol v 3NOn 0,12 = mol.

Phng trnh ion:

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Ban u: 0,1 0,24 0,12 mol

Phn ng: 0,09 0,24 0,06 0,06 mol

Sau phn ng: 0,01 (d) (ht) 0,06 (d) VNO = 0,06 22,4 = 1,344 lt. (p n A)

V d 3: Dung dch X cha dung dch NaOH 0,2M v dung dch Ca(OH) 2 0,1M. Sc7,84 lt kh CO2 (ktc) vo 1 lt dung dch X th lng kt ta thu c l


Hng dn gii

2COn = 0,35 mol ; nNaOH = 0,2 mol; 2Ca(OH)n = 0,1 mol.

Tng: OHn = 0,2 + 0,1 2 = 0,4 mol v2Can + = 0,1 mol.

Phng trnh ion rt gn:

CO2 + 2OH CO32 + H2O

0,35 0,4

0,2 0,4 0,2 mol

2CO ( )

n d = 0,35 0,2 = 0,15 mol

tip tc xy ra phn ng:

CO32 + CO2 + H2O 2HCO3

Ban u: 0,2 0,15 mol

Phn ng: 0,15 0,15 mol

23CO

n cn li bng 0,15 mol

3

CaCOn = 0,05 mol


31/91

3CaCO

m = 0,05 100 = 5 gam. (p n B)

V d 4: Ha tan ht hn hp gm mt kim loi kim v mt kim loi kim th trong ncc dung dch A v c 1,12 lt H2 bay ra ( ktc). Cho dung dch cha 0,03 molAlCl3 vo dung dch A. khi lng kt ta thu c l


Hng dn gii

Phn ng ca kim loi kim v kim loi kim th vi H2O:

M + nH2O M(OH)n + 2n

H2

T phng trnh ta c:

2HOHn 2n = = 0,1mol.

Dung dch A tc dng vi 0,03 mol dung dch AlCl3:Al3+ + 3OH Al(OH)3

Ban u: 0,03 0,1 mol

Phn ng: 0,03 0,09 0,03 mol

OH ( )n d = 0,01mol

tip tc ha tan kt ta theo phng trnh:

Al(OH)3 + OH AlO2 + 2H2O

0,01 0,01 mol

Vy:3Al(OH)

m = 78 0,02 = 1,56 gam. (p n B)

V d 5: Dung dch A cha 0,01 mol Fe(NO3)3 v 0,15 mol HCl c kh nng ha tan ti abao nhiu gam Cu kim loi? (Bit NO l sn phm kh duy nht)


Hng dn gii

Phng trnh ion:

Cu + 2Fe3+ 2Fe2+ + Cu2+

0,005 0,01 mol3Cu + 8H+ + 2NO3

3Cu2+ + 2NO + 4H2O

Ban u: 0,15 0,03 mol H+ d

Phn ng: 0,045 0,12 0,03 mol

mCu ti a = (0,045 + 0,005) 64 = 3,2 gam. (p n C)


32/91

V d 6: Cho hn hp gm NaCl v NaBr tc dng vi dung dch AgNO3 d thu ckt ta c khi lng ng bng khi lng AgNO3 phn ng. Tnh phntrm khi lng NaCl trong hn hp u.

A. 23,3% B. 27,84%. C. 43,23%. D. 31,3%.

Hng dn giiPhng trnh ion:

Ag+ + Cl AgCl

Ag+ + Br AgBrt: nNaCl = x mol ; nNaBr = y mol

mAgCl + mAgBr = 3( )AgNOm p.

3Cl Br NO

m m m + =

35,5x + 80y = 62(x + y) x : y = 36 : 53

Chn x = 36, y = 53 NaCl58,5 36 100

%m58,5 36 103 53

=

+ = 27,84%. (p n B)

V d 7: Trn 100 ml dung dch A (gm KHCO3 1M v K2CO3 1M) vo 100 ml dungdch B (gm NaHCO3 1M v Na2CO3 1M) thu c dung dch C.

Nh t t 100 ml dung dch D (gm H2SO4 1M v HCl 1M) vo dung dch Cthu c V lt CO2 (ktc) v dung dch E. Cho dung dch Ba(OH)2 ti d vo

dung dch E th thu c m gam kt ta. Gi tr ca m v V ln lt lA. 82,4 gam v 2,24 lt. B. 4,3 gam v 1,12 lt.

C. 43 gam v 2,24 lt. D. 3,4 gam v 5,6 lt.

Hng dn gii

Dung dch C cha: HCO3 : 0,2 mol ; CO32

: 0,2 mol.

Dung dch D c tng: Hn + = 0,3 mol.

Nh t t dung dch C v dung dch D:

CO32 + H+ HCO3

0,2 0,2 0,2 mol

HCO3 + H+ H2O + CO2

Ban u: 0,4 0,1 mol

Phn ng: 0,1 0,1 0,1 mol

D: 0,3 mol


33/91

Tip tc cho dung dch Ba(OH)2 d vo dung dch E:

Ba2+ + HCO3 + OH BaCO3 + H2O

0,3 0,3 mol

Ba2+ + SO42 BaSO4

0,1 0,1 mol

2CO

V = 0,1 22,4 = 2,24 lt.

Tng khi lng kt ta:

m = 0,3 197 + 0,1 233 = 82,4 gam. (p n A)

V d 8: Ha tan hon ton 7,74 gam mt hn hp gm Mg, Al bng 500 ml dung dchgm H2SO4 0,28M v HCl 1M thu c 8,736 lt H2 (ktc) v dung dch X.

Thm V lt dung dch cha ng thi NaOH 1M v Ba(OH) 2 0,5M vo dungdch X thu c lng kt ta ln nht.

a) S gam mui thu c trong dung dch X l

A. 38,93 gam. B. 38,95 gam.

C. 38,97 gam. D. 38,91 gam.

b) Th tch V l

A. 0,39 lt. B. 0,4 lt.

C. 0,41 lt. D. 0,42 lt.

c) Lng kt ta l

A. 54,02 gam. B. 53,98 gam.C. 53,62 gam. D. 53,94 gam.

Hng dn gii

a) Xc nh khi lng mui thu c trong dung dch X:

2 4H SOn = 0,28 0,5 = 0,14 mol

24SO

n = 0,14 mol v Hn + = 0,28 mol.

nHCl = 0,5 mol

Hn + = 0,5 mol v Cln = 0,5 mol.Vy tng Hn + = 0,28 + 0,5 = 0,78 mol.

M2H

n = 0,39 mol. Theo phng trnh ion rt gn:

Mg0 + 2H+ Mg2+ + H2(1)


34/91

Al + 3H+ Al3+ +32

H2(2)

Ta thy 2HH (p-)n 2n+ = H+ ht.

mhh mui = mhh k.loi +2

4SO Cl

m m +

= 7,74 + 0,14 96 + 0,5 35,5 = 38,93gam. (p n A)

b) Xc nh th tch V:

2

NaOH

Ba(OH)

n 1V mol

n 0,5V mol

= =

Tng OHn = 2V mol v 2Ban + = 0,5V mol.

Phng trnh to kt ta:

Ba

2+

+ SO42

BaSO4 (3)0,5V mol 0,14 mol

Mg2+ + 2OH Mg(OH)2 (4)

Al3+ + 3OH Al(OH)3 (5)

kt ta t ln nht th s mol OH kt ta ht cc ion Mg2+ v Al3+. Theo ccphng trnh phn ng (1), (2), (4), (5) ta c:

Hn + = OHn = 0,78 mol

2V = 0,78 V = 0,39 lt. (p n A)

c) Xc nh lng kt ta:

2Ban + = 0,5V = 0,5 0,39 = 0,195 mol > 0,14 mol Ba2+ d.

4BaSO

m = 0,14 233 = 32,62 gam.

