Phương pháp giải nhanh bài toán hóa học - Trắc nghiệm Vô cơ + Hữu cơ + 40 Đề thi - Nguyễn Minh Tuấn

NGUYN MINH TUN Gio vin trng THPT Chuyn Hng Vng

Gii thiu cc chuyn

PHNG PHP GII NHANH BI TP HA HC

Sch dnh cho hc sinh lp 10, 11, 12 Ni dung :

C s xy dng cc phng php gii bi tp ha hc Phn dng bi tp c trng i vi tng phng php Cc v d minh ha dnh cho tng i tng hc sinh lp 10, 11, 12 Cc bi tp p dng v p n hc sinh rn luyn v kim tra kt qu hc tp

Vit Tr, ngy 28 thng 03 nm 2011

Bn in gc r nt nht ch c ti hiu Photo Thanh B nh SN 59 (Gn cng trng CHV)

Photo Thanh Bnh T: 0914235215 hoc 01698001858 1

Li ni u

Cc em , sng trn i iu quan trng l phi hiu c ai l ngi yuthng mnh nht. Cc em th i tm cu tr li thng qua cu truyn di ynh!

Nhng chic v lon bia ca b(Dn tr) - Nm ti ln 10 tui, b mua v cho mt con bp b. l mn qu u tin trong i tinhn c khi kt thc lp 4 vi kt qu hc sinh gii.

B lm ngh li xe m, quanh nm i nng i ma ch khch kim thm thu nhp cho gia nhngoi ci ngh thu mua ng nt ca m. Tuy chy xe m, nh ng u thy nhng lon bia hay lonnc ngt nm ln lc l b dng xe nht, mc cho khch ngi sau xe t ra kh chu hay cm thngcho b. Nhiu hm c nhng v khch khi thy h nh ng ca b khen b l ngi chng t t, lngi b bit chm lo gia nh, khi thanh ton tin xe h trnh trng th ng thm cho b t tinv bo: Ti khm phc nhng ngi nh anh.

M ti vi vnh nn l t ti trn vai i quang gnh i t ng con hm mua nhm ng, nha v nhng th ngi ta chun b vt i mong bn li kim cht li.

Gia nh ti sng trong con hm cht chi nm ngoi th nh. B m cht chiu mi vn khng xyni cn nh ng hong che ma che nng, ch c cn lu che tm trong ci xm ng c . T cingy b m bn i cn nh tranh xiu vo, khng chng chi ni vi nhng c n bo di qu sng tm b ni khu chut ny, khun mt b ngy cng gy thm, i tay m ngy mt en smv nng.

Thnh ph lc ny l khng gian thch th i vi anh em chng ti, nhng vi b m l mt s vtln kim tng ng thu nhp nh nhoi, thm lng v lao lc. Tui th l ng ti cha h tmti.


Photo Thanh Bnh T: 0914235215 hoc 016980018582

Cuc sng thiu thn l th, nhng gia nh 5 thnh vin ca chng ti lun rn r ting ci a vs chm sc chu o, ht mc ca ba m.

Nm ti hc lp 12, trong mt ln cng bn b i hc v, tnh c thy b ang ch khch trn ngbt ng dng xe li trc mt ng v bia chng 20 lon, v khch y ngi tr n xe, b ly baonilon nht chng vo trc nhng con mt ngc nhin ca m bn. Ti mt, tai khi trong mbn c a ln ting bo: C phi b ca con P. y khng? Ti chy mt mch v nh , b lim bn, b li mt cht t ti, mc cm rt ngy th ri m mt khc nc n.

Ti b cm ti, nm l trn ging. M n d xt mi ti mi chu ni ra l v sao mnh khc. Khiti ni ra nhng iu , m m ti bt khc. M k cho ti nghe b tng ti con bp b hi hc lp4 l kt qu t vic nht lon bia, ri tin mua ci o nhn lc sinh nht, v ng eo tay, sch v ttc u l tin b ki cp t m c. V tt c tin chy xe m b u a cho m trang tri cucsng, duy ch c thu nhp t i t nhng lon bia l b gi li i n lc cn thit s mua qu bnhcho ti.

Tt c nhng th con c u l t lon bia y con . M ti bo nh th. Nghe xong nhng cu nitrong nc mt ca m, ti khc nh cha bao gi c khc ri mong b i chuyn xe m cuicng v nh sm, ti s chy o ra m b v bo: B i con thng b nhiu, con c li vi b.

Yn M Sn



MC LC

Trang

Phn 1: Gii thiu cc chuyn phng php gii nhanh bi tp ha hc 4-196Chuyn 1 : Phng php ng cho 4

Chuyn 2 : Phng php t chn lng cht 25

Chuyn 3 : Phng php bo ton nguyn t 40

Chuyn 4 : Phng php bo ton khi lng 52

Chuyn 5 : Phng php tng gim khi lng, s mol,th tch kh 71

Chuyn 6 : Phng php bo ton electron 88

Chuyn 7 : Phng php quy i 123

Chuyn 8 : Phng php s dng phng trnh ion rt gn 135

Chuyn 9 : Phng php bo ton in tch 159

Chuyn 10 : Phng php s dng cc gi tr trung b nh 170

Chuyn 11 : Gii thiu mt s cng thc tnh nhanh s ng phn v gii nhanh bi tp ha hc 187

Phn 2 : p n 197 - 199



PHN 1: GII THIU CC CHUYN PHNG PHP GII NHANH BI TP HA HCCHUYN 1 : PHNG PHP NG CHO

I. Nguyn tc :- Cc gi tr trung bnh nh : Khi lng mol trung bnh; s nguyn t cacbon trung bnh; s

nguyn t hiro trung bnh; s lin kt pi trung bnh; nng mol trung bnh; nng % trungbnh; s khi trung bnh ca cc ng v lun c mi quan h vi khi lng mol; s nguyn tcacbon; s nguyn t hiro; s lin kt pi; nng mol; nng %; s khi ca cc cht hocnguyn t bng cc ng cho.

- Trong phn ng axit baz : Th tch ca dung dch axit, baz ; nng mol ca H+, OH- banu v nng mol ca H+, OH- d lun c mi quan h vi nhau bng cc ng cho.II. Cc trng hp s dng s ng cho1. Trn ln hai cht kh, hai cht tan hoc hai cht rn khng tc dng vi nhau

Ta c s ng cho :

BA AB B A

M M n Vn V M M

Trong :- nA, nB l s mol ca : Cc cht A, B hoc cc ng v A, B ca mt nguyn t ha hc.- VA, VB l th tch ca cc cht kh A, B.- MA, MB l khi lng mol ca : Cc cht A, B hoc s khi ca cc ng v A, B ca mt

nguyn t ha hc.

- M l khi lng mol trung bnh ca cc cht A, B hoc s khi trung bnh ca cc ng v A, Bca mt nguyn t ha hc.2. Trn ln hai dung dch c cng cht tan :

- Dung dch 1 : c khi lng m1, th tch V1, nng C1 (nng phn trm hoc nng mol), khi lng ring d1.

- Dung dch 2 : c khi lng m2, th tch V2, nng C2 (C2 > C1 ), khi lng ring d2.- Dung dch thu c : c khi lng m = m1 + m2, th tch V = V1 + V2, nng C

(C1 < C < C2) v khi lng ring d. S ng cho v cng thc tng ng vi mi trng hp l :a. i vi nng % v khi lng :

m1 C1 |C2 - C|C

m2 C2 |C1 - C| (1)|CC|

|CC|m

m

1

2

2

1

n M M MBA AM

n M M MB B A



b. i vi nng mol/lt :

V1 C1 |C2 - C|C

V2 C2 |C1 - C| (2)|CC|

|CC|VV

1

2

2

1

c. i vi khi lng ring :

V1 d1 |d2 - d|d

V2 d2 |d1 - d| (3)|dd|

|dd|VV

1

2

2

1

Lu : Mt s cng thc lin quan n bi ton c cn, pha long dung dch- Dung dch 1 : c khi lng m1, th tch V1, nng C1 (nng phn trm hoc nng

mol).- Sau khi c cn hay pha long dung dch bng nc, dung dch thu c c khi lng

m2 = m1 2H Om ; th tch V2 = V1 2H OV nng C (C1 > C2 hay C1 < C2 ). i vi nng % v khi lng :

mct = m1C1 = m2C2 1 22 1

m Cm C

i vi nng mol/lt :

nct = V1C1 = V2C2 1 22 1

V CV C

3. Phn ng axit - baza. Nu axit d : Ta c s ng cho :

b dA

B b d

OH + HVV H H

- VA, VA l th tch ca dung dch axit v baz.- b

OH l nng OH- ban u.- b

H , dH l nng H+ ban u v nng H+ d.b. Nu baz d : Ta c s ng cho :

V H O H HA d b bH d

V O H H HB d b b

V H O H O HA d b bO H d

V O H H O HB d b b



b dA

B b d

OH OHVV H + OH

- VA, VA l th tch ca dung dch axit v baz.- b

OH , dOH l nng OH- ban u v nng OH- d.- b

H l nng H+ ban u.III. Cc v d minh ha

Dng 1: Pha trn hai dung dch c cng cht tan hoc pha nc vo dung dchcha 1 cht tan

Phng php gii Nu pha trn hai dung dch c nng phn trm khc nhau th ta dng cng thc :

(1)|CC||CC|

m

m

1

2

2

1

Nu pha trn hai dung dch c nng mol khc nhau th ta dng cng thc :

(2)|CC||CC|

VV

1

2

2

1

Nu pha trn hai dung dch c khi l ng ring khc nhau th ta dng cng th c :

(3)|dd||dd|

VV

1

2

2

1

Cc v d minh ha i vi dng 1 Dnh cho hc sinh lp 10V d 1: thu c dung dch HCl 25% cn ly m 1 gam dung dch HCl 35% pha vi m2 gam dungdch HCl 15%. T l m1/m2 l :

A. 1:2. B. 1:3. C. 2:1. D. 1:1.Hng dn gii

p dng s ng cho ta c :21

2 1

C Cmm C C

1

2

15 25m 10 1m 35 25 10 1

.p n D.



