[tuituhoc.com] tổng hợp phương pháp giải hóa 12

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PHNG PHP S DNG CNG THC KINH NGHIM ............................................................. 2 PHNG PHP BO TON KHI LNG ................................................................................... 7 PHNG PHP TNG GIM KHI LNG ............................................................................... 18 PHNG PHP BO TON IN TCH....................................................................................... 34 PHNG PHP BO TON ELECTRON ...................................................................................... 40 PHNG PHP TRUNG BNH ....................................................................................................... 56 PHNG PHP QUY I .............................................................................................................. 73 PHNG PHP NG CHO ..................................................................................................... 84 PHNG PHP PHN TCH H S ............................................................................................ 100 PHNG PHP S DNG PHNG TRNH ION THU GN ................................................... 109 PHNG PHP KHO ST TH ........................................................................................... 119 PHNG PHP KHO ST T L S MOL CO2 V H2O.......................................................... 126 PHNG PHP CHIA HN HP THNH HAI PHN KHNG U NHAU ............................ 137 PHNG PHP S DNG MI QUAN H GIA CC I LNG....................................... 142 PHNG PHP CHN I LNG THCH HP ...................................................................... 150 K THUT SO SNH PHN TCH .............................................................................................. 161
http://www.tuituhoc.com/


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PHNG PHP S DNG CNG THC KINH NGHIM I. PHNG PHP GII

1. N dung phng php Xt bi ton tng qut quen thuc: M0 hn hp rn (M, MxOy) M+n + N (S)

m gam m1 gam (n: max)

Gi: S mol kim loi l a S oxi ha cao nht (max) ca kim loi l n S mol electron nhn (2) l t mol

Ta c:

M ne M+n a mol na mol

Mt khc: ne nhn = ne (oxi) + ne (2)

= 1m m

16

. 2 + t = 1m m

8

+ t

Theo nh lut bo ton electron: ne nhng = ne nhn na = 1m m

8

+ t

Nhn c 2 v vi M ta c:

(M.a)n = 1M.(m m)

8

+ M.t m.n = 1M.m8

- M.n8

+ M.t

Cui cng ta c:

ng vi M l Fe (56), n = 3 ta c: m = 0,7.m1 + 5,6.t (2) ng vi M l Cu (64), n = 2 ta c: m = 0,8.m1 + 6,4.t (3)

T (2, 3) ta thy: Bi ton c 3 i lng: m, m1 v en nhn (hoc Vkh (2))

Khi bit 2 trong 3 i lng trn ta tnh c ngay i lng cn li. giai on (2) bi c th cho s mol, th tch hoc khi lng ca mt kh hoc nhiu

kh; giai on (1) c th cho s lng cht rn c th l cc oxit hoc hn hp gm kim loi d v cc oxit. 2. Phm vi p dng v mt s ch

Ch dng khi HNO3 (hoc H2SO4 c nng) ly d hoc va . Cng thc kinh nghim trn ch p dng vi 2 kim loi Fe v Cu.

3. Cc bc gii Tm tng s mol electron nhn giai on kh N+5 hoc S+6.

O2 + HNO3 (H2SO4 c, nng) (1) (2)

en nhng = na (mol)

m = 1

M.m M.t

8M

n8

(1)


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Tm tng khi lng hn hp rn (kim loi v oxit kim loi): m1 p dng cng thc (2) hoc (3).

II TH D MINH HA Th d 1. t chy hon ton 5,6 gam bt Fe trong bnh O2 thu c 7,36 gam hn hp X gm Fe2O3, Fe3O4 v mt phn Fe cn d. Ha tan hon ton lng hn hp X trn vo dung dch HNO3 thu c V lt hn hp kh Y gm NO2 v NO c t khi so vi H2 bng 19. Gi tr ca V l A. 0,896. B. 0,672. C. 1,792 D. 0,448

Hng dn gii: p dng cng thc (2): 5,6 = 0,7. 7,36 + 5,6 n

H QKDQ nH QKDQ = 0,08 T

2Y/Hd = 19

2NOn = nNO = x

5

2 N

+ 4e 4

N

+ 2

N

4x x x

Vy: V = 22,4. 0,02. 2 = 0,896 lt p n A. Th d 2. m gam bt Fe trong khng kh mt thi gian thu dc 11,28 gam hn hp X gm 4 cht. Ha tan ht X trong lng d dung dch HNO3 thu c 672ml kh NO (sn phm kh duy nht, ktc). Gi tr ca m l: A. 5,6. B. 11,2. C. 7,0. D. 8,4.