Vy mkt ta = 4BaSOm + m 2 k.loi + OHm

= 32,62 + 7,74 + 0,78 17 = 53,62 gam. (p n C)

V d 9: (Cu 40 - M 182 - TS i Hc - Khi A 2007)

Cho m gam hn hp Mg, Al vo 250 ml dung dch X cha hn hp axit HCl

1M v axit H2SO4 0,5M, thu c 5,32 lt H2 ( ktc) v dung dch Y (coi thtch dung dch khng i). Dung dch Y c pH l

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

Hng dn gii

nHCl = 0,25 mol ; 2 4H SOn = 0,125.


35/91

Tng: Hn + = 0,5 mol ;

2H ( )n tothnh = 0,2375 mol.

Bit rng: c 2 mol ion H+ 1 mol H2

vy 0,475 mol H+

0,2375 mol H2

H ( )n d+ = 0,5 0,475 = 0,025 mol

0,025

H0,25

+ = = 0,1 = 101M pH = 1. (p n A)

V d 10: (Cu 40 - M 285 - Khi B - TSH 2007)

Thc hin hai th nghim:

1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO 3 1M thot ra V1 ltNO.

2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO40,5 M thot ra V2 lt NO.

Bit NO l sn phm kh duy nht, cc th tch kh o cng iu kin. Quanh gia V1 v V2 l

A. V2 = V1. B. V2 = 2V1. C. V2 = 2,5V1. D. V2 = 1,5V1.

Hng dn gii

TN1:

3

Cu

HNO

3,84n 0,06 mol

64

n 0,08 mol

= = =

3

H

NO

n 0,08 mol

n 0,08 mol

+

=

=

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Ban u: 0,06 0,08 0,08 mol H+ phn ng ht

Phn ng: 0,03 0,08 0,02 0,02 mol

V1 tng ng vi 0,02 mol NO.

TN2: nCu = 0,06 mol ; 3HNOn = 0,08 mol ; 2 4H SOn = 0,04 mol.

Tng: Hn + = 0,16 mol ;

3NOn = 0,08 mol.

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Ban u: 0,06 0,16 0,08 mol Cu v H+ phn ng ht

Phn ng: 0,06 0,16 0,04 0,04 mol

V2 tng ng vi 0,04 mol NO.


36/91

Nh vy V2 = 2V1. (p n B)

V d 11: (Cu 33 - M 285 - Khi B - TSH 2007)

Trn 100 ml dung dch (gm Ba(OH)2 0,1M v NaOH 0,1M) vi 400 ml dungdch (gm H2SO4 0,0375M v HCl 0,0125M), thu c dung dch X. Gi tr pH

ca dung dch X lA. 7. B. 2. C. 1. D. 6.

Hng dn gii

2Ba(OH)

NaOH

n 0,01 mol

n 0,01 mol

=

= Tng OHn = 0,03 mol.

2 4H SO

HCl

n 0,015 mol

n 0,005 mol

=

= Tng Hn + = 0,035 mol.

Khi trn hn hp dung dch baz vi hn hp dung dch axit ta c phng trnh ionrt gn:

H+ + OH H2O

Bt u 0,035 0,03 mol

Phn ng: 0,03 0,03

Sau phn ng: H ( )n d+ = 0,035 0,03 = 0,005 mol.

Tng: Vdd (sau trn) = 500 ml (0,5 lt).

0,005H

0,5

+ = = 0,01 = 102 pH = 2. (p n B)

V d 12: (Cu 18 - M 231 - TS Cao ng - Khi A 2007)

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

A. 150 ml. B. 75 ml. C. 60 ml. D. 30 ml.

Hng dn gii

Na + H2O NaOH +1

2

H2

Ba + 2H2O Ba(OH)2 + H2

2Hn = 0,15 mol, theo phng trnh tng s 2 2HOH (d X)n 2n = = 0,3 mol.

Phng trnh ion rt gn ca dung dch axit vi dung dch baz l

H+ + OH H2O


37/91

Hn + = OHn = 0,3 mol 2 4H SOn = 0,15 mol

2 4H SO

0,15V

2= = 0,075 lt (75 ml). (p n B)

V d 13: Ha tan hn hp X gm hai kim loi A v B trong dung dch HNO 3 long. Ktthc phn ng thu c hn hp kh Y (gm 0,1 mol NO, 0,15 mol NO2 v0,05 mol N2O). Bit rng khng c phn ng to mui NH4NO3. S molHNO3 phn ng l:


Hng dn gii

Ta c bn phn ng:

NO3 + 2H+ + 1e NO2 + H2O (1)

2 0,15 0,15

NO3 + 4H+ + 3e NO + 2H2O (2)4 0,1 0,1

2NO3 + 10H+ + 8e N2O + 5H2O (3)

10 0,05 0,05

T (1), (2), (3) nhn c:

3HNO Hn n

p+= = 2 0,15 4 0,1 10 0,05 + + = 1,2 mol. (p n D)

V d 14: Cho 12,9 gam hn hp Al v Mg phn ng vi dung dch hn hp hai axit HNO3

v H2SO4 (c nng) thu c 0,1 mol mi kh SO 2, NO, NO2. C cn dungdch sau phn ng khi lng mui khan thu c l:


Hng dn gii

Ta c bn phn ng:

2NO3 + 2H+ + 1e NO2 + H2O + NO3 (1)

0,1 0,1

4NO3 + 4H+ + 3e NO + 2H2O + 3NO3 (2)

0,1 3 0,12SO42

+ 4H+ + 2e SO2 + H2O + SO42 (3)

0,1 0,1

T (1), (2), (3) s mol NO3 to mui bng 0,1 + 3 0,1 = 0,4 mol;

s mol SO42 to mui bng 0,1 mol.


38/91

mmui = mk.loi + 3NOm + 24SOm

= 12,9 + 62 0,4 + 96 0,1 = 47,3. (p n C)

V d 15: Ha tan 10,71 gam hn hp gm Al, Zn, Fe trong 4 lt dung dch HNO 3 aMva thu c dung dch A v 1,792 lt hn hp kh gm N 2 v N2O c t lmol 1:1. C cn dung dch A thu c m (gam.) mui khan. gi tr ca m, a l:

A. 55,35 gam. v 2,2M B. 55,35 gam. v 0,22M

C. 53,55 gam. v 2,2M D. 53,55 gam. v 0,22M

Hng dn gii

2 2N O N

1,792n n 0,04

2 22,4= = =

mol.

Ta c bn phn ng:

2NO3 + 12H+ + 10e N2 + 6H2O

0,08 0,48 0,04

2NO3 + 10H+ + 8e N2O + 5H2O

0,08 0,4 0,04

3HNO H

n n 0,88+= = mol.

0,88

a 0,224

= = M.

S mol NO3 to mui bng 0,88 (0,08 + 0,08) = 0,72 mol.

Khi lng mui bng 10,71 + 0,72 62 = 55,35 gam. (p n B)V d 16: Ha tan 5,95 gam hn hp Zn, Al c t l mol l 1:2 bng dung dch HNO 3

long d thu c 0,896 lt mt sn shm kh X duy nht cha nit. X l:

A. N2O B. N2 C. NO D. NH4+

Hng dn gii

Ta c: nZn = 0,05 mol; nAl = 0,1 mol.