V d 2: pha c 500 ml dung dch NaCl 0,9M cn ly V ml dung dch NaCl 3M pha vi n cct. Gi tr ca V l :

A. 150 ml. B. 214,3 ml. C. 285,7 ml. D. 350 ml.Hng dn gii

Gi th tch ca dung dch NaCl (C1 = 3M) v th tch ca H2O (C2 = 0M) ln lt l V1 v 2V .


2 1

C CV 0,9 0 0,9V C C 3 0,9 2,1

V1 = 0,9 .5002,1 0,9 = 150 ml.

p n A. Ch : Cng c th p dng cng thc pha long dung dch :

1 2

2 1

V CV C

2 211

V C 500.0,9V 150C 3

ml.

V d 3: Cn bao nhiu lt axit H2SO4 (D = 1,84 gam/ml) v bao nhiu lt nc ct pha thnh 9 ltdung dch H2SO4 c D = 1,28 gam/ml ? Bit khi lng ring ca nc l 1 gam/ml

A. 2 lt v 7 lt. B. 3 lt v 6 lt. C. 4 lt v 5 lt. D. 6 lt v 3 lt.Hng dn gii

S ng cho:

22 4

H O

H SO

V 0,56 2V 0,28 1

Mt khc2H O

V +2 4H SO

V = 9

2H O

V = 6 lt v2 4H SO

V = 3 lt.

p n B.

2

2 4

HO: 1 |1,84 1,28| 0,561,28

HSO : 1,84 |1,28 1| 0,28



Dng 2 : Ha tan mt kh (HCl, NH3), mt oxit (SO3, P2O5) hoc mt tinh th(CuSO4.5H2O, FeSO4.7H2O, NaCl) vo dung dch cha mt cht tan c mtdung dch mi cha mt cht tan duy nht

Phng php gii Trng hp ha tan tinh th vo dung dch th ta coi tinh th l mt dung dch c nng

phn trm l : C% = chat tantinh the

m.100%

m, sau p dng cng thc :

(1)|CC||CC|

m

m

1

2

2

1

Trng hp ha tan kh hoc oxit vo dung dch th ta vit phng trnh phn ng ca khhoc oxit vi nc (nu c) trong dung dch , sau tnh khi lng ca cht tan thu c. Coi

oxit l mt dung dch cht tan c nng phn trm l : C% = chat tanoxit

m.100%

m(C% >100%), sau

p dng cng thc :

(1)|CC||CC|

m

m

1

2

2

1

Cc v d minh ha i vi dng 2 Dnh cho hc sinh lp 10V d 4: Cn ly bao nhiu gam tinh th CuSO4.5H2O v bao nhiu gam dung dch CuSO4 8% pha thnh 280 gam dung d ch CuSO4 16% ?

A. 180 gam v 100 gam. B. 330 gam v 250 gam.C. 60 gam v 220 gam. D. 40 gam v 240 gam.

Hng dn gii

4 2

160

250

CuSO .5H O

Ta coi CuSO4.5H2O nh l dung dch CuSO4 c: C% = 160.100250 64%.

Gi m1 l khi lng ca CuSO4.5H2O v m2 l khi lng ca dung dch CuSO4 8%. Theo s ng cho :

21

2 1

C Cmm C C

1

2

m 8 1m 48 6

. Mt khc m1 + m2 = 280 gam.

Vy khi lng CuSO4.5H2O l : m1 =280

.11 6 = 40 gam m2 = 280 40 = 240 gam.

p n D.



V d 5: Ha tan 200 gam SO3 vo m2 gam dung dch H2SO4 49% ta c dung dch H2SO4 78,4%.Gi tr ca m2 l :

A. 133,3 gam. B. 146,9 gam. C. 272,2 gam. D. 300 gam.Hng dn gii

Phng trnh phn ng : SO3 + H2O H2SO4

200 gam SO3 98.20080 = 245 gam H2SO4

Coi SO3 l dung dch H2SO4 c nng phn trm l : C% =245

.100% 122,5%200

Gi m1, m2 ln lt l khi lng ca SO3 v dung dch H2SO4 49% cn ly.p dng s ng cho ta c :

21

2 1

C Cmm C C

1

2

49 78,4m 29,4m 122,5 78,4 44,1

2 44,1m .20029,4 = 300 gam.

p n D.

Dng 3: Xc nh % s nguyn t (% s mol nguyn t) ca cc ng v ca mtnguyn t ha hc

Phng php gii S dng cng thc :

BA

B A

M M nn M M

Trong :- nA, nB l s mol ca : Cc ng v A, B ca mt nguyn t ha hc.- MA, MB l khi lng mol ca : S khi ca cc ng v A, B ca mt nguyn t ha hc.

- M l s khi trung bnh ca cc ng v A, B ca mt nguyn t ha hc.

Cc v d minh ha i vi dng 3 Dnh cho hc sinh lp 10V d 6: Nguyn t khi trung bnh ca clo l 35,5. Trong t nhin clo c hai ng v bn l 35Cl v37 Cl . Thnh phn % s nguyn t ca 35Cl l :

A. 75. B. 25. C. 80. D. 20.Hng dn gii


35

Cl

Cl

n 35,5 35 1n 37 35,5 3



% 35 Cl = 3 .100%4

= 75%.

p n A.V d 7: Trong nc, hiro tn ti hai ng v 1H v 2H. Bit nguyn t khi trung bnh ca hiro l1,008; ca oxi l 16. S nguyn t ng v ca 2H c trong 1 ml nc nguyn cht (d = 1 gam/ml) l:

A. 5,53.1020. B. 5,35.1020. C. 3,35.1020. D. 4,85.1020.Hng dn gii

p dng s ng cho ta c : 12

H

H

n 2 1,008 0,992 99,2%n 1 1,008 0,008 0,8%

S mol nc l : 1

18 mol ; S mol H l : 2. 1

18 ; S mol 2H l : 2. 1

18. 0,8%

S nguyn t ng v 2H trong 1 gam nc l : 2. 118

. 0,8%.6,02.1023 = 5,35.1020.

p n B.

Dng 4: Xc nh % v s mol (phn trm v th tch) ca hn hp cht kh hocphn trm v s mol ca hn hp cht rn

Phng php gii S dng cng thc :

BA A

B B A

M M n Vn V M M

Trong :- nA, nB l s mol ca cc cht A, B.- VA, VB l th tch ca cc cht kh A, B.- MA, MB l khi lng mol ca cc cht A, B.

- M l khi lng mol trung bnh ca cc cht A, B.

Cc v d minh ha i vi dng 4 Dnh cho hc sinh lp 10V d 8: Hn hp hai kh NO v N2O c t khi hi i vi hiro bng 16,75. T l s mol hoc t lth tch ca NO v N2O trong hn hp ln lt l :

A. 1 : 3. B. 3 : 1. C. 1 : 1. D. 2 : 3.Hng dn gii

2(NO,N O)M =16,75.2 =33,5

p dng s ng cho ta c : 2N ONO

V 33,5 30 1V 44 33,5 3

p n B.



V d 9: t chy hon ton m gam cacbon trong V lt kh oxi (ktc), thu c hn hp kh A c tkhi so vi oxi l 1,25. Xc nh thnh phn phn trm theo th tch ca cc kh trong A.

Hng dn gii

Theo gi thit : AM 1, 25.32 40 C th xy ra cc trng hp sau : Trng hp 1 : Hn hp A gm hai kh l O2 d v CO2 p dng s ng cho ta c :

2 2

2 2

CO CO

O O

V n 40 32 8 2V n 44 40 4 1

%CO2 = 2 .100% 66,67%2 1 ; %O2 = (100 66,67)% = 33,33%.

Trng hp 2 : Hn hp A gm hai kh l CO v CO2 p dng s ng cho ta c :

2 2CO CO

CO CO

V n 40 28 12 3V n 44 40 4 1

%CO2 = 3 .100% 75%3 1 ; %CO = (100 75)% = 25%.

V d 10: Mt hn hp gm O2, O3 iu kin tiu chun c t khi hi vi hiro l 19,2. Thnhphn % v khi lng ca O3 trong hn hp l :

A. 66,67%. B. 33,33%. C. 35%. D. 75%.Hng dn gii

p dng s ng cho :

3 32 2

O O

O O

V n 19,2.2 32 6,4 2V n 48 19,2.2 9,6 3

3 2.48%O .100 50%2.48 3.32 .

p n B.V d 11: Hn hp gm NaCl v NaBr. Cho hn hp tc dng vi dung dch AgNO 3 d th to ra ktta c khi lng bng khi lng ca AgNO3 tham gia phn ng. Thnh phn % theo khi lngca NaCl trong hn hp u l :

A. 25,84%. B. 27,84%. C. 40,45%. D. 27,48%.Hng dn gii

Cc phng trnh phn ng :NaCl + AgNO3 AgCl + NaNO3 (1)NaBr + AgNO3 AgBr + NaNO3 (2)

Khi lng kt ta (gm AgCl v AgBr) bng khi lng AgNO3, do khi lng mol trungbnh ca hai mui kt ta

3A g C l A g B r A g N OM M 1 7 0 (v 3A g C l A g B r A g N On n ). Do :

Cl ,BrM = 170 108 = 62. Khi lng mol trung bnh ca hai mui ban u NaCl,NaBrM = 23 + 62 = 85

p dng s ng cho ta c :



NaCl

NaBr

n 103 85 18n 85 58,5 26,5

NaC l

N aBr N aC l

m 18.58, 5% NaC l .100% 27, 84%m m (26, 5 .103) (18 .58, 5).

p n B. Dnh cho hc sinh lp 11V d 12: Cho hn hp gm N2, H2 v NH3 c t khi so vi hiro l 8. Dn hn hp i qua dungdch H2SO4 c, d th th tch kh cn li mt na. Thnh phn phn trm (%) theo th tch ca mikh trong hn hp ln lt l :

A. 25% N2, 25% H2 v 50% NH3. B. 25% NH3, 25% H2 v 50% N2.C. 25% N2, 25% NH3 v 50% H2. D. 15% N2, 35% H2 v 50% NH3.

Hng dn giiKhi i qua dung dch H2SO4 c, d ton b NH3 b hp th, do thnh phn ca NH3 l 50%.