Hng dn gii: p dng cng thc (2): N+5 + 3e N+2 0,09 0,03

p n D. Th d 3. Cho 11,36 gam hn hp X gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch HNO3 long, d thu c 1,344 lt kh NO (sn phm 1 kh duy nht, o ktc) v dung dch Y. C cn dung dch Y thu c m gam mui khan. Gi tr ca m l A. 49,09. B. 35,50. C. 38,72. D. 34,36.

Hng dn gii p dng cng thc (2): N+5 + 3e N+3 0,18 0,06

3 3Fe(NO )n = nFe =

0,7.11,36 5,6.0,1856

= 0,16

m = 242 . 0,16 = 38,72gam p n C. Th d 4. Cho 11,6 gam hn hp X gm Fe, FeO, Fe2O3 vo dung dch HNO3 long, d thu c V lt kh Y gm NO v NO2 c t khi so vi H2 bng 19. Mt khc, nu cho cng lng

4x = 0,08 x = 0,02

en nhn = 0,09 m = 0,7. 11,28 + 5,6.0,09 = 8,4gam

en nhn = 0,18


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hn hp X trn tc dng vi kh CO nng d th sau khi phn ng xy ra hon ton thu c 9,52 gam Fe. Gi tr ca V l A. 1,40. B. 2,80. C. 5,60. D. 4,20.

Hng dn gii: T

2Y/Hd = 19

2NO NO en n x n nhn = 4x

p dng cng thc: 9,52 = 0,7. 11,6 + 5,6. 4x x = 0,0625 V = 22,4. 0,0625. 2 = 2,80 lt p n B. Th d 5. Nung m gam bt Cu trong oxi thu c 24,8 gam hn hp cht rn X gm Cu, CuO v Cu2O. Ho tan hon ton X trong H2SO4 c nng 1 thot ra 4,48 lt kh SO2 (sn phm kh duy nht, ktc). Gi tr ca m l A. 9,6. B. 14,72. C. 21,12. D. 22,4.

Hng dn gii: S ha bi ton: Cu X Cu+2 + S+4

p dng cng thc (3): m = 0,8.mrn + 6 4.ne nhn (2) m = 0,8.24,8 + 6,4.0,2.2 = 22,4gam p n D.

III. BI TP P DNG 1. m gam bt st ngoi khng kh, sau mt thi gian thy khi lng ca hn hp thu c l 12 gam. Ha tan hn hp ny trong dung dch HNO3 thu c 2,24 lt kh NO (sn phm kh duy nht, ktc). Gi tr ca m l A. 5,6 gam. B. 10,08 gam. C. 11,84 gam. D. 14,95 gam.

2. Ha tan hon ton 10 gam hn hp X (Fe, Fe2O3) trong dung dch HNO3 va c 1,12 lt NO ( ktc, sn phm kh duy nht) v dung dch Y. Cho Y tc dng vi dung dch NaOH d c kt ta Z. Nung Z trong khng kh n khi lng khng i c m gam cht rn. Gi tr ca m l A. 12 gam. B. 16 gam. C. 11,2 gam. D. 19,2 gam.

3. Ha tan ht m gam hn hp Fe, Fe2O3, Fe3O4 trong dung dch HNO3 c, nng d c 448 ml kh NO2 ( ktc). C cn dung dch sau phn ng c 14,52 gam mui khan. Gi tr ca m l

A. 3,36 gam. B. 4,28 gam. C. 4,64 gam. D. 4,80 gam.

4. t chy hon ton 5,6 gam bt Fe trong mt bnh oxi thu c 7,36 gam hn hp X gm Fe2O3, Fe3O4 v mt phn Fe d. Ha tan hon ton hn hp X bng dung dch HNO3 thu c V lt hn hp kh Y gm NO2 v NO c t khi so vi H2 bng 19. Gi tr ca V A. 0,896 lt. B. 0,672 lt. C. 0,448 lt. D. 1,08 lt.

5. Cho lung kh CO i qua ng s ng m gam Fe2O3 nung nng. Sau mt thi gian thu c 13,92 gam hn hp X gm 4 cht. Ha tan ht X bng HNO3 c, nng d c 5,824 lt NO2 (sn phm kh duy nht, ktc). Gi tr ca m l A. 16 gam. B. 32 gam. C. 48 gam. D. 64 gam.