Gi a l s mol ca NxOy, ta c:

Zn Zn2+ + 2e Al Al3+ + 3e

0,05 0,1 0,1 0,3xNO3

+ (6x 2y)H+ + (5x 2y)e NxOy + (3x 2y)H2O

0,04(5x 2y) 0,04

0,04(5x 2y) = 0,4 5x 2y = 10

Vy X l N2. (p n B)


39/91

V d 17: Cho hn hp gm 0,15 mol CuFeS2 v 0,09 mol Cu2FeS2 tc dng vi dungdch HNO3 d thu c dung dch X v hn hp kh Y gm NO v NO2. ThmBaCl2 d vo dung dch X thu c m gam kt ta. Mt khc, nu thmBa(OH)2 d vo dung dch X, ly kt ta nung trong khng kh n khilng khng i thu c a gam cht rn. Gi tr ca m v a l:

A. 111,84g v 157,44g B. 111,84g v 167,44g

C. 112,84g v 157,44g A. 112,84g v 167,44g

Hng dn gii

Ta c bn phn ng:

CuFeS2 + 8H2O 17e Cu2+ + Fe3+ + 2SO42 + 16+

0,15 0,15 0,15 0,3

Cu2FeS2 + 8H2O 19e 2Cu2+ + Fe3+ + 2SO42 + 16+

0,09 0,18 0,09 0,18

24SO

n 0,48 = mol;

Ba2+ + SO42 BaSO4

0,48 0,48

m = 0,48 233 = 111,84 gam.

nCu = 0,33 mol; nFe = 0,24 mol.

Cu CuO 2Fe Fe2O30,33 0,33 0,24 0,12

a = 0,33 80 + 0,12 160 + 111,84 = 157,44 gam. (p n A).

V d 18: Ha tan 4,76 gam hn hp Zn, Al c t l mol 1:2 trong 400ml dung dchHNO3 1M va , dc dung dch X cha m gam mui khan v thy c khthot ra. Gi tr ca m l:

A. 25.8 gam. B. 26,9 gam. C. 27,8 gam. D. 28,8 gam.

Hng dn gii

nZn = 0,04 mol; nAl = 0,08 mol.

- Do phn ng khng to kh nn trong dung dch to NH4NO3. Trong dung dch c:

0,04 mol Zn(NO3)2 v 0,08 mol Al(NO3)3

Vy s mol NO3 cn li to NH4NO3 l:

0,4 0,04 2 0,08 3 = 0,08 mol

- Do trong dung dch to 0,04 mol NH4NO3

m = 0,04 189 + 0,08 213 + 0,04 80 = 27,8 gam. (p n C)


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Phng php 5

S DNG CC GI TR TRUNG BNH

y l mt trong mt s phng php hin i nht cho php gii nhanh chng v ngin nhiu bi ton ha hc v hn hp cc cht rn, lng cng nh kh.

Nguyn tc ca phng php nh sau: Khi lng phn t trung bnh (KLPTTB) (khiu M ) cng nh khi lng nguyn t trung bnh (KLNTTB) chnh l khi lng camt mol hn hp, nn n c tnh theo cng thc:

M =tng khi l ng hn h p (tnh theo gam)

tng s mol c c cht trong hn h p.

i i1 1 2 2 3 3

1 2 3 i

M nM n M n M n ...M

n n n ... n

+ + += =

+ + +

(1)

trong M1, M2,... l KLPT (hoc KLNT) ca cc cht trong hn hp; n 1, n2,... l s moltng ng ca cc cht.

Cng thc (1) c th vit thnh:

1 2 31 2 3

i i i

n n nM M . M . M . ...

n n n= + + +

1 1 2 2 3 3M M x M x M x ...= + + + (2)

trong x1, x2,... l % s mol tng ng (cng chnh l % khi lng) ca cc cht. c

bit i vi cht kh th x1, x2, ... cng chnh l % th tch nn cng thc (2) c th vitthnh:

i i1 1 2 2 3 3

1 2 3 i

M VM V M V M V ...M

V V V ... V

+ + += =

+ + + (3)

trong V1, V2,... l th tch ca cc cht kh. Nu hn hp ch c 2 cht th cc cngthc (1), (2), (3) tng ng tr thnh (1), (2), (3) nh sau:


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1 1 2 1M n M (n n )Mn

+ = (1)

trong n l tng s s mol ca cc cht trong hn hp,

1 1 2 1M M x M (1 x )= + (2)

trong con s 1 ng vi 100% v

1 1 2 1M V M (V V )MV

+ = (3)

trong V1 l th tch kh th nht v V l tng th tch hn hp.

T cng thc tnh KLPTTB ta suy ra cc cng thc tnh KLNTTB.

Vi cc cng thc:

x y z 1

x y z 2

C H O ; n mol

C H O ; n mol

ta c:

- Nguyn t cacbon trung bnh:

1 1 2 2

1 2

x n x n ...x

n n ...

+ +=

+ +

- Nguyn t hiro trung bnh:

1 1 2 2

1 2

y n y n ...y

n n ...

+ +=

+ +

v i khi tnh c c s lin kt

, s nhm chc trung bnh theo cng thc trn.

V d 1: Ha tan hon ton 2,84 gam hn hp hai mui cacbonat ca hai kim loi phnnhm IIA v thuc hai chu k lin tip trong bng tun hon bng dung dchHCl ta thu c dung dch X v 672 ml CO2 ( ktc).

1. Hy xc nh tn cc kim loi.

A. Be, Mg. B. Mg, Ca. C. Ca, Ba. D. Ca, Sr.

2. C cn dung dch X th thu c bao nhiu gam mui khan?

A. 2 gam. B. 2,54 gam. C. 3,17 gam. D. 2,95 gam.

Hng dn gii1. Gi A, B l cc kim loi cn tm. Cc phng trnh phn ng l

ACO3 + 2HCl ACl2 + H2O + CO2 (1)

BCO3 + 2HCl BCl2 + H2O + CO2 (2)

(C th gi M l kim loi i din cho 2 kim loi A, B lc ch cn vit mt phng trnh phn ng).


42/91

Theo cc phn ng (1), (2) tng s mol cc mui cacbonat bng:

2CO

0,672n 0,03

22,4= = mol.

Vy KLPTTB ca cc mui cacbonat l

2,84M 94,670,03

= = v A,BM 94,67 60 34,67= =

V thuc 2 chu k lin tip nn hai kim loi l Mg (M = 24) v Ca (M = 40). (pn B)

2. KLPTTB ca cc mui clorua:

M 34,67 71 105,67= + =mui clorua .

Khi lng mui clorua khan l 105,67 0,03 = 3,17 gam. (p n C)

V d 2: Trong t nhin, ng (Cu) tn ti di hai dng ng v 6329

Cu v 6529

Cu . KLNT

(xp x khi lng trung bnh) ca Cu l 63,55. Tnh % v khi lng ca miloi ng v.

A. 65Cu: 27,5% ; 63Cu: 72,5%.

B. 65Cu: 70% ; 63Cu: 30%.

C. 65Cu: 72,5% ; 63Cu: 27,5%.

D. 65Cu: 30% ; 63Cu: 70%.

Hng dn gii

Gi x l % ca ng v 6529

Cu ta c phng trnh:

M = 63,55 = 65.x + 63(1 x)

x = 0,275

Vy: ng v 65Cu chim 27,5% v ng v 63Cu chim 72,5%. (p n C)

V d 3: Hn hp kh SO2 v O2 c t khi so vi CH4 bng 3. Cn thm bao nhiu lt O2vo 20 lt hn hp kh cho t khi so vi CH4 gim i 1/6, tc bng 2,5.Cc hn hp kh cng iu kin nhit v p sut.

A. 10 lt. B. 20 lt. C. 30 lt. D. 40 lt.

Hng dn giiCch 1: Gi x l % th tch ca SO2 trong hn hp ban u, ta c:

M = 16 3 = 48 = 64.x + 32(1 x)

x = 0,5

Vy: mi kh chim 50%. Nh vy trong 20 lt, mi kh chim 10 lt.

Gi V l s lt O2 cn thm vo, ta c:


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64 10 32(10 V)M 2,5 16 40

20 V

+ + = = =+

.

Gii ra c V = 20 lt. (p n B)

Cch 2:

Ghi ch: C th coi hn hp kh nh mt kh c KLPT chnh bng KLPT trung bnhca hn hp, v d, c th xem khng kh nh mt kh vi KLPT l 29.