2 2 3(N ,H ,NH )M = 8.2 = 16p dng s ng cho ta c :

2, 23

2 2

(N H )NH

(H ,N )

n 16 M 1n 17 16 1

2, 2(N H )M = 15

2, 2(N H )M = 15 l khi lng mol trung bnh ca hn hp ca N2 v H2. Tip tc p dng s ng cho ta c :

2

2

H

N

n 28 15 1n 15 2 1

%N2 = %H2 = 25%.

p n A.

Dng 5 : Xc nh nng mol, th tch ca dung dch axit, baz hoc t l thtch ca chng trong phn ng gia cc dung dch axit v dung dch baz

Phng php gii Nu axit d ta s dng cng thc :

b dA

B b d

OH + HVV H H

Nu baz d ta s dng cng thc :

b dA

B b d

OH OHVV H + OH



Cc v d minh ha i vi dng 5 Dnh cho hc sinh lp 11V d 13: Trn 250 ml dung dch cha hn hp HCl 0 ,08M v H2SO4 0,01 M vi 250 ml dung dchNaOH aM thu c 500 ml dung dch c pH = 12. Gi tr a l :

A. 0,13M. B. 0,12M. C. 0,14M. D. 0.10M.Hng dn gii

Nng H+ ban u l : 0,08 + 0,01.2 = 0,1M. Nng OH- ban u l : aM.

Dung dch sau phn ng c pH = 12, suy ra OH - d, pOH = 2. Nng OH- d l : 10-2 = 0,01M.

p dng s ng cho cho trng hp OH- d ta c :

b dA

B b d

OH OHVV H + OH

=a 0,01 1

a 0,120,1 0,01 1 .

p n B.V d 14: Trn ln 3 dung dch H2SO4 0,1M, HNO3 0,2M v HCl 0,3M vi nhng th tch bngnhau thu c dung dch A. Ly 300 ml dung dch A cho phn ng vi V lt dung dch B gmNaOH 0,2M v KOH 0,29M thu c dung dch C c pH = 2. G i tr V l :

A. 0,134 lt. B. 0,214 lt. C. 0,414 lt. D. 0,424 lt.Hng dn gii

Nng H+ ban u l : (0,1.2.0,1 + 0,2.0,1 + 0,3.0,1) : 0,3 = 0,73

M.

Nng OH- ban u l : (0,2 + 0,29) = 0,49M. Dung dch sau phn ng c pH = 2, suy ra H + d. Nng H+ d l : 10-2 = 0,01M.

p dng s ng cho cho trng hp H+ d ta c :

b dA

B b d

OH + HVV H H

=0, 49 0,01 0,3 V 0,1340,7 V0,01

3

.

p n A.V d 15: Dung dch A gm HCl 0,2M ; HNO3 0,3M ; H2SO4 0,1M ; HClO4 0,3M, dung dch Bgm KOH 0,3M ; NaOH 0,4M ; Ba(OH)2 0,15M. Cn trn A v B theo t l th tch l bao nhiu c dung dch c pH = 13 ?

A. 11: 9. B. 9 : 11. C. 101 : 99. D. 99 : 101.Hng dn gii

Nng H+ ban u l : (0,2 + 0,3 + 0,1.2 + 0,3) = 1M. Nng OH- ban u l : (0,3 + 0,4 + 0,15.2) = 1M. Dung dch sau phn ng c pH = 13 pOH = 1 Nng OH- d l : 10-1 = 0,1M.

p dng s ng cho cho trng hp OH- d ta c :

b dA

B b d

OH OHVV H + OH

=1 0,1 91 0,1 11 .

p n B.



Dng 6: Xc nh % v s mol (% v th tch - i vi cc cht kh) ; % v khilng ca hn hp cc cht hoc xc nh cng thc ca mt hp cht hay mtn cht

Phng php gii S dng cng thc ng cho :

2A A

B B 1

n n n Vn V n n

hocBA A

B B A

M M n Vn V M M

Trong :- nA, nB l s mol ca cc cht A, B.- VA, VB l th tch ca cc cht A, B.- n1, n2 l s nguyn t cacbon ca cc cht hu c A, B.

- n l s cacbon trung bnh ca cc cht hu c A, B.- MA, MB l khi lng mol ca cc cht A, B.

- M l khi lng mol trung bnh ca cc cht A, B.

Cc v d minh ha i vi dng 6 Dnh cho hc sinh lp 11V d 16: Hn hp kh A gm etan v propan. t chy hn hp A thu c kh CO2 v hi H2Otheo t l th tch 11:15. a. Thnh phn % theo th tch ca hn hp l :

A. 18,52% ; 81,48%. B. 45% ; 55%. C. 28,13% ; 71,87%. D. 25% ; 75%. b. Thnh phn % theo khi lng ca hn hp l :

A. 18,52% ; 81,48%. B. 45% ; 55%. C. 28,13% ; 71,87%. D. 25% ; 75%.Hng dn gii

a. t CTPT trung bnh ca etan v propan l :n 2n 2C H

Phn ng chy :n 2n 2C H + O2 n CO2 + ( n +1)H2O

Theo gi thit ta c : n 1 15 n 2,7511n

p dng s ng cho cho s nguyn t cacbon trung bnh ca hai cht ta c :

2 6

3 8

C H

C H

n 3 2,75 0,25 25%n 2,75 2 0,75 75%

p n B.b. Thnh phn phn trm v khi lng ca cc cht l :

%C2H6 =0,25.30

.100%0,25.30 0,75.44

18,52% %C3H8 = 81,48%.p n A.



V d 17: Hn hp kh A iu kin tiu chun gm hai olefin. t chy 7 th tch A cn 31 thtch O2 (ktc). Bit olefin cha nhiu cacbon chim khong 40% 50% th tch hn hp A. Cngthc phn t ca hai elefin l :

A. C2H4 v C3H6. B. C3H6 v C4H8. C. C2H4 v C4H8. D. A hoc C ng.Hng dn gii

t cng thc trung bnh ca hai olefin l :n 2nC H

Phng trnh phn ng :

n 2nC H +3n2

O2 n CO2 + n H2O (1)

Th tch: 7 7. 3n2

Theo (1) v gi thit ta c : 7. 3n2

= 31 n 2,95 Trong hai olefin phi c mt cht l C2H4 v cht cn li c cng thc l n 2nC H

V olefin cha nhiu cacbon chim khong 40% 50% th tch hn hp A nn

n 2n

2 4 n 2n

C H

C H C H

n40% 50%

n n (2)

p dng s ng cho i vi s cacbon ca hai olefin ta c :n 2n

2 4

C H

C H

n 2,95 2n n 2,95

n 2n

2 4 n 2n

C H

C H C H

n 2,95 2 0,95n n n 2,95 2,95 2 n 2

(3)

Kt hp gia (2) v (3) ta c : 3,9 < n < 4,375 n = 4p n C.V d 18: Cn trn hai th tch metan vi mt th tch ng ng X ca metan thu c hn hpkh c t khi hi so vi hiro bng 15. X l :

A. C3H8. B. C4H10. C. C5H12. D. C6H14.Hng dn gii

p dng s ng cho :4CH X

XX

V M 30 2 M 58V 30 16 1

14n + 2 = 58 n = 4 X l C4H10.

p n B.V d 19: Hn hp gm hirocacbon X v oxi c t l s mol tng ng l 1:10. t chy hon tonhn hp trn thu c hn hp kh Y. Cho Y qua dung dch H2SO4 c, thu c hn hp kh Z c tkhi i vi hiro bng 19. Cng thc phn t ca X l :


ZM 38 Z gm CO2 v O2 p dng s ng cho ta c : 2

2

O

CO

n 44 38 1n 38 32 1



Phng trnh phn ng :

CxHy + (x+ y4 ) O2 xCO2 +y2

H2O

b: 1 10

p: 1 (x+ y4

) x

sp: 0 10 (x+ y4

) x

10 (x+ y4

) = x 40 = 8x + y x = 4 v y = 8p n C.V d 20: Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung nng. Saukhi phn ng hon ton, khi lng cht rn trong bnh gim 0,32 gam. Hn hp hi thu c c tkhi i vi hiro l 15,5. Gi tr ca m l :

A. 0,92. B. 0,32. C. 0,64. D. 0,46.Hng dn gii

t cng thc phn t ca ancol no, n chc X l : CnH2n + 2OPhng trnh phn ng :

CnH2n + 2O + CuO CnH2nO + H2O + Cu (1)mol : x x x x x

Khi lng cht rn gim = mCuO mCu = 80x 64x = 0,32 x = 0,02Hn hp hi gm CnH2nO v H2O c khi lng mol trung bnh l : 15,5.2 = 31 gam/mol.p dng s ng cho ta c :

n 2n

2

C H O

H O

n 31 18 13 1n 2

n (14n 16) 31 14n 15 1

Vy khi lng ca X l : m = (14n + 18).0,02 = (14.2 + 18).0,02 = 0,92 gam.p n A.Nhn xt : Bi tp ny nn lm theo phng php bo ton khi lng th ngn gn hn (Xem ligii chuyn phng php bo ton khi lng). Dnh cho hc sinh lp 12V d 21: Cho hn hp X gm 2 este c CTPT l C4H8O2 v C3H6O2 tc dng vi NaOH d thuc 6,14 gam hn hp 2 mui v 3,68 gam ancol B duy nht c t khi so vi oxi l 1,4375. S gamca C4H8O2 v C3H6O2 trong A ln lt l :

A. 3,6 gam v 2,74 gam. B. 3,74 gam v 2,6 gam.C. 6,24 gam v 3,7 gam. D. 4,4 gam v 2,22 gam.

Hng dn giiMB = 1,4375.32 = 46 ancol B l C2H5OH. nB = nmui = 3,6846 = 0,08 (mol) mui

6,14M 76,75 (gam / mol)0,08 p dng s ng cho ta c :

3

HCOONa

CH COONa

n 82 76,75 3n 76,75 68 5



3 2 5 4 8 23

2 5 3 6 2

CH COOC H C H OCH COONaHCOOC H C H OHCOONa

n 0,05 m 4,4 gamn 0,05n 0,03 m 2,22 gamn 0,03

p n D.V d 22: Cho hn hp X gm hai cht hu c c cng cng thc phn t C2H7NO2 tc dng va vi dung dch NaOH v un nng, thu c dung dch Y v 4,48 lt hn hp Z ( ktc) gm hai kh(u lm xanh giy qu m). T khi hi ca Z i vi H2 bng 13,75. C cn dung dch Y thu ckhi lng mui khan l :