6. Cho 11,6 gam hn hp X gm Fe, FeO, Fe2O3 vo dung dch HNO3 long, d c V lt kh Y gm NO v NO2 c t khi hi so vi H2 l 19. Mt khc, nu cho cng lng hn hp X trn tc dng vi kh CO d th sau khi phn ng hon ton c 9,52 gam Fe. Gi tr ca V l A. 2,8 lt. B. 5,6 lt. C. 1,4 lt. D. 1,344 lt.

7. Nung m gam bt ng kim loi trong oxi thu c 24,8 gam hn hp rn X gm Cu, CuO v

O2 (1)

H2SO4 (2)


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Cu2O. Ha tan hon ton X trong H2SO4 c nng thot ra 4,48 lt kh SO2 (sn phm kh duy nht, ktc). Gi tr ca m l A. 9,6 gam. B. 14,72 gam. C. 21,12 gam. D. 22,4 gam.

8. Ha tan hon ton 18,16 gam hn hp X gm Fe v Fe3O4 trong 2 lt dung dch HNO3 2M thu c dung dch Y v 4,704 lt kh NO (sn phm kh duy nht, ktc). Phn trm khi lng Fe trong hn hp X l A. 38,23%. B. 61,67%. C. 64,67%. D. 35,24%.

9. Cho m gam hn hp X gm Fe, Fe3O4 tc dng vi 200 ml dung dch HNO3 3,2M. Sau khi phn ng hon ton c 0,1 mol kh NO (sn phm kh duy nht) v cn li 1,46 gam kim loi khng tan. Gi tr ca m l A. 17,04 gam. B. 19,20 gam. C. 18,50 gam. D. 20,50 gam.

10. m gam Fe trong khng kh mt thi gian c 7,52 gam hn hp X gm 4 cht. Ha tan ht X trong dung dch H2SO4 c, nng d c 0,672 lt kh SO2 (sn phm kh duy nht, ktc) v dung dch Y. C cn cn thn dung dch Y c m1 gam mui khan. Gi tr ca m v m1 ln rt l A. 7 gam v 25 gam. C. 4,48 gam v 16 gam.

B. 4,2 gam v 1,5 gam. D. 5,6 gam v 20 gam.

11. Cho 5,584 gam hn hp bt Fe v Fe3O4 tc dng va vi 500 ml dung dch HNO3 long. Sau khi phn ng xy ra hon ton c 0,3136 lt kh NO (sn phm kh duy nht, ktc) v dung dch X. Nng mol/l ca dung dch HNO3 l A. 0,472M. B. 0,152M C. 3,04M. D. 0,304M.

12. kh hon ton 9,12 gam hn hp cc oxit: FeO, Fe3O4 v Fe2O3 cn 3,36 lt kh H2 (ktc). Nu ha tan 9,12 gam hn hp trn bng H2SO4 c, nng d th th tch kh SO2 (sn phm kh duy nht, ktc) thu c ti a l A. 280 ml. B. 560 ml. C. 672 ml. D. 896 ml.

13. Cho kh CO i qua ng s ng 16 gam Fe2O3 un nng, sau khi phn ng thu c hn hp X gm Fe, FeO, Fe3O4 v Fe2O3 Ha tan hon ton X bng H2SO4, c, nng thu c dung dch Y. Khi lng mui trong Y l: A. 20 gam. B. 32 gam. C. 40 gam. D. 48 gam.

14. Ha tan 11,2 gam kim loi M trong dung dch HCI (d), thu c 4,48 lt ( ktc) H2. Cn nu ho tan hn hp X gm 11,2 gam kim loi M v 69,6 gam oxit MxOy trong lng d dung dch HNO3 th c 6,72 lt kh NO (sn phm kh duy nht, ktc). Cng thc ca oxit kim loi l

A. Fe3O4. B. FeO. C. Cr2O3 D. CrO

15. Cho 37 gam hn hp X gm Fe, Fe3O4 tc dng vi 640 ml dung dch HNO3 2M long, un nng. Sau khi cc phn ng xy ra hon ton thu c V lt kh NO (sn phm kh duy nht, ktc), dung dch Y v cn li 2,92 gam kim loi. Gi tr ca V l A. 2,24 lt. B. 4,48 lt. C. 3,36 lt. D. 6,72 lt.

16. Cho lung kh CO i qua ng s cha 0,12 mol hn hp gm FeO v Fe2O3 nung nng, phn ng to ra 0,138 mol CO2. Hn hp cht rn cn li trong ng nng 14,352 gam gm 4 cht. Ha tan ht hn hp 4 cht ny vo dung dch HNO3 d thu c V lt kh NO (sn phm kh duy nht ktc). Gi tr ca V l A. 0,244 lt. B. 0,672 lt. C. 2,285 lt. D. 6,854 lt.