Hn hp kh ban u coi nh kh th nht (20 lt c M = 16 3 = 48), cn O2 thmvo coi nh kh th hai, ta c phng trnh:

48 20 32VM 2,5 16 40

20 V

+= = =

+,

Rt ra V = 20lt. (p n B)

V d 4: C 100 gam dung dch 23% ca mt axit n chc (dung dch A). Thm 30 gam

mt axit ng ng lin tip vo dung dch ta c dung dch B. Trung ha 1/10dung dch B bng 500 ml dung dch NaOH 0,2M (va ) ta c dung dch C.

1. Hy xc nh CTPT ca cc axit.

A. HCOOH v CH3COOH.

B. CH3COOH v C2H5COOH.

C. C2H5COOH v C3H7COOH.

D. C3H7COOH v C4H9COOH.

2. C cn dung dch C th thu c bao nhiu gam mui khan?

A. 5,7 gam.

B. 7,5 gam. C. 5,75 gam. D. 7,55 gam.Hng dn gii

1. Theo phng php KLPTTB:

RCOOH

1 23m 2,3

10 10= = gam,

2RCH COOH

1 30m 3

10 10= = gam.

2,3 3M 53

0,1

+= = .

Axit duy nht c KLPT < 53 l HCOOH (M = 46) v axit ng ng lin tip phi lCH3COOH (M = 60). (p n A)

2. Theo phng php KLPTTB:

V Maxit = 53 nn M = 53+ 23 1 75 =mui . V s mol mui bng s mol axit bng 0,1

nn tng khi lng mui bng 75 0,1 = 7,5 gam. (p n B)


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V d 5: C V lt kh A gm H2 v hai olefin l ng ng lin tip, trong H2 chim60% v th tch. Dn hn hp A qua bt Ni nung nng c hn hp kh B. tchy hon ton kh B c 19,8 gam CO2 v 13,5 gam H2O. Cng thc ca haiolefin l

A. C2H4 v C3H6. B. C3H6 v C4H8.

C. C4H8 v C5H10. D. C5H10 v C6H12.

Hng dn gii

t CTTB ca hai olefin l n 2nC H .

cng iu kin nhit v p sut th th tch t l vi s mol kh.

Hn hp kh A c:

n 2n

2

C H

H

n 0,4 2

n 0,6 3= = .

p dng nh lut bo ton khi lng v nh lut bo ton nguyn t t chyhn hp kh B cng chnh l t chy hn hp kh A. Ta c:

n 2nC H + 23n

O2

n CO2 + n H2O (1)

2H2 + O2 2H2O (2)

Theo phng trnh (1) ta c:

2 2CO H On n= = 0,45 mol.

n 2nC H

0,45n

n

= mol.

Tng:2H O

13,5n

18= = 0,75 mol

2H O (pt2)

n = 0,75 0,45 = 0,3 mol

2H

n = 0,3 mol.

Ta c: n 2n

2

C H

H

n 0,45 2

n 0,3 n 3= =

n = 2,25

Hai olefin ng ng lin tip l C2H4 v C3H6. (p n B)

V d 6: t chy hon ton a gam hn hp hai ru no, n chc lin tip trong dyng ng thu c 3,584 lt CO2 ktc v 3,96 gam H2O. Tnh a v xc nhCTPT ca cc ru.

A. 3,32 gam ; CH3OH v C2H5OH.


45/91

B. 4,32 gam ; C2H5OH v C3H7OH.

C. 2,32 gam ; C3H7OH v C4H9OH.

D. 3,32 gam ; C2H5OH v C3H7OH.

Hng dn gii

Gi n l s nguyn t C trung bnh v x l tng s mol ca hai ru.

CnH2n+1OH + 23n

O2

2n CO + 2(n 1)H O+

x mol n x mol (n 1)+ x mol

2CO

3,584n n.x 0,16

22,4= = = mol (1)

2H O

3,96n (n 1)x 0,22

18= + = = mol (2)

T (1) v (2) gii ra x = 0,06 v n = 2,67.

Ta c: a = (14 n + 18).x = (14 2,67) + 18 0,06 = 3,32 gam.

n = 2,672 5

3 7

C H OH

C H OH(p n D)

V d 7: Hn hp 3 ru n chc A, B, C c tng s mol l 0,08 v khi lng l 3,38gam. Xc nh CTPT ca ru B, bit rng B v C c cng s nguyn t cacbon

v s mol ru A bng 5 3 tng s mol ca ru B v C, MB > MC.

A. CH3OH. B. C2H5OH.

C. C3H7OH. D. C4H9OH.Hng dn gii

Gi M l nguyn t khi trung bnh ca ba ru A, B, C. Ta c:

3,38M 42,2

0,08= =

Nh vy phi c t nht mt ru c M < 42,25. Ch c CH3OH c (M = 32)

Ta c: A0,08 5

n 0,055 3

= =

+;

mA = 32 0,05 = 1,6 gam.mB + C = 3,38 1,6 = 1,78 gam;

B C

0,08 3n 0,03

5 3+

= =+

mol ;

B C

1,78M 59,33

0.03+= = .


46/91

Gi y l s nguyn t H trung bnh trong phn t hai ru B v C. Ta c:

x yC H OH 59,33= hay 12x + y +17 = 59,33

12x + y = 42,33

Bin lun:

x 1 2 3 4

y 30,33 18,33 6,33 < 0

Ch c nghim khi x = 3. B, C phi c mt ru c s nguyn t H < 6,33 v mtru c s nguyn t H > 6,33.

Vy ru B l C3H7OH.

C 2 cp nghim: C3H5OH (CH2=CHCH2OH) v C3H7OH

C3H3OH (CH CCH2OH) v C3H7OH (p n C)

V d 8: Cho 2,84 gam hn hp 2 ru n chc l ng ng lin tip nhau tc dngvi mt lng Na va to ra 4,6 gam cht rn v V lt kh H2 ktc. Tnh V.

A. 0,896 lt. B. 0,672 lt. C. 0,448 lt. D. 0,336 lt.

Hng dn gii

t R l gc hirocacbon trung bnh v x l tng s mol ca 2 ru.

ROH + Na RONa + 21

H2

x mol x x

2 .

Ta c:( )

( )

R 17 x 2,84

R 39 x 4,6

+ =

+ = Gii ra c x = 0,08.

Vy :2H

0,08V 22,4 0,896

2= = lt. (p n A)

V d 9: (Cu 1 - M 182 - Khi A - TSH nm 2007)

Cho 4,48 lt hn hp X ( ktc) gm 2 hirocacbon mch h li t t qua bnh

cha 1,4 lt dung dch Br2 0,5M. Sau khi phn ng hon ton, s mol Br2 gimi mt na v khi lng bnh tng thm 6,7 gam. Cng thc phn t ca 2hirocacbon l

A. C2H2 v C4H6. B. C2H2 v C4H8.

C. C3H4 v C4H8. D. C2H2 v C3H8.

Hng dn gii


47/91

hh X

4,48n 0,2

22,4= = mol

n 1,4 0,5 0,72Br ban u

= = mol

0,7

n 22Br p.ng = = 0,35 mol.

Khi lng bnh Br2 tng 6,7 gam l s gam ca hirocabon khng no. t CTTB

ca hai hirocacbon mch h l n 2n 2 2aC H + ( a l s lin kt trung bnh).

Phng trnh phn ng:

n 2n 2 2aC H + + 2aBr n 2n 2 2a 2aC H Br+

0,2 mol 0,35 mol

0,35

a0,2

= = 1,75

6,7

14n 2 2a0,2

+ = n = 2,5.