A. 16,5 gam. B. 14,3 gam. C. 8,9 gam. D. 15,7 gam.Hng dn gii

Hn hp Z gm 2 kh c tnh baz l NH3 v CH3NH2. Vy hn hp X gm CH3COONH4 vHCOOH3NCH3

Phng trnh phn ng :CH3COONH4 + NaOH CH3COONa + NH3 + H2O (1)

mol: x x xHCOOH3NCH3 + NaOH HCOONa + CH3NH2 + H2O (2)

mol: y y y p dng s ng cho ta c:

3

3 2

NH

CH NH

n 31 13,5.2 1n 13,5.2 17 3

Theo (1), (2) v gi thit ta c h :x y 0,2

x 0,05x 1

y 0,15y 3

m = 68.0,15 + 82.0,05 = 14,3 gam.

p n B.Nhn xt : Bi tp ny nn lm theo phng php bo ton khi lng th ngn gn hn (Xem ligii chuyn phng php bo ton khi lng).V d 23: Cho 3,6 gam hn hp gm K v mt kim loi kim A tc dng va ht vi n c, thu c2,24 lt H2 0,5 atm v 0oC. Bit s mol kim loi (A) trong hn hp ln hn 10% tng s mol 2 kimloi. A l kim loi :

A. K. B. Na. C. Li. D. Rb.Hng dn gii

Gi khi lng mol trung bnh ca hai kim loi kim l : M Phng trnh phn ng :

2K + 2H2O 2KOH + H2 (1)2A + 2H2O 2AOH + H2 (2)

Theo cc phn ng ta thy :2(K,A) H

PVn 2.n 2. 0,1

RT mol M = 3,6

0,1= 36 gam/mol.

V M < MK nn M > MA A c th l Na hoc Li. p dng s ng cho ta c :

A

K A A

n 39 36 3n 36 M 36 M

A A

hh A K A

n n 3n n n 3 36 M

(3)



Theo gi thit nA > 10%.nhh Ahh

n 10% 0,1n

(4) T (3) v (4) MA > 9 A l Nap n B.V d 24: A l qung hematit cha 60% Fe2O3. B l qung manhetit cha 69,6% Fe 3O4. Trn mA tnqung A vi mB tn qung B thu c qung C, m t 1 tn qung C c th iu ch c 0,5 tngang cha 4% cacbon. T l mA/mB l :

A. 5:2. B. 3:4. C. 4:3. D. 2:5.Hng dn gii

S kg Fe c trong 1 tn ca mi qung l :

Qung A cha: 60 1121000 420100 160

kg

Qung B cha: 69,6 1681000 504100 232

kg

Qung C cha: 4500 1 480100

kg

S ng cho:mA 420 |504 - 480| = 24

480mB 504 |420 - 480| = 60

52

6024

m

m

B

A p n D.

Dng 7: Tnh nng mol ; nng % ; th tch ca n c cn pha thm hay ccn bt ; th tch ca dung dch cht tan tr c hay sau khi pha long, c cn dungdch

Phng php giiS dng cc cng thc pha long, c cn dung dch Lu : Khi pha long hay c cn dung dch th lng cht tan khng i nn :

- i vi nng % v khi lng ta c :

mct = m1C1 = m2C2 1 22 1

m Cm C

- i vi nng mol/lt ta c :

nct = V1C1 = V2C2 1 22 1

V CV C



Cc v d minh ha i vi dng 7 Dnh cho hc sinh lp 10, 11V d 25: Lm bay hi 500 ml dung d ch cht A 20% (D = 1,2 g/ml) ch c n 300 gam dung dch.Nng % ca dung dch ny l :

A. 30%. B. 40%. C. 50%. D. 60%.Hng dn gii

p dng cng thc c cn, pha long dung dch ta c :1 2

2 1

m Cm C

1 122

m C 500.1, 2.20%C 40%m 300

.p n B.V d 26: pha c 500 ml (V2 = 500) dung dch KCl 0,9M cn ly V ml (V 1) dung dch KCl 3Mpha vi nc ct. Gi tr ca V l :

A. 150 ml. B. 214,3 ml. C. 285,7 ml. D. 350 ml.Hng dn gii


2 1

V CV C

2 211

V C 500.0,9V 150C 3

ml.p n A.V d 27: S ml H2O cn thm vo 1 lt dung dch HCl 2M thu c dung dch mi c nng 0,8M l :

A. 1,5 lt. B. 2 lt. C. 2,5 lt. D. 3 lt.Hng dn gii


2 1

V CV C

1 122

V C 1.2V 2,5C 0,8

lt.

M V2 =V1 + 2H OV 2H OV = 2,5 1 = 1,5 lt.p n A.V d 28: Pha long dung dch HCl c pH = 3 bao nhiu ln c dung dch mi c pH = 4 ? A. 5. B. 4. C. 9. D. 10.

Hng dn giip dng cng thc c cn, pha long dung dch ta c :

1 2

2 1

V CV C

3

1 1 12 14

2

V C V 10V 10VC 10

Vy phi pha long dung dch HCl c pH = 3 10 ln c dung dch HCl c pH = 4p n D.V d 29: Pha long 1 lt dung d ch NaOH c pH = 13 bng bao nhi u lt nc c dung dchmi c pH = 11 ? A. 9. B. 99. C. 10. D. 100.

Hng dn giiDung dch NaOH c pH = 13 pOH = 1 C1 = [OH-] = 10-1

Dung dch NaOH sau khi pha long c pH = 11 pOH = 3 C2 = [OH-] = 10-3




2 1

V CV C

1

1 12 3

2

V C 1.10V 100C 10

lt 2H O 2 1V V V 100 1 99 lt.

p n B.

IV. Cc bi tp p dng1. Bi tp dnh cho hc sinh lp 10Cu 1: Nguyn t khi trung bnh ca ng 63,54. ng c hai ng v bn: 6329 Cu v 6529 Cu . Thnhphn % s nguyn t ca 6329 Cu l :

A. 73,0%. B. 34,2%. C. 32,3%. D. 27,0%.Cu 2: Nguyn t khi trung bnh ca clo l 35,5. Trong t nhin clo c hai ng v bn l 35Cl(75%) v A Cl (25%). Gi tr ca A l :

A. 36. B. 37. C. 38. D. 39.Cu 3: Nguyn t khi trung bnh ca clo l 35,5. Clo trong t nhin c 2 ng v l 3517 Cl v 3717 Cl .Phn trm v khi lng ca 3717 Cl cha trong HClO4 (vi hiro l ng v 11H , oxi l ng v 168 O ) lgi tr no sau y ?

A. 9,20%. B. 8,95%. C. 9,67%. D. 9,40%.Cu 4: Trong nc, hiro tn ti hai ng v 1H v 2H. Bit nguyn t khi trung bnh ca hiro l1,008; ca oxi l 16. S nguyn t ng v ca 2H c trong 1 ml nc nguyn cht (d = 1 gam/ml) l:

A. 5,53.1020. B. 5,35.1020. C. 3,35.1020. D. 4,85.1020.Cu 5: Ha tan m gam Al bng dung dch HNO3 long thu c hn hp kh NO v N2O c t khiso vi H2 bng 18,5. T l th tch kh NO v N2O trong hn hp ln lt l :

A. 2 : 3. B. 1 : 2. C. 1 : 3. D. 1 : 1.Cu 6: Mt hn hp gm H2, CO iu kin tiu chun c t khi hi vi hiro l 3,6. a. Thnh phn % v th tch ca CO trong hn hp l :

A. 66,67%. B. 20%. C. 35%. D. 75%. b. Thnh phn % v khi lng ca CO trong hn hp l :

A. 77,77%. B. 33,33%. C. 35%. D. 75%.Cu 7: Mt hn hp 104 lt (ktc) gm H 2 v CO c t khi hi i vi metan bng 1,5 th 2HV vVCO trong hn hp l :

A. 16 lt v 88 lt. B. 88 lt v 16 lt. C. 14 lt v 90 lt. D. 10 lt v 94 lt.Cu 8: Cho m gam hn hp bt Zn v Fe vo lng d dung dch CuSO4. Sau khi kt thc cc phnng, lc b phn dung dch thu c m gam bt rn. Thnh phn phn trm theo khi lng ca Zntrong hn hp bt ban u l :

A. 85,30%. B. 90,27%. C. 82,20%. D. 12,67%.Cu 9: S ml H2O cn thm vo 1 lt dung dch HCl 2M thu c dung dch mi c nng 0,8M l :

A. 1,5 lt. B. 2 lt. C. 2,5 lt. D. 3 lt.Cu 10: Th tch nc v dung dch CuSO4 2M cn pha c 100 ml dung dch CuSO4 0,4M lnlt l :

A. 50 ml v 50 ml. B. 40 ml v 60 ml. C. 80 ml v 20 ml. D. 20 ml v 80 ml.



Cu 11: Mt dung dch KOH nng 2M v mt dung dch KOH khc nng 0,5M. c dungdch mi nng 1M th cn phi pha ch v th tch gia 2 dung dch theo t l l :

A. 1 : 2 B. 2 : 1 C. 1 : 3 D. 3 : 1Cu 12: Cn trn V1 ml dung dch HCl 2M vi V2 ml dung dch HCl 0,5M thu c 300 ml dungdch HCl 1M. Gi tr V1, V2 ln lt l :

A. V1 = V2 = 150. B. V1 = 100, V2 = 200.C. V1 = 200, V2 = 100. D. V1 = 50, V2 = 250.

Cu 13: Mt dung dch HCl nng 35% v mt dung dch HCl khc c nng 15%. thu cdung dch mi c nng 20% th cn phi pha ch 2 dung dch ny theo t l khi lng l :

A. 1:3. B. 3:1. C. 1:5. D. 5:1.Cu 14: Khi lng dung dch NaCl 15% cn trn vi 200 gam dung dch NaCl 30% thu cdung dch NaCl 20% l :

A. 250 gam. B. 300 gam. C. 350 gam. D. 400 gam.Cu 15: Cn ly bao nhiu gam tinh th CuSO4.5H2O v bao nhiu gam dung d ch CuSO4 8% pha thnh 420 gam dung d ch CuSO4 16% ?