17. Cho lung kh CO i qua ng s ng 5,8 gam FexOy nung nng trong mt thi gian thu c hn hp kh X v cht rn Y. Cho Y tc ng vi dung dch HNO3 d c dung dch Z


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v 0,784 lt kh NO (sn phm kh duy nht, ktc). C cn dung dch Z c 18,15 gam mui khan. Ha tan Y bng HCl d thy c 0,672 lt kh ( ktc). Phn trm khi lng ca st trong Y l

A. 67,44%. B. 32,56%. C. 40,72%. D. 59,28%.

18. Cho lung kh CO i qua ng s ng 30,4 gam hn hp X gm Fe2O3 v FeO nung nng trong mt thi gian thu c hn hp cht rn Y. Ha tan ht Y trong HNO3 va c dung dch Z. Nhng thanh ng vo dung dch Z n phn ng hon ton thy khi lng thanh ng gim 12,8 gam. Phn trm khi lng ca cc cht trong hn hp X ln lt bng A. 33,3% v 66,7%. B. 61,3% v 38,7%.

C. 52,6% v 47,4%. D. 75% v 25%.

19. Ha tan hon ton m gam Fe3O4 trong dung dch HNO3, ton b lng kh NO thot ra em trn vi lng O2 va hn hp hp th hon ton trong nc c dung dch HNO3. Bit th tch oxi tham gia vo qu trnh trn l 336 ml ( ktc). Gi tr ca m l A. 34,8 gam. B. 13,92 gam. C. 23,2 gam. D. 20,88 gam.

20. Thi t t V lt hn hp kh CO v H2 c t khi hi so vi H2 l 7,5 qua mt ng s ng 16,8 gam hn hp 3 oxit CuO, Fe3O4, Al2O3 nung nng. Sau phn ng thu c hn hp kh v hi c t khi so vi H2 l 15,5. Dn hn hp kh ny vo dung dch Ca(OH)2 d thy c 5 gam kt ta. Th tch V ( ktc) v khi lng cht rn cn li trong ng s ln lt l

A. 0,448 lt; 16,48 gam. C. 1,568 lt; 15,68 gam

B. 1,12 lt; 16 gam. D. 2,24 lt; 15,2 gam.

III. P N

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

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


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PHNG PHP BO TON KHI LNG PHNG PHP GII

1. Ni dung phng php

- p dng nh lut bo ton khi lng (BTKL): Tng khi lng cc cht tham gia phn ng 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 H2SO4 c, 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]

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.

mui kim loi anion to mui

rn oxit


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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 tm m = 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 n

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

mdung dch = mK + OH2m - 2Hm = 3,9 + 36,2 - 0,05 u2 = 40 gam

C%KOH = 40

560,1u100u % = 14% p n C

V d 2: in phn dung dch cha hn hp CuSO4 v KCl vi in cc tr n 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 o Cu p + Cl2 n + K2SO4 (1)

0,01 m0,01

Dung dch sau in phn ho tan c MgO L dung dch axit, chng t sau phn ng (1)

CuSO4 d

2CuSO4 + 2H2O o 2Cu p + O2 n + H2SO4 (2)

n + 2O

n = 22400480

= 0,02 (mol)

H2SO4 + MgO o MgSO4 + H2O (3)

0,02 m0,02 (mol)

Cl2

trc sau

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


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mdung dch gim = mCu + 2Cl

m + 2O

m = 0,03 64u + 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 o BaCO3 p + 2NaCl

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

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

Na2CO3 + H2SO4 o Na2SO4 + CO2 + H2O

0,02 0,02

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

C% = 100

10607,0 u%100u = 7,42%

LBTKL: mdd sau cng = 50 + 100 + 50 - m p - m

= 50 + 100 + 50 - 0,05.197 - 0,02.44 = 189,27 gam

p n B

V d 4: X l mt D- 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 o HOOC -R-NH3Cl

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

Maminoxit = 01,089,0

= 89

Mt khc X l D-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.

H2SO4 BaCl2

Na2CO3

Na2CO3

CO2


10

Gii:

2 OHR + 2Na o 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 =

4,06,15

= 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 =

405,06,15

= 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 trng i hc, Cao ng khi B, 2008)

Gii:

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

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 trng i hc, Cao ng khi B, 2008)

Gii:

RCOOH + KOH RCOOK + H2O

H2

H2

M

M

M

0,06 o 0,02 (mol)

[tuituhoc.com] tổng hợp phương pháp giải hóa 12

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