Do hai hirocacbon mch h phn ng hon ton vi dung dch Br2 nn chng ul hirocacbon khng no. Vy hai hirocacbon l C2H2 v C4H8. (p n B)

V d 10: Tch nc hon ton t hn hp X gm 2 ancol A v B ta c hn hp Y gmcc olefin. Nu t chy hon ton X th thu c 1,76 gam CO 2. Khi t chyhon ton Y th tng khi lng H2O v CO2 to ra l


Hng dn giiHn hp X gm hai ancol A v B tch nc c olefin (Y) hai ancol l ru

no, n chc.

t CTTB ca hai ancol A, B l n 2n 1C H OH+ ta c cc phng trnh phn ng sau:

n 2n 1C H OH+ + 23n

O2

2nCO + 2(n 1)H O+

n 2n 1C H OH+ 2

o

H SO

170 C

4 n 2nC H + H2O

(Y)

n 2nC H + 23n

O2

2nCO + 2n H O

Nhn xt:

- Khi t chy X v t chy Y cng cho s mol CO2 nh nhau.

- t chy Y cho2 2CO H O

n n= .

Vy t chy Y cho tng


48/91

( )2 2CO H O

m m 0,04 (44 18) 2,48+ = + = gam. (p n B)

MT S BI TP VN DNG GII THEP PHNG

PHP TRUNG BNH

01. t chy hon ton 0,1 mol hn hp hai axit cacboxylic l ng ng k tip thuc 3,36 lt CO2 (ktc) v 2,7 gam H2O. S mol ca mi axit ln lt l

A. 0,05 mol v 0,05 mol. B. 0,045 mol v 0,055 mol.

C. 0,04 mol v 0,06 mol. D. 0,06 mol v 0,04 mol.

02. C 3 ancol bn khng phi l ng phn ca nhau. t chy mi cht u c s molCO2 bng 0,75 ln s mol H2O. 3 ancol l

A. C2H6O; C3H8O; C4H10O. B. C3H8O; C3H6O2; C4H10O.

C. C3H8O; C3H8O2; C3H8O3. D. C3H8O; C3H6O; C3H8O2.

03. Cho axit oxalic HOOCCOOH tc dng vi hn hp hai ancol no, n chc, ngng lin tip thu c 5,28 gam hn hp 3 este trung tnh. Thy phn lng estetrn bng dung dch NaOH thu c 5,36 gam mui. Hai ru c cng thc

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

C. C3H7OH v C4H9OH. D. C4H9OH v C5H11OH.

04. Nitro ha benzen c 14,1 gam hn hp hai cht nitro c khi lng phn t hnkm nhau 45 vC. t chy hon ton hn hp hai cht nitro ny c 0,07 mol N 2.Hai cht nitro l

A. C6 H5NO2 v C6H4(NO2)2.

B. C6 H4(NO2)2 v C6H3(NO2)3.C. C6 H3(NO2)3 v C6H2(NO2)4.

D. C6 H2(NO2)4 v C6H(NO2)5.

05. Mt hn hp X gm 2 ancol thuc cng dy ng ng c khi lng 30,4 gam. ChiaX thnh hai phn bng nhau.

- Phn 1: cho tc dng vi Na d, kt thc phn ng thu c 3,36 lt H2 (ktc).

- Phn 2: tch nc hon ton 180oC, xc tc H2SO4 c thu c mt anken chohp th vo bnh ng dung dch Brom d thy c 32 gam Br2 b mt mu. CTPT haiancol trn l

A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.C. CH3OH v C3H7OH. D. C2H5OH v C4H9OH.

06. Chia hn hp gm 2 anehit no n chc lm hai phn bng nhau:

- Phn 1: em t chy hon ton thu c 1,08 gam nc.

- Phn 2: tc dng vi H2 d (Ni, to) th thu c hn hp A. em A t chy honton th th tch kh CO2 (ktc) thu c l


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A. 1,434 lt. B. 1,443 lt. C. 1,344 lt. D. 1,444 lt.

07. Tch nc hon ton t hn hp Y gm hai ru A, B ta c hn hp X gm ccolefin. Nu t chy hon ton Y th thu c 0,66 gam CO 2. Vy khi t chy honton X th tng khi lng H2O v CO2 to ra l

A. 0,903 gam. B. 0,39 gam. C. 0,94 gam. D. 0,93 gam.08. Cho 9,85 gam hn hp 2 amin n chc no bc 1 tc dng va vi dung dch HCl

th thu c 18,975 gam mui. Vy khi lng HCl phi dng l


09. Cho 4,2 gam hn hp gm ru etylic, phenol, axit fomic tc dng va vi Nathy thot ra 0,672 lt kh (ktc) v mt dung dch. C cn dung dch thu c hnhp X. Khi lng ca X l


10. Hn hp X gm 2 este A, B ng phn vi nhau v u c to thnh t axit nchc v ru n chc. Cho 2,2 gam hn hp X bay hi 136,5oC v 1 atm th thuc 840 ml hi este. Mt khc em thu phn hon ton 26,4 gam hn hp X bng100 ml dung dch NaOH 20% (d = 1,2 g/ml) ri em c cn th thu c 33,8 gamcht rn khan. Vy cng thc phn t ca este l

A. C2H4O2. B. C3H6O2. C. C4H8O2. D. C5H10O2.

p n cc bi tp trc nghim vn dng:

1. A 2. C 3. A 4. A 5. C

6. C 7. D 8. B 9. B 10. C

Phng php 6

TNG GIM KHI LNG

Nguyn tc ca phng php l xem khi chuyn t cht A thnh cht B (khng nhtthit trc tip, c th b qua nhiu giai on trung gian) khi lng tng hay gim baonhiu gam thng tnh theo 1 mol) v da vo khi lng thay i ta d dng tnh cs mol cht tham gia phn ng hoc ngc li. V d trong phn ng:

MCO3 + 2HCl MCl2 + H2O + CO2


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Ta thy rng khi chuyn 1 mol MCO3 thnh MCl2 th khi lng tng

(M + 2 35,5) (M + 60) = 11 gam

v c 1 mol CO2 bay ra. Nh vy khi bit lng mui tng, ta c th tnh lng CO 2 bayra.

Trong phn ng este ha:CH3COOH + ROH CH3COOR + H2O

th t 1 mol ROH chuyn thnh 1 mol este khi lng tng

(R + 59) (R + 17) = 42 gam.

Nh vy nu bit khi lng ca ru v khi lng ca este ta d dng tnh cs mol ru hoc ngc li.

Vi bi tp cho kim loi A y kim loi B ra khi dung dch mui di dng t do:

- Khi lng kim loi tng bng

mB (bm)mA (tan).- Khi lng kim loi gim bng

mA (tan)mB (bm).

Sau y l cc v d in hnh:

V d 1: C 1 lt dung dch hn hp Na2CO3 0,1 mol/l v (NH4)2CO3 0,25 mol/l. Cho 43gam hn hp BaCl2 v CaCl2 vo dung dch . Sau khi cc phn ng kt thcta thu c 39,7 gam kt ta A v dung dch B.

Tnh % khi lng cc cht trong A.

A. 3BaCO%m = 50%, 3CaCO%m = 50%.

B.3BaCO

%m = 50,38%,3CaCO

%m = 49,62%.

C. 3BaCO%m = 49,62%, 3CaCO%m = 50,38%.

D. Khng xc nh c.

Hng dn gii

Trong dung dch:

Na2CO3 2Na+ + CO32

(NH4)2CO3 2NH4+ + CO32

BaCl2 Ba2+ + 2Cl

CaCl2 Ca2+ + 2Cl

Cc phn ng:

Ba2+ + CO32 BaCO3 (1)


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Ca2+ + CO32 CaCO3 (2)

Theo (1) v (2) c 1 mol BaCl2, hoc CaCl2 bin thnh BaCO3 hoc CaCO3 th khi

lng mui gim (71 60) = 11 gam. Do tng s mol hai mui BaCO3 v CaCO3bng:

43 39,711 = 0,3 mol

m tng s mol CO32 = 0,1 + 0,25 = 0,35, iu chng t d CO32

.