A. 140 gam v 280 gam. B. 330 gam v 90 gam.C. 60 gam v 360 gam. D. 180 gam v 240 gam.

Cu 16: Lng SO3 cn thm vo dung dch H2SO4 10% c 100 gam dung dch H2SO4 20%l:

A. 2,5 gam. B. 8,88 gam. C. 6,66 gam. D. 24,5 gam.Cu 17: Lm bay hi 500 ml dung d ch cht A 20% (D = 1,2 g/ml) ch cn 300 gam dung dch.Nng % ca dung dch ny l :

A. 30%. B. 40%. C. 50%. D. 60%.Cu 18: Trn V1 ml dung dch NaOH (d = 1,26 g/ml) vi V2 ml dung dch NaOH (d = 1,06 g/ml) thuc 1 lt dung dch NaOH (d = 1,16 g/ml). Gi tr V1, V2 ln lt l :

A. V1 = V2 = 500. B. V1 = 400, V2 = 600.C. V1 = 600, V2 = 400. D. V1 = 700, V2 = 300.

Cu 19: T 200 gam dung dch KOH 30% c dung dch 50% cn th m vo s gam KOH nguyncht l :

A. 70 gam. B. 80 gam. C. 60 gam. D. 90 gam.Cu 20: Th tch nc nguyn cht cn thm vo 1 lt dung dch H2SO4 98% (d = 1,84 g/ml) c dung dch mi c nng 10% l :

A. 14,192 ml. B. 15,192 ml. C. 16,192 ml. D. 17,192 ml.Cu 21: Cn cho m gam H2O vo 100 gam dung d ch H2SO4 90% c dung dch H2SO450%.Gi tr m l :

A. 90 gam. B. 80 gam. C. 60 gam. D. 70 gam.Cu 22: Trn 1 lt dung dch KCl C 1 M (dung dch A) vi 2 lt dung dch KCl C 2 M (dung dch B)c 3 lt dung dch KCl (dung dch C). Cho dung dch C tc dng va vi dung dch AgNO 3 thuc 86,1 gam kt ta. Nu C1 = 4C2 th C1 c gi tr l :

A. 1 M. B. 1,2 M. C. 1,4 M. D.1,5 M.Cu 23: Mt hn hp kh X gm SO 2 v O2 c t khi so vi metan bng 3. Thm V lt O2 vo 20 lthn hp X thu c hn hp Y c t khi so vi metan bng 2,5. Gi tr ca V l :

A. 20. B. 30. C. 5. D. 10.



2. Bi tp dnh cho hc sinh lp 11Cu 24: Pha long dung dch KOH c pH = 13 bao nhiu ln c dung dch mi c pH = 12 ? A. 5. B. 4. C. 9. D. 10.Cu 25: Pha long 1 lt dung d ch HCl c pH = 3 bng bao nhiu lt nc c dung dch mi cpH = 4 ? A. 9. B. 99. C. 10. D. 100.Cu 26: Hn hp Kh X gm N2 v H2 c t khi hi so vi He l 1,8. un nng X mt thi giantrong bnh kn c xc tc thu c hn hp kh Y c t khi h i so vi He l 2. Hiu sut ca phnng tng hp NH3 l :

A. 25%. B. 50%. C. 60%. D. 75%.Cu 27: Mt dung dch HNO3 nng 60% v mt dung dch HNO3 khc c nng 20%. c200 gam dung dch mi c nng 45% th cn phi pha ch v khi lng gia 2 dung dch HNO 360%, 20% ln lt l :

A. 75 gam ; 125 gam. B. 125 gam ; 75 gam.C. 80 gam ; 120 gam. D. 100 gam ; 100 gam.

Cu 28: Thm 150 ml dung dch KOH 2M vo 120 ml dung dch H3PO4 1M. Khi lng cc muithu c trong dung dch l :

A. 10,44 gam KH2PO4; 8,5 gam K3PO4. B. 10,44 gam K2HPO4; 12,72 gam K3PO4.C. 10,44 gam K2HPO4; 13,5 gam KH2PO4. D. 13,5 gam KH2PO4; 14,2 gam K3PO4.

Cu 29: Ha tan hon ton 12 gam hn hp Fe v Cu (t l mol 1:1) bng axit HNO 3, 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 caX i vi H2 bng 19. Gi tr ca V l :

A. 2,24. B. 4,48. C. 5,60. D. 3,36.Cu 30: Cho 6,12 gam Mg tc dng vi dung dch HNO3 thu c dung dch X ch c mt mui v hn hp kh Y gm NO v N2O c t khi hi i vi hiro bng 16,75. Th tch NO v N2O ( ktc)thu c ln lt l :

A. 2,24 lt v 6,72 lt. B. 2,016 lt v 0,672 lt.C. 0,672 lt v 2,016 lt. D. 1,972 lt v 0,448 lt.

Cu 31: Trn ln 3 dung dch H2SO4 0,1M, HNO3 0,2M v HCl 0,3M vi nhng th tch bng nhauthu c dung dch A. Ly 450 ml dung dch A cho phn ng vi V lt dung dch B gm NaOH0,2M v KOH 0,29M thu c dung dch C c pH = 2. Gi tr V l :

A. 0,201 lt. B. 0,321 lt. C. 0,621 lt. D. 0,636 lt.Cu 32: Trn 100 ml dung dch c pH = 1 gm HC l v HNO3 vi 100 ml dung dch NaOH nng aM thu c 200 ml dung dch c pH = 12. Gi tr ca a l ([H+][OH-] = 10-14) :

A. 0,15. B. 0,30. C. 0,03. D. 0,12.Cu 33: Trn 200 ml dung dch cha hn hp HCl 0,1M v H2SO4 0,05M vi 300 ml dung dchBa(OH)2 c nng a mol/lt thu c m gam kt ta v 500 ml dung dch c pH = 13. Gi tr a vm ln lt l :

A. 0,15 M v 2,33 gam. B. 0,15 M v 4,46 gam.C. 0,2 M v 3,495 gam. D. 0,2 M v 2,33 gam.

Cu 34: Ly dung dch axit A c pH = 5 v dung dch baz B c pH = 9 theo t l no thu cdung dch c pH = 8 ?

A. BA

V 11V 9

. B. BA

V 9V 11

. C. VB = VA. D. Khng xc nh c.



Cu 35: Trn dd X cha NaOH 0,1M, Ba(OH) 2 0,2M vi dd Y cha HCl 0,2M, H2SO4 0,1M theo tl VX: VY no dd thu c c pH = 13 ?

A. 5/4. B. 4/5. C. 5/3. D. 3/2.Cu 36: Trn mt th tch metan vi mt th tch hirocacbon X thu c hn hp kh (ktc) c tkhi so vi H2 bng 22. Cng thc phn t ca X l :

A. C2H6. B. C3H8. C. C4H10. D. C5H12.Cu 37: Khi tin hnh phn ng th gia ankan X vi kh clo c chiu sng ng i ta thu c hnhp Y ch cha hai cht sn phm. T khi h i ca Y so vi hiro l 35,75. Tn ca X l : A. 2,2-imetylpropan. B. metan.

C. etan. D. 2,2,3,3 - tetrametylbutan.Cu 38: t chy 2,24 lt (ktc) hn hp X gm metan v etan trong oxi d, thu c 3,36 lt khCO2 (ktc). Thnh phn phn trm v s mol ca metan v etan ln lt l :

A. 77,77% v 22,23%. B. 33,33% v 66,67%.C. 35% v 65%. D. 50% v 50%.

Cu 39: Hn hp kh A iu kin tiu chun gm hai olefin. t chy 7 th tch A cn 31 thtch O2 (ktc). Bit olefin cha nhiu cacbon chim khong 40% - 50% th tch hn hp A. Cngthc phn t ca hai elefin l :

A. C2H4 v C3H6. B. C3H6 v C4H8.C. C2H4 v C4H8. D. A hoc C ng.

Cu 40: Hn hp X gm C2H2 v C3H8 c t khi so vi hiro l 15,25. t chy ht 4,48 lt(ktc) hn hp X th th tch O2 (ktc) ti thiu cn dng l : A. 14 lt. B. 15,6 lt. C. 22,4 lt. D. 28 lt.Cu 41: Dung dch ru etylic 13,8o c d (g/ml) bng bao nhiu ? Bit

2 5C H OH(ng.cht)d = 0,8 g/ml ;2H Od 1 g ml .

A. 0,805. B. 0,855. C. 0,972. D. 0,915.Cu 42: Cho m gam mt ancol (ru) no, n chc X qua bnh ng CuO (d), nung nng. Sau khiphn ng hon ton, khi lng cht rn trong bnh gim 0,64 gam. Hn hp hi thu c c t khii vi hiro l 15,5. Gi tr ca m l :

A. 1,84. B. 0,64. C. 1,48. D. 0,92.Cu 43: Cho m gam hn hp X gm hai ancol no, n chc, k tip nhau trong dy ng ng tcdng vi CuO (d) nung nng, thu c mt hn hp rn Z v mt hn hp hi Y (c t khi hi sovi H2 l 13,75). Cho ton b Y phn ng vi mt lng d Ag2O (hoc AgNO3) trong dung dchNH3 un nng, sinh ra 64,8 gam Ag. Gi tr ca m l :

A. 7,8. B. 7,4. C. 9,2. D. 8,8.3. Bi tp dnh cho hc sinh lp 12Cu 44: Cho hn hp X gm 2 este c CTPT l C4H8O2 v C3H6O2 tc dng vi NaOH d thu c12,28 gam hn hp 2 mui v 7,36 gam ancol B duy nht c t khi so vi oxi l 1,4375. S gam caC4H8O2 v C3H6O2 trong A ln lt l :

A. 3,6 gam v 2,74 gam. B. 8,8 gam v 4,44 gam.C. 6,24 gam v 3,7 gam. D. 4,4 gam v 2,22 gam.

Cu 45: Cho hn hp X gm hai cht hu c c cng cng thc phn t C2H7NO2 tc dng va vi dung dch NaOH v un nng, thu c dung dch Y v 2,24 lt hn hp Z ( ktc) gm hai kh(u lm xanh giy qu m). T khi hi ca Z i vi H2 bng 13,75. C cn dung dch Y thu ckhi lng mui khan l :

A. 7,15 gam. B. 14,3 gam. C. 8,9 gam. D. 15,7 gam.



Cu 46: Cn ha tan bao nhiu gam P2O5 vo 500 gam dung dch H3PO415% thu c dung dchH3PO4 30%?

A. 73,1 gam. B. 69,44 gam. C. 107,14 gam. D. 58,26 gam.Cu 47: C 200 ml dung dch H2SO4 98% (D = 1,84 g/ml). Ngi ta mun pha long th tch H2SO4trn thnh dung dch H2SO4 40% th th tch nc cn pha long l bao nhiu ?

A. 711,28cm3. B. 533,60 cm3. C. 621,28cm3. D. 731,28cm3.Cu 48: Ha tan hon ton m gam Na 2O nguyn cht vo 40 gam dung dch NaOH 12% thu cdung dch NaOH 51%. Gi tr ca m l :

A. 11,3. B. 20,0. C. 31,8. D. 40,0.Cu 49: Ha tan 3,164 gam hn hp 2 mui CaCO3 v BaCO3 bng dung dch HCl d, sau phnng thu c 448 ml kh CO2 (ktc). Thnh phn % s mol ca BaCO3 trong hn hp l :

A. 50%. B. 55%. C. 60%. D. 65%.Cu 50: Ha tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dch HCl (d) thu c0,672 lt kh iu kin tiu chun. Thnh phn % khi lng ca MgCO3 trong hn hp l :

A. 33,33%. B. 29,58%. C. 54,45%. D. 66,67%.Cu 51: Cho 4,48 lt CO (ktc) t t i qua ng s nung nng ng 8 gam mt ox it st n khi phnng xy ra hon ton. Kh thu c sau phn ng c t khi so vi hiro bng 20. Cng thc ca oxitst v phn trm th tch ca kh CO 2 trong hn hp kh sau phn ng l :

A. FeO; 75%. B. Fe2O3; 75%. C. Fe2O3; 65%. D. Fe3O4; 65%.Cu 52: A l khong vt cuprit cha 45% Cu2O. B l khong vt tenorit cha 70% CuO. Cn trn Av B theo t l khi lng T = mA : mB no c qung C m t 1 tn qung C c th iu chc ti a 0,5 tn ng nguyn cht ?

A. 5 : 3 B. 5 : 4 C. 4 : 5 D. 3 : 5Cu 53: T 1 tn qung hematit A iu ch c 420 kg Fe. T 1 tn qung manhetit B iu chc 504 kg Fe. Hi phi trn hai qung tr n vi t l khi lng (mA : mB) l bao nhiu c 1tn qung hn hp m t 1 tn qung hn hp ny iu ch c 480 kg Fe ?

A. 1 : 3. B. 2 : 5. C. 2 : 3. D. 1 : 1.Cu 54: Ha tan 12 gam hn hp Fe, Cu (t l mol 1:1) bng axit HNO 3, thu c V lt ( ktc) hnhp kh X (gm NO v NO2), dung dch Y (ch cha hai mui) v 3,2 gam cht rn khng tan. Tkhi ca X i vi H2 bng 19. Gi tr ca V l :

A. 2,24. B. 4,48. C. 5,60. D. 3,36.



CHUYN 2 : PHNG PHP T CHN LNG CHT

1. Nguyn tc p dng :Khi gp cc bi tp ha hc c lng cht m bi cho di dng tng qut (dng t l mol, t l

% theo th tch, khi lng, hoc cc lng cht m bi cho u c cha chung mt tham s: mgam, V lt, x mol ) th cc bi ton ny s c kt qu khng ph thuc vo lng cht cho. Do ta t chn mt lng cht c th theo hng c li cho vic tnh ton, bin b i ton t phc tptr nn n gin. Sau khi chn lng cht thch hp th bi ton tr thnh mt dng rt c bn,vic gii ton lc ny s thun li hn rt nhiu.2. Phn dng bi tp : T nhng yu cu c th ca bi ton, ta phn chia ra thnh cc d ng bi tp gii bngphng php t chn lng cht thng gp nh sau :

Dng 1: i lng t chn l mol Dng 2: i lng t chn l gam (thng chn l 100 gam) Dng 3: i lng t chn ph thuc vo cho, nhm trit tiu biu thc ton hc phc tp

thnh s c th3. Cc v d minh ha

Dng 1: i lng t chn l molPhng php gii

- Ta thng la chn s mol ca mt cht hoc ca hn hp cc cht l 1 mol.- i vi bi ton v cht kh ta thng chn s mol ca cc kh ban u bng ng t l mol

hoc t l th tch ca cc kh m bi cho.

Cc v d minh ha i vi dng 1 V d dnh cho hc sinh lp 10V d 1: Ho tan hon ton mt lng kim loi R ha tr n bng dung dch H 2SO4 long ri c cndung dch sau phn ng thu c mt lng mui khan c khi lng gp 5 ln khi lng kim loiR ban u em ho tan. Kim loi R l

A. Al. B. Ba. C. Zn. D. MgHng dn gii

Chn s mol ca R tham gia phn ng l 1 mol S phn ng :

2 4H SO2 4 n2R R (SO )

mol: 1 0,5 Theo gi thit ta c :

0,5.(2R + 96n) =5R R= 12n Nu n = 1 M = 12 (loi) Nu n = 2 M = 24 (nhn) Nu n = 3 M = 36 (loi)

Vy M l magie (Mg)p n D.



V d 2: Ho tan mt lng oxit ca kim loi R vo trong dung dch H2SO4 4,9% (va ) th thuc mt dung dch mui c nng 5,87%. Cng thc ca oxit kim l oi l :

A. CuO. B. FeO. C. MgO. D. ZnO.Hng dn gii

t cng thc tng qut ca oxit l R2Ox ( x l ho tr ca R ) Gi s ho tan 1 mol R2Ox

R2Ox + xH2SO4 R2 (SO4)x + xH2Omol: 1 x 1gam: (2MR + 16x) 98x (2MR + 96x)Theo nh lut bo ton khi lng ta c :

d d s a u p R R9 8 .xm (2 M 1 6 x ) 1 0 0 (2 M 2 0 1 6 x ) g a m4 , 9Phng trnh nng % ca dung dch mui l :

R

R

2 M 9 6 x1 0 0 % 5 , 8 7

2 M 2 0 1 6 x MR = 12x

R

x 2M 24

Vy kim loi R l Mg ; oxit kim loi l : MgO

p n B.V d 3: Ho tan x gam kim loi M trong y gam dung dch HCl 7,3% (l ng axit va ) thu cdung dch A c nng 11,96%. Kim loi M l :

A. Cu. B. Fe. C. Mg. D. Mn.Hng dn gii

Gi s s mol ca kim loi M (c ho tr n) phn ng l 1 mol Phng trnh phn ng :

2M + 2nHCl 2MCln + nH2mol: 1 n 1 0,5ngam: M 36,5n (M + 35,5n) n

Theo gi thit ta c

dd HCl

36,5n.100m 500n7,3

p dng nh lut bo ton khi lng ta c:

n 2dd MCl M dd HCl Hm m m m M 500n n M 499n

2MClM 35,5nC% .100 11,96M 499n M = 27,5 n

Nu n = 1 M = 27,5 (loi) Nu n = 2 M = 55 (nhn) Nu n = 3 M = 72,5 (loi)

Vy M l mangan (Mn)p n D.



V d 4: Cho hn hp gm NaBr v NaI ho tan hon ton vo n c c dung dch A. Cho vodung dch A mt lng brom va thu c mui X c khi lng nh hn khi lng ca muiban u l a gam. Ho tan X vo nc thu c dung dch B. Sc kh clo vo dung dch B thu cmui Y c khi lng nh hn khi lng ca mui X l 2a gam. Phn trm theo khi lng caNaBr v NaI trong hn hp mui ban u l (coi clo, brom, iot khng tc dng vi H2O) :

A. 50%; 50%. B. 65%; 45%.C. 43,31%; 56,69%%. D. 56,69%; 43,31% .

Hng dn gii Gi s trong 1 mol hn hp c x mol NaI v (1 x) mol NaBr Cho dung dch A tc dung vi brom

2 NaI + Br2 2 NaBr + I2 (1)mol : x x

mui X ch c NaBr vi s mol l x + (1 x) = 1 mol mNaBr = 103 . 1 = 103 gam mhh u = 103 + a gam

Cho dung dch B tc dng vi clo2 NaBr + Cl2 2 NaCl + Br2 (2)

mol : 1 1 m

NaCl = 58,5 . 1 = 58,5 gam Theo gi thit: mNaBr = mNaCl + 2a

103 = 58,5 + 2a a = 22,25 Vy mhh u = 103 + 22,25 = 125,25 gam

M m hh u = mNaI + mNaBr = 150x + 103(1 x) = 125,25 x = 0,4734

NaI

NaBr

0,4734.150%m .100 56,69%125,26

%m 100 56,69 43,31%

p n C.V d 5: Hn hp gm NaCl v NaBr. Cho hn hp tc dng vi dung dch AgNO 3 d th to ra ktta c khi lng bng khi lng ca AgNO3 tham gia phn ng. Thnh phn % theo khi lngca NaCl trong hn hp u l :

A. 25,84%. B. 27,84%. C. 40,45%. D. 27,48%.Hng dn gii

Cc phng trnh phn ng :NaCl + AgNO3 AgCl + NaNO3 (1)

mol : x x xNaBr + AgNO3 AgBr + NaNO3 (2)

mol : y y yTheo gi thit ta c :

170(x+y) = 143,5x + 188y x 18y 26,5

Chn NaCl NaIn 18 mol ; n 26,5 mol



NaC l

N aBr N aC l

m 18.58, 5% N aC l .100% 27, 84%m m (26, 5 .103) (18 .58, 5 ).