Gi x, y l s mol BaCO3 v CaCO3 trong A ta c:

x y 0,3

197x 100y 39,7

+ = + =

x = 0,1 mol ; y = 0,2 mol.

Thnh phn ca A:

3BaCO

0,1 197%m 100

39,7

= = 49,62%;

3CaCO%m = 100 49,6 = 50,38%. (p n C)

V d 2: Ho tan hon ton 23,8 gam hn hp mt mui cacbonat ca kim loi ho tr (I)v mt mui cacbonat ca kim loi ho tr (II) bng dung dch HCl thy thot ra4,48 lt kh CO2 (ktc). C cn dung dch thu c sau phn ng th khi lngmui khan thu c l bao nhiu?

A. 26,0 gam. B. 28,0 gam. C. 26,8 gam. D. 28,6 gam.Hng dn gii

C 1 mol mui cacbonat to thnh 1 mol mui clorua cho nn khi lng mui khan

tng (71 60) = 11 gam, m

2COn = nmui cacbonat = 0,2 mol.

Suy ra khi lng mui khan tng sau phn ng l 0,2 11 = 2,2 gam.

Vy tng khi lng mui khan thu c l 23,8 + 2,2 = 26 gam. (p n A)

V d 3: Cho 3,0 gam mt axit no, n chc A tc dng va vi dung dch NaOH. Ccn dung dch sau phn ng thu c 4,1 gam mui khan. CTPT ca A l

A. HCOOH B. C3H7COOH

C. CH3COOH D. C2H5COOH.

Hng dn gii


52/91

C 1 mol axit n chc to thnh 1 mol mui th khi lng tng (23 1) = 22 gam,m theo u bi khi lng mui tng (4,1 3) = 1,1 gam nn s mol axit l

naxit =1,1

22= 0,05 mol. Maxit =

3

0,05= 60 gam.

t CTTQ ca axit no, n chc A l CnH2n+1COOH nn ta c:

14n + 46 = 60 n = 1.

Vy CTPT ca A l CH3COOH. (p n C)

V d 4: Cho dung dch AgNO3 d tc dng vi dung dch hn hp c ha tan 6,25 gamhai mui KCl v KBr thu c 10,39 gam hn hp AgCl v AgBr. Hy xcnh s mol hn hp u.


Hng dn gii

C 1 mol mui halogen to thnh 1 mol kt ta

khi lng tng: 108 39 = 69 gam;

0,06 mol khi lng tng: 10,39 6,25 = 4,14 gam.

Vy tng s mol hn hp u l 0,06 mol. (p n B)

V d 5: Nhng mt thanh graphit c ph mt lp kim loi ha tr (II) vo dung dchCuSO4 d. Sau phn ng khi lng ca thanh graphit gim i 0,24 gam. Cngthanh graphit ny nu c nhng vo dung dch AgNO3 th khi phn ng xongthy khi lng thanh graphit tng ln 0,52 gam. Kim loi ha tr (II) l kim

loi no sau y?A. Pb. B. Cd. C. Al. D. Sn.

Hng dn gii

t kim loi ha tr (II) l M vi s gam l x (gam).

M + CuSO4 d MSO4 + Cu

C M gam kim loi tan ra th s c 64 gam Cu bm vo. Vy khi lng kim loigim (M 64) gam;

Vy: x (gam) =

0,24.M

M 64 khi lng kim loi gim 0,24 gam.Mt khc: M + 2AgNO3 M(NO3)2 + 2Ag

C M gam kim loi tan ra th s c 216 gam Ag bm vo. Vy khi lng kim loitng (216 M) gam;

Vy: x (gam) =0,52.M

216 M khi lng kim loi tng 0,52 gam.


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Ta c:0,24.MM 64

=0,52.M

216 M M = 112 (kim loi Cd). (p n B)

V d 6: Ho tan hon ton 104,25 gam hn hp X gm NaCl v NaI vo nc cdung dch A. Sc kh Cl2 d vo dung dch A. Kt thc th nghim, c cn dung

dch thu c 58,5 gam mui khan. Khi lng NaCl c trong hn hp X lA. 29,25 gam. B. 58,5 gam.

C. 17,55 gam. D. 23,4 gam.

Hng dn gii

Kh Cl2 d ch kh c mui NaI theo phng trnh

2NaI + Cl2 2NaCl + I2C 1 mol NaI to thnh 1 mol NaCl

Khi lng mui gim 127 35,5 = 91,5 gam.

Vy: 0,5 mol Khi lng mui gim 104,25 58,5 = 45,75 gam.

mNaI = 150 0,5 = 75 gam

mNaCl = 104,25 75 = 29,25 gam. (p n A)

V d 7: Ngm mt vt bng ng c khi lng 15 gam trong 340 gam dung dchAgNO3 6%. Sau mt thi gian ly vt ra thy khi lng AgNO3 trong dungdch gim 25%. Khi lng ca vt sau phn ng l


Hng dn gii

3AgNO ( )

340 6n =

170 100ban u

= 0,12 mol;

3AgNO ( )

25n = 0,12

100ph.ng = 0,03 mol.

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

0,015 0,03 0,03 mol

mvt sau phn ng = mvt ban u + mAg (bm)mCu (tan)

= 15 + (108 0,03) (64 0,015) = 17,28 gam.(p n C)

V d 8: Nhng mt thanh km v mt thanh st vo cng mt dung dch CuSO 4. Saumt thi gian ly hai thanh kim loi ra thy trong dung dch cn li c nng mol ZnSO4 bng 2,5 ln nng mol FeSO4. Mt khc, khi lng dung dchgim 2,2 gam.

Khi lng ng bm ln thanh km v bm ln thanh st ln lt l


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A. 12,8 gam; 32 gam. B. 64 gam; 25,6 gam.

C. 32 gam; 12,8 gam. D. 25,6 gam; 64 gam.

Hng dn gii

V trong cng dung dch cn li (cng th tch) nn:

[ZnSO4] = 2,5 [FeSO4]

4 4ZnSO FeSOn 2,5n=

Zn + CuSO4 ZnSO4 + Cu (1)

2,5x 2,5x 2,5x mol

Fe + CuSO4 FeSO4 + Cu (2)

x x x x mol

T (1), (2) nhn c gim khi lng ca dung dch l

mCu (bm)mZn (tan)mFe (tan)

2,2 = 64 (2,5x + x) 65 2,5x 56x

x = 0,4 mol.

Vy: mCu (bm ln thanh km) = 64 2,5 0,4 = 64 gam;

mCu (bm ln thanh st) = 64 0,4 = 25,6 gam. (p n B)

V d 9: (Cu 15 - M 231 - TSC - Khi A 2007)

Cho 5,76 gam axit hu c X n chc, mch h tc dng ht vi CaCO3 thuc 7,28 gam mui ca axit hu c. Cng thc cu to thu gn ca X l

A. CH2=CHCOOH. B. CH3COOH.

C. HC CCOOH. D. CH3CH2COOH.

Hng dn gii

t CTTQ ca axit hu c X n chc l RCOOH.

2RCOOH + CaCO3 (RCOO)2Ca + CO2 + H2O

C 2 mol axit phn ng to mui th khi lng tng (40 2) = 38 gam.

x mol axit (7,28 5,76) = 1,52 gam.

x = 0,08 mol RCOOH 5,76M 720,08= = R = 27

Axit X: CH2=CHCOOH. (p n A)

V d 10: Nhng thanh km vo dung dch cha 8,32 gam CdSO 4. Sau khi kh hon tonion Cd2+ khi lng thanh km tng 2,35% so vi ban u. Hi khi lngthanh km ban u.


55/91


Hng dn gii

Gi khi lng thanh km ban u l a gam th khi lng tng thm l2,35a

100gam.