p n B. V d dnh cho hc sinh lp 11V d 6: Sau qu trnh tng hp NH3 t H2 v N2

2 2H N(n : n 1: 3) , p sut trong bnh gim i 10% so

vi p sut lc u. Bit nhit ca phn ng gi khng i tr c v sau phn ng. Phn trm theoth tch ca N2, H2, NH3 trong hn hp kh thu c sau phn ng ln lt l :

A. 25%; 25%; 50%. B. 30%; 25%; 45%.C. 22,22%; 66,67%; 11,11%. D. 20%; 40%; 40% .

Hng dn gii Gi s lc u ta ly 1 mol N2 v 3 mol H2 Trong mt bnh kn c nhit khng i th p sut t l thun vi s mol hn hp kh :

1 1 1

2 2 1

n p pn p 90%p

nhn hp kh sau phn ng = 2n = 904. 3,6 mol100 Gi s c x mol N2 phn ng Phng trnh ho hc:

N2 + 3H2 2NH3 S mol ban u: 1 3 0 S mol phn ng: x 3x 2xS mol sau phn ng: 1x 33x 2x

nhn hp kh sau phn ng = (1x) + (33x) + 2x = 4 2x = 3,6 x = 0,2

2

2

3

N

H

NH

1 0,2%V .100 22,22%3,6

3 3.0,2%V .100 66,67%3,6

0,2.2%V .100 11,11%3,6

p n D.V d 7: Hn hp kh X gm N2 v H2 c t khi so vi He bng 1,8. un nng X mt thi giantrong bnh kn (c bt Fe lm xc tc), thu c hn hp kh Y c t khi so vi He bng 2. Hiusut ca phn ng tng hp NH3 l :

A. 50%. B. 36%. C. 40%. D. 25%.Hng dn gii

p dng phng php ng cho ta c :2

2

H

N

n 28 1,8.4 20,8 4n 1,8.4 2 5,2 1

Chn 2 2N Hn 1 mol ; n 4 mol

Phng trnh phn ng ho hc:N2 + 3H2 2NH3

S mol ban u: 1 4 0 S mol phn ng: x 3x 2xS mol sau phn ng: 1x 43x 2x



nhn hp kh sau phn ng = (1x) + (43x) + 2x = 5 2x p dng nhlut bo ton khi lng ta c :

mX = mY nX. XM = nY. YM YXXY

n M 2.4 5x 0,25

n 1,8.4 5 2xM

Hiu sut phn ng tnh theo N 2 v H2 d : H =0,25 .100% 25%

1

p n D.V d 8: Hn hp kh X gm H2 v C2H4 c t khi so vi He l 3,75. Dn X qua Ni nung nng, thuc hn hp kh Y c t khi so vi He l 5. Hiu sut ca phn ng hiro ho l :

A. 20%. B. 40%. C. 50%. D. 25%.Hng dn gii


2 4

H

C H

n 28 15 1n 15 2 1

C th tnh hiu sut phn ng theo H2 hoc theo C2H4 Phng trnh phn ng :

H2 + C2H4oNi,t C2H6

p dng nh lut bo ton khi lng ta c :


n M 5.4 4n 3,75.4 3M

Chn nX = 4 mol 2H

n =2 4C H

n = 2 mol ;2( p )H X Y

n n n 1 mol.

Hiu sut phn ng : H = 1 .100% 50%2

.p n C.V d 9: Cho hn hp X gm anken v hiro c t khi so vi heli bng 3,33. Cho X i qua btniken nung nng n khi phn ng xy ra hon ton, thu c hn hp Y c t khi so vi heli l 4.CTPT ca X l :


V YM = 4.4 = 16 nn suy ra sau phn ng H2 cn d, CnH2n phn ng ht. p dng nh lut bo ton khi lng ta c :


n M 4.4 1,2n 3,33.4 1M

Chn nX = 1,2 v nY =1 2( p ) n 2nH C H X Y

n n n n 0,2 mol.

Ban u trong X c 0,2 mol CnH2n v 1 mol H2 Ta c : XM = 0, 2.14n 1.2 3,33.4 n 5

1, 2

p n D.



V d dnh cho hc sinh lp 12V d 10: A l hn hp 2 kim loi kim X v Y thuc 2 chu k k tip. Nu cho A tc dng va vi dung dch HCl th thu c a gam 2 mui, cn nu cho A tc dng va vi dung d ch H2SO4th thu c 1,1807a gam 2 mui. X v Y l

A. Li v Na. B. Na v K. C. K v Rb. D. Rb v Cs.Hng dn gii

t cng thc phn t trung b nh ca hai kim loi kim l MChn s mol ca M tham gia phn ng l 1 molS phn ng :

HClM MClmol: 1 1

2 4H SO 2 42M M SOmol: 1 0,5

Theo gi thit ta c :

( M +35,5) = a (1) v 0,5.(2 M +96) =1,1807a (2) T (1) v (2) ta c: M =33,67 Nhn xt : MNa < M < MK X v Y l Na v Kp n D.V d 11: Hn hp X gm Na v Al. Cho m gam X vo mt lng d nc th thot ra V lt kh. Nucng cho m gam X vo dung dch NaOH d th c 1,75V lt kh. Thnh phn phn trm theo khilng ca K trong X l (bit cc th tch kh o trong cng iu kin) A. 41,94%. B. 77,31%. C. 49,87%. D. 29,87%.

Hng dn gii X tc dng vi dung dch NaOH d thu c lng kh nhiu hn so vi khi X tc dng viH2O, chng t khi X tc dng vi H 2O th Al cn d, dung dch sau phn cha NaAlO2. i vi cc cht kh th t l th tch cng chnh l t l mol nn cn c vo gi thit ta chn smol H2 gii phng hai trng hp ln lt l 1 mol v 1,75 mol t s mol ca Na v Al tham gia phn ng vi H2O l x mol

p dng nh lut bo ton electron ta c :2Na Al H

1.n 3.n 2.n 1.x + 3.x = 2.1 x = 0,5 t s mol Al ban u l y, khi X tc dng vi dung dch NaOH d th Al phn ng ht p dng nh lut bo ton electron ta c :

2Na Al H1.n 3.n 2.n 1.0,5 + 3.y = 2.1,75 y = 1

Thnh phn phn trm theo khi lng ca Na trong X l : 0,5.23 .100% 29,87%0,5.23 1.27

.p n D.



V d 12: Ho tan hon ton hn hp X gm Fe v Mg bng mt lng va dung dch HCl 20%,thu c dung dch Y. Nng ca FeCl 2 trong dung dch Y l 15,76%. Nng phn trm caMgCl2 trong dung dch Y l :

A. 15,76%. B. 28,21%. C. 11,79%. D. 24,24%.Hng dn gii

Chn s mol ca Fe bng 1 mol ; s mol ca Mg bng x mol Phng trnh phn ng :

Fe + 2HCl FeCl2 + H2 (1)mol: 1 2 1 1

Mg + 2HCl MgCl2 + H2 (2)mol: x 2x x x

p dng nh lut bo ton khi lng ta c :2 2 2

2 2

(Mg, Fe) dd HCl dd (FeCl , MgCl ) H

dd (FeCl , MgCl )

m m m m

(2x 2).36,5m (1.56 24x) 2.(1 x) (419 387x) gam20%

2FeClm 1.(56 71) 127 gam

2(FeCl )

127C% 15,76% x 1419 387x

2(MgCl )

1.(24 71)C% .100% 11,79%419 387.1

p n C.V d 13: Cho cng mt lng kh clo ln lt tc dng hon ton vi kim loi R (ho tr I) v kimloi X (ho tr II) th khi lng kim loi R phn ng gp 3,375 ln khi lng ca kim loi X phn ng. Khi lng mui clorua ca R thu c gp 2,126 ln khi lng mui clorua ca X to thnh. Hai kim loi R v X l :

A. Ag v Cu. B. Ag v Zn. C. Na v Cu. D. Na v Zn.Hng dn gii

Gi s c 1 mol clo tham gia phn ng Phng trnh phn ng:

Cl2 + 2R 2RCl mol: 1 2 2

Cl2 + X XCl2mol: 1 1 1

Theo gi thit ta c :

R RR X

X X

m 2M3,375 2M 3,375M

m M (1)

2

RCl RR X

XCl X

m 2M 712,126 2M 2,126M 79,946

m M 71 (2)

T (1) v (2) ta c X l Cu (MX = 64)R l Ag (MR = 108)

p n A.



Dng 2: i lng t chn l khi lngPhng php gii

- Ta thng la chn khi lng ca mt cht, mt dung dch hay mt hn hp l 100 gam- Dng ny thng gp vi bi tp cho i lng tng qut l khi lng ca mt hn hp, l

phn trm khi lng, hoc nng phn trm.

Cc v d minh ha i vi dng 2 V d dnh cho hc sinh lp 10V d 14: Cho x gam dung dch H2SO4 nng y% tc dng ht vi mt l ng d hn hp khilng Na, Mg. Lng H2 (kh duy nht) thu c bng 0,05x gam. Nng phn trm ca dungdch H2SO4 l :

A. 15,5%. B. 15,81%. C. 18,5%. D. 8,45% .Hng dn gii

Gi s khi lng dung dch H2SO4 ban u x = 100 gam

2 4H SO

m y gam ;2H

0,05.100n 2,5 mol2

Phng trnh phn ng :

H2SO4 + 2Na Na2SO4 + H2 (1)H2SO4 + Mg MgSO4 + H2 (2)

Do Na v Mg cn d nn c phn ng :2Na + 2H2O 2NaOH + H2 (3)2NaOH + MgSO4 Na2SO4 + Mg(OH)2 (4)

Theo phng trnh (1) v (2) 2 2 4H (1), (2) H SO

yn n mol98

Theo phng trnh (3) 2 2H (3) H O

1 1 100 y 100 yn n . mol2 2 18 36

Vy tng s mol H2 thu c cc phng trnh trn l :

2H

y 100 yn 2,598 36

y = 15,81=>

2 4(H SO )C% 15,81%

p n B. V d dnh cho hc sinh lp 11V d 15: Cho dung dch NaOH 20,00% tc dng va vi dung dch FeCl 2 10,00%. un nngtrong khng kh cho cc phn ng xy ra hon ton. Nng phn trm ca mui to th nh trongdung dch sau phn ng l (coi nc bay hi khng ng k) :