Zn + CdSO4 ZnSO4 + Cd

65 1 mol 112, tng (112 65) = 47 gam

8,32

208(=0,04 mol)

2,35a

100gam

Ta c t l:1 47

2,35a0,04100

= a = 80 gam. (p n C)

V d 11: Nhng thanh kim loi M ho tr 2 vo dung dch CuSO4, sau mt thi gian ly

thanh kim loi ra thy khi lng gim 0,05%. Mt khc nhng thanh kim loitrn vo dung dch Pb(NO3)2, sau mt thi gian thy khi lng tng 7,1%.Xc nh M, bit rng s mol CuSO4 v Pb(NO3)2 tham gia 2 trng hp nhnhau.

A. Al. B. Zn. C. Mg. D. Fe.

Hng dn gii

Gi m l khi lng thanh kim loi, M l nguyn t khi ca kim loi, x l s molmui phn ng.

M + CuSO4

MSO4 + CuM (gam) 1 mol 64 gam, gim (M 64)gam.

x mol gim0,05.m

100gam.

x =0,05.m

100M 64

(1)

M + Pb(NO3)2M(NO3)2 + Pb

M (gam) 1 mol 207, tng (207 M) gam

x mol tng7,1.m

100gam

x =7,1.m100

207 M(2)


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T (1) v (2) ta c:0,05.m

100M 64

=7,1.m100

207 M(3)

T (3) gii ra M = 65. Vy kim loi M l km. ( p n B)

V d 12: Cho 3,78 gam bt Al phn ng va vi dung dch mui XCl3 to thnhdung dch Y. Khi lng cht tan trong dung dch Y gim 4,06 gam so vidung dch XCl3. xc nh cng thc ca mui XCl3.

A. FeCl3. B. AlCl3. C. CrCl3. D. Khng xc nh.

Hng dn gii

Gi A l nguyn t khi ca kim loi X.

Al + XCl3 AlCl3 + X

3,78

27

= (0,14 mol) 0,14 0,14 mol.

Ta c : (A + 35,5 3) 0,14 (133,5 0,14) = 4,06

Gii ra c: A = 56. Vy kim loi X l Fe v mui FeCl3. (p n A)

V d 13: Nung 100 gam hn hp gm Na2CO3 v NaHCO3 cho n khi khi lng hn hpkhng i c 69 gam cht rn. Xc nh phn trm khi lng ca mi chttng ng trong hn hp ban u.

A. 15,4% v 84,6%. B. 22,4% v 77,6%.

C. 16% v 84%. D. 24% v 76%.

Hng dn giiCh c NaHCO3 b phn hy. t x l s gam NaHCO3.

2NaHCO3ot Na2CO3 + CO2

+ H2O

C nung 168 gam khi lng gim: 44 + 18 = 62 gam

x khi lng gim: 100 69 = 31 gam

Ta c:168 62

x 31= x = 84 gam.

Vy NaHCO3 chim 84% v Na2CO3 chim 16%. (p n C)

V d 14: Ha tan 3,28 gam hn hp mui CuCl2 v Cu(NO3)2 vo nc c dung dchA. Nhng Mg vo dung dch A cho n khi mt mu xanh ca dung dch. Lythanh Mg ra cn li thy tng thm 0,8 gam. C cn dung dch sau phn ngthu c m gam mui khan. Tnh m?

A. 1.28 gam. B. 2,48 gam. C. 3,1 gam. D. 0,48 gam.

Hng dn gii


57/91

Ta c:

mtng = mCumMg phn ng = ( )2 2 2Cu Mg Mgm m 3,28 m m 0,8gc axit+ + + = + =

m = 3,28 0,8 = 2,48 gam. (p n B)

V d 15: Ha tan 3,28 gam hn hp mui MgCl2 v Cu(NO3)2 vo nc c dung dchA. Nhng vo dung dch A mt thanh st. Sau mt khong thi gian ly thanhst ra cn li thy tng thm 0,8 gam. C cn dung dch sau phn ng thuc m gam mui khan. Gi tr m l


Hng dn gii

p dng nh lut bo ton khi lng: Sau mt khong thi gian tng khilng ca thanh Fe bng gim khi lng ca dung dch mui. Do :

m = 3,28 0,8 = 2,48 gam. (p n B)

MT S BI TP VN DNG GII THEO PHNG PHP TNG

GIM KHI LNG

01. Cho 115 gam hn hp gm ACO3, B2CO3, R2CO3 tc dng ht vi dung dch HClthy thot ra 22,4 lt CO2 (ktc). Khi lng mui clorua to ra trong dung dch l


02. Ngm mt l st trong dung dch CuSO4. Nu bit khi lng ng bm trn l st l9,6 gam th khi lng l st sau ngm tng thm bao nhiu gam so vi ban u?


03. Cho hai thanh st c khi lng bng nhau.

- Thanh 1 nhng vo dung dch c cha a mol AgNO3.

- Thanh 2 nhng vo dung dch c cha a mol Cu(NO3)2.

Sau phn ng, ly thanh st ra, sy kh v cn li thy s cho kt qu no sau y?

A. Khi lng hai thanh sau nhng vn bng nhau nhng khc ban u.

B. Khi lng thanh 2 sau nhng nh hn khi lng thanh 1 sau nhng.

C. Khi lng thanh 1 sau nhng nh hn khi lng thanh 2 sau nhng.

D. Khi lng hai thanh khng i vn nh trc khi nhng.

04. Cho V lt dung dch A cha ng thi FeCl3 1M v Fe2(SO4)3 0,5M tc dng vi dungdch Na2CO3 c d, phn ng kt thc thy khi lng dung dch sau phn ng gim69,2 gam so vi tng khi lng ca cc dung dch ban u. Gi tr ca V l:

A. 0,2 lt. B. 0,24 lt. C. 0,237 lt. D.0,336 lt.


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05. Cho lung kh CO i qua 16 gam oxit st nguyn cht c nung nng trong mt cing. Khi phn ng thc hin hon ton v kt thc, thy khi lng ng gim 4,8gam.

Xc nh cng thc v tn oxit st em dng.

06. Dng CO kh 40 gam oxit Fe2O3 thu c 33,92 gam cht rn B gm Fe2O3, FeOv Fe. Cho

1B

2tc dng vi H2SO4 long d, thu c 2,24 lt kh H2 (ktc).

Xc nh thnh phn theo s mol cht rn B, th tch kh CO (ktc) ti thiu cc kt qu ny.

07. Nhng mt thanh st nng 12,2 gam vo 200 ml dung dch CuSO 4 0,5M. Sau mt thigian ly thanh kim loi ra, c cn dung dch c 15,52 gam cht rn khan.

a) Vit phng trnh phn ng xy ra, tm khi lng tng cht c trong 15,52 gamcht rn khan.

b) Tnh khi lng thanh kim loi sau phn ng. Ha tan hon ton thanh kim loiny trong dung dch HNO3 c nng, d thu c kh NO2 duy nht, th tch V lt(o 27,3 oC, 0,55 atm). Vit cc phng trnh phn ng xy ra. Tnh V.

08. Ngm mt thanh ng c khi lng 140,8 gam vo dung dch AgNO3 sau mt thigian ly thanh ng em cn li thy nng 171,2 gam. Tnh thnh phn khi lngca thanh ng sau phn ng.

09. Ngm mt l km nh trong mt dung dch c cha 2,24 gam ion kim loi c intch 2+. Phn ng xong, khi lng l km tng thm 0,94 gam.

Hy xc nh tn ca ion kim loi trong dung dch.

10. C hai l kim loi cng cht, cng khi lng, c kh nng to ra hp cht c s oxi ha+2. Mt l c ngm trong dung dch Pb(NO 3)2 cn l kia c ngm trong dung dchCu(NO3)2.

Sau mt thi gian ngi ta ly l kim loi ra khi dung dch, ra nh. Nhn thy khilng l kim loi c ngm trong mui ch tng thm 19%, khi lng l kim loikia gim 9,6%. Bit rng, trong hai phn ng trn, khi lng cc kim loi b ha tannh nhau.

Hy xc nh tn ca hai l kim loi ang dng.


01. B 02. D. 03. B. 04. A.

05. Fe2O3. 06. VCO = 8,512 lt ; %nFe = 46,51% ; %nFeO = 37,21% ;

2 3Fe O%n 16,28%.=

07. a) 6,4 gam CuSO4 v 9,12 gam FeSO4.


59/91

b) mKL = 12,68 gam ; 2NOV 26,88= lt.

08. Thanh Cu sau phn ng c mAg (bm) = 43,2 gam v mCu (cn li) = 128 gam.

09. Cd2+

10. Cd

Phng php 7

QUI I HN HP NHIU CHT V S LNG CHT THN

Mt s bi ton ha hc c th gii nhanh bng cc phng php bo ton electron,bo ton nguyn t, bo ton khi lng song phng php quy i cng tm ra p srt nhanh v l phng php tng i u vit, c th vn dng vo cc bi tp trcnghim phn loi hc sinh.

Cc ch khi p dng phng php quy i:

1. Khi quy i hn hp nhiu cht (hn hp X) (t ba cht tr ln) thnh hn hp haicht hay ch cn mt cht ta phi bo ton s mol nguyn t v bo ton khi lng hnhp.

2. C th quy i hn hp X v bt k cp cht no, thm ch quy i v mt cht. Tuy

nhin ta nn chn cp cht no n gin c t phn ng oxi ha kh nht n gin victnh ton.

3. Trong qu trnh tnh ton theo phng php quy i i khi ta gp s m l dos b tr khi lng ca cc cht trong hn hp. Trong trng hp ny ta vn tnh tonbnh thng v kt qu cui cng vn tha mn.

4. Khi quy i hn hp X v mt cht l Fe xOy th oxit FexOy tm c ch l oxit ginh khng c thc.

V d 1: Nung 8,4 gam Fe trong khng kh, sau phn ng thu c m gam cht rn Xgm Fe, Fe2O3, Fe3O4, FeO. Ha tan m gam hn hp X vo dung dch HNO 3 d

thu c 2,24 lt kh NO2 (ktc) l sn phm kh duy nht. Gi tr ca m lA. 11,2 gam. B. 10,2 gam. C. 7,2 gam. D. 6,9 gam.

Hng dn gii

Quy hn hp X v hai cht Fe v Fe2O3:

Ha tan hn hp X vo dung dch HNO3 d ta c


60/91

Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O

0,1

3 0,1 mol

S mol ca nguyn t Fe to oxit Fe2O3 l

Fe

8,4 0,1 0,35n56 3 3

= = 2 3Fe O

0,35n3 2

=

Vy:2 3X Fe Fe O

m m m= +

X0,1 0,35

m 56 1603 3

= + = 11,2 gam.

Quy hn hp X v hai cht FeO v Fe2O3:

FeO + 4HNO3 Fe(NO3)3 + NO2 + 2H2O

0,1 0,1 mol

ta c:

2

2 2 3

2Fe O 2FeO

0,1 0,1 mol0,15 mol

4Fe 3O 2Fe O

0,05 0,025 mol

+

+

2h Xm = 0,1 72 + 0,025 160 = 11,2 gam. (p n A)

Ch : Vn c th quy hn hp X v hai cht (FeO v Fe3O4) hoc (Fe v FeO),hoc (Fe v Fe3O4) nhng vic gii tr nn phc tp hn (c th l ta phi t n s mol

mi cht, lp h phng trnh, gii h phng trnh hai n s). Quy hn hp X v mt cht l FexOy:

FexOy + (6x2y)HNO3 Fe(NO3)3 + (3x2y) NO2 + (3xy)H2O

0,1

3x 2ymol 0,1 mol.

Fe8,4 0,1.x

n56 3x 2y

= =

x 6

y 7= mol.

Vy cng thc quy i l Fe6O7 (M = 448) v

6 7Fe O0,1n

3 6 2 7=

= 0,025 mol.

mX = 0,025 448 = 11,2 gam.

Nhn xt: Quy i hn hp gm Fe, FeO, Fe2O3, Fe3O4 v hn hp hai cht l FeO,Fe2O3 l n gin nht.


61/91

V d 2: Ha tan ht m gam hn hp X gm FeO, Fe2O3, Fe3O4 bng HNO3 c nng thuc 4,48 lt kh NO2 (ktc). C cn dung dch sau phn ng thu c 145,2gam mui khan gi tr ca m l


Hng dn gii

Quy hn hp X v hn hp hai cht FeO v Fe2O3 ta c

FeO + 4HNO3 Fe(NO3)3 + NO2 + 2H2O

0,2 mol 0,2 mol 0,2 mol

Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O

0,2 mol 0,4 mol

3 3Fe(NO )

145,2n

242

= = 0,6 mol.

mX = 0,2 (72 + 160) = 46,4 gam. (p n B)

V d 3:Ha tan hon ton 49,6 gam hn hp X gm Fe, FeO, Fe2O3, Fe3O4 bng H2SO4c nng thu c dung dch Y v 8,96 lt kh SO2 (ktc).

a) Tnh phn trm khi lng oxi trong hn hp X.

A. 40,24%. B. 30,7%. C. 20,97%. D. 37,5%.

b) Tnh khi lng mui trong dung dch Y.

A. 160 gam.

B.140 gam. C. 120 gam. D. 100 gam.Hng dn gii

Quy hn hp X v hai cht FeO, Fe2O3, ta c:

2 4 2 4 3 2 2

2 3 2 4 2 4 3 2

2FeO 4H SO Fe (SO ) SO 4H O

0,8 0,4 0,4 mol49,6 gam

Fe O 3H SO Fe (SO ) 3H O

0,05 0,05 mol

+ + +

+ +

2 3Fe O

m

= 49,6 0,8 72 = 8 gam (0,05 mol) nO (X) = 0,8 + 3 (0,05) = 0,65 mol.

Vy: a) O0,65 16 100

%m49,9

= = 20,97%. (p n C)

b)2 4 3Fe (SO )

m = [0,4 + (-0,05)] 400 = 140 gam. (p n B)


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V d 4: kh hon ton 3,04 gam hn hp X gm FeO, Fe 2O3, Fe3O4 th cn 0,05 molH2. Mt khc ha tan hon ton 3,04 gam hn hp X trong dung dch H 2SO4 cnng th thu c th tch kh SO2 (sn phm kh duy nht ktc) l.

A. 224 ml. B. 448 ml. C. 336 ml. D. 112 ml.

Hng dn giiQuy hn hp X v hn hp hai cht FeO v Fe2O3 vi s mol l x, y, ta c:

FeO + H2ot Fe + H2O

x y

Fe2O3 + 3H2ot 2Fe + 3H2O

x 3y

x 3y 0,05

72x 160y 3,04

+ = + =

x 0,02 mol

y 0,01 mol

= =2FeO + 4H2SO4 Fe2(SO4)3 + SO2 + 4H2O

0,02 0,01 mol

Vy:2SO

V = 0,01 22,4 = 0,224 lt (hay 224 ml). (p n A)

V d 5: Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ha tan hthn hp X trong dung dch HNO3 (d) thot ra 0,56 lt NO ( ktc) (l snphm kh duy nht). Gi tr ca m l


Hng dn giiQuy hn hp cht rn X v hai cht Fe, Fe2O3:

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O

0,025 0,025 0,025 mol

2 3Fe O

m = 3 56 0,025 = 1,6 gam

2 3Fe (trong Fe O )

1,6m 2

160= = 0,02 mol

mFe = 56

(0,025 + 0,02) = 2,52 gam. (p n A)

V d 6: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol, ha tanht vo dung dch Y gm (HCl v H2SO4 long) d thu c dung dch Z. Nht t dung dch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot kh NO.Th tch dung

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10 PHUONG PHAP GIAI HOA