A. 7,5%. B. 7,45%. C. 8,5%. D. 8,45% .Hng dn gii

Gi s c 100 gam dung dch NaOH tham gia phn ng

NaOH

100.20n 0,5 mol100.40



Phng trnh phn ng:FeCl2 + 2NaOH Fe(OH)2 + 2NaCl

mol: 0,25 0,5 0,25 0,54Fe(OH)2 + O2 + 2H2O 4Fe(OH)3

mol: 0,25 0,0625 0,25Theo gi thit ta c :

2dd FeCl

0,25.127m 317,5 gam10%

S gam kt ta :

3Fe(OH)m 0,25.107 26,75 gam

p dng nh lut bo ton khi lng ta c :2 2 3dd NaCl dd FeCl dd NaOH O Fe(OH)

m m m m m mdd NaCl = 317,5 + 100 + 32. 0,0625 26,75 = 392,75 gam

Khi lng mui trong dung dch sau phn ng: mNaCl = 0,5 . 58,5 = 29,25 gam

29,25C%(NaCl) .100 7,45%392,75

p n B.V d 16: Mt loi phn supephotphat kp c cha 69,62% mui canxi ihirophotphat, c n li gmcc cht khng cha photpho. dinh dng ca loi phn ln ny l

A. 48,52%. B. 42,25%. C. 39,76%. D. 45,75%.Hng dn gii

dinh dng ca phn ln c nh gi bng hm lng ca P2O5 tng ng vi lngphotpho c trong thnh phn ca n. Gi s c 100 gam supephotphat kp th khi lng ca canxi ihirophotphat l 69,62 gam

Ta c s :

Ca(H2PO4)2 P2O5gam: 234 142gam: 69,62 69,62.142x 42,25

234

Vy dinh dng ca ca loi phn ln ny l 42,25%.p n B. V d dnh cho hc sinh lp 12V d 17: X l 1 loi vi cha 80% CaCO 3, phn cn li l tp cht tr. Nung 50 gam X mt thigian, thu c 39 gam cht rn. % CaCO 3 b phn hu l

A. 50,5%. B. 60%. C. 62,5%. D. 65%.Hng dn gii

Gi s c 100 gam vi th khi lng ca CaCO3 l 80 gam. Do trong 50 gam X c 40 gamCaCO3. Phng trnh phn ng ha hc :

CaCO3 CaO + CO2mol: x x



Theo phng trnh v theo gi thit ta c :100x 56x = 50 39 = 11 x = 0,25

Vy % CaCO3 b phn hy l0,25.100 62,5%

40

p n C.V d 18: Nung nng m gam PbS ngoi khng kh sau m t thi gian di, thu c hn hp rn (ccha mt oxit) nng 0,95m gam. Phn trm khi l ng PbS b t chy l:

A. 74,69%. B. 95,00%. C. 25,31%. D. 64,68%.Hng dn gii

Phng trnh phn ng :

PbS + 32

O2 PbO + SO2 (1)mol : a a

T PbS PbO th khi lng gim = 16a = 0,05m. Chn m =100 a = 0,3125 % PbS b t chy : % PbS = 239.0,3125 .100%

100= 74,69%.

p n A.

Dng 3: i lng t chn ph thuc vo cho nhm trit tiu biu thc tonhc phc tp thnh s c th

Cc v d minh ha i vi dng 3 V d dnh cho hc sinh lp 11

V d 19: t chy hon ton a gam mt hp cht A ca photpho cn a17

mol O2 sn phm ch thu

c P2O5 v13,5a gam

17H2O. Xc nh cng thc phn t ca A bit M A< 65

Hng dn gii Gi s a = 17

2On = 1 mol

V sn phm ch c P2O5 v H2O trong A c H, P v c th c O Gi cng thc ca A l HxPyOz

4 HxPyOz + (x + 5y 2z) O2 2x H2O + 2y P2O5 p dng nh lut bo ton khi lng ta c

2 5P Om = 17 + 32 13,5 = 35,5

Ta c :2 2 5H H O P P O

13,5.2 35,5.2n 2n 1,5 mol ; n 2n 0,5 mol18 142

O 17 (1,5 0,5.31)n 016 . Vy trong A khng c oxi

x : y = nH : nP = 1,5 : 0,5 = 3 : 1 V MA < 65 nn cng thc ca A l PH3



V d 20: t chy hon ton m gam ancol X, sn phm thu c cho i qua bnh ng dung dchnc vi trong d thy khi lng bnh tng thm p gam v c t gam k t ta. Cng thc ca X l

(Bit p = 0,71t ; t = m p1,02 ) :

A. C2H5OH. B. C3H5(OH)3. C. C2H4(OH)2. D. C3H5OH.Hng dn gii

Chn t = m p1,02

= 100 =3CaCO

m

p = 71 gam ; m = 31 gam Gi cng thc tng qut ca ancol R l CxHyOz

x y zC H O + 2 2 2y z y(x )O xCO H O4 2 2

(1)CO2 + Ca(OH)2 CaCO3 + H2O (2)

Theo phng trnh (2) 2 3C CO CaCO

n n n 1 mol Khi lng bnh tng ln: p =

2 2CO H Om m

2 2H O H O

m 71 44 27 gam n 1,5 mol V

2 2H O COn n nn ancol X l ancol no

O

31 (12 1,5.2)n 1 mol16

Vy ta c x : y : z = nC : nH : nO = 1 : 3 : 1

Cng thc ca ancol X c dng (CH 3O)n = CnH3nOn = CnH2n(OH)n V X l ancol no nn: s nguyn t H = 2.s nguyn t C + 2 s nhm OH

2n = 2n + 2 n n = 2 Vy cng thc ca ancol R l: C2H4(OH)2p n C.V d 21: Hn hp X gm 2 hirocacbon A v B c khi lng a gam. Nu em t chy ho n ton

X th thu c 132a41

gam CO2 v45a41

gam H2O. Nu thm vo X mt na lng A c trong X ri

t chy hon ton th thu c 165a41

gam CO2 v60,75a

41 gam H2O. Tm cng thc phn t ca A

v B. Bit X khng lm mt mu dung dch nc brom v A, B thuc loi hirocacbon hc.Hng dn gii

Gi s a = 41

Khi t chy X:2 2CO H O

132 45n 3 mol ; n 2,5 mol44 18

Khi t chy X +21 A:

2 2CO H O

165 60,75n 3,75 mol ; n 3,375 mol44 18

Vy khi t chy21 A ta thu c:

2 2CO H On 0,75 mol ; n 0,875 mol

V2 2CO H O

n n A l hirocacbon no



Gi cng thc ca A l CnH2n + 22CnH2n + 2 + (3n + 1) O2 2n CO2 + 2(n+1) H2O

Ta c 22

H O

CO

n 2(n 1) 0,875 n 6n 2n 0,75

Vy cng thc phn t ca A l C6H14 Khi t chy B ta thu c s mol ca H2O v CO2 l :

2

2

CO C

H O H

n 3 0,75.2 1,5 mol n 1,5 mol

n 2,5 0,875.2 0,75 mol n 1,5 mol

nC : nH = 1,5 : 1,5 = 1 : 1 Vy cng thc n gin nht ca B l (CH)n = CnHn Theo gi thit B khng lm mt mu dung dch nc brom B ch c th l aren s nguyn t H = 2 s nguyn t C 6

Hay n = 2n 6 n = 6 Vy cng thc ca B l C6H6

4. Bi tp p dng4.1. Bi tp dnh cho hc sinh lp 10Cu 1: t chy hon ton hn hp Mg, Fe trong oxi d, sau phn ng thy khi lng cht rnnng gp 1,5 ln so vi khi lng cht rn ban u. Phn trm khi l ng ca Mg v Fe trong hnhp ln lt l :

A. 30% v 70%. B. 40% v 60%. C. 70% v 30%. D. 60% v 40%.Cu 2: Ho tan a gam mt oxit kim loi M ho tr II (khng i) bng mt l ng va dung dchH2SO4 4,9% ngi ta thu c mt dung dch mui c nng 5,88%. Kim loi M l :

A. Cu. B. Fe. C. Mg. D. Zn.Cu 3: Khi ha tan hiroxit kim loi M(OH)2 bng mt lng va dung dch H2SO4 20% thuc dung dch mui trung ho c nng 27,21%. Kim loi M l :

A. Cu. B. Zn. C. Fe. D. Mg.Cu 4: Ho tan a gam M2(CO3)n bng mt lng va dung dch H2SO4 10% thu c dung dchmui c nng 15,09%. Cng thc ca mui cacbonat l :

A. K2CO3. B. FeCO3. C. SrCO3. D. CuCO3.Cu 5: Cho a gam dung dch H2SO4 long nng C% tc dng hon ton vi hn hp 2 kim loi Kv Fe (Ly d so vi lng phn ng). Sau phn ng, khi l ng kh sinh ra l 0,04694a gam. TmC%

A. 24,5%. B. 14,5%. C. 13%. D. 25%.Cu 6: Hn hp A gm NaCl v NaBr khi tc dng vi AgNO3 d to ra mt lng cht kt tabng lng AgNO3 phn ng. Phn trm khi lng ca NaCl v NaBr ln lt l :

A. 30 % v 70%. B. 25% v 75%.C. 27,84% v 72,16%. D. 22,2% v 77,8%.

Cu 7: Mui A to bi kim loi M (ho tr II) v phi kim X (ho tr I). Ho tan 1 lng A vo ncc dung dch A1. Nu thm AgNO3 d vo A1 th lng kt ta tch ra bng 188% l ng A. Nuthm Na2CO3 d vo dung dch A1 th lng kt ta tch ra bng 50% lng A. Cng thc ca muiA l :

A. CaBr2. B. CaCl2. C. BaBr2. D. BaCl2.



Cu 8: Phng in qua O2 c hn hp kh O2, O3 c M 33 gam. Hiu sut phn ng l :A. 7,09%. B. 9,09%. C. 11,09%. D. 13,09%.

4.2. Bi tp dnh cho hc sinh lp 11Cu 9: Cho dung dch NaHSO3 26% phn ng va ht vi dung dch H 2SO4 19,6%, thu c dungdch A. Nng phn trm ca cht tan trong dung dch A l :

A. 21,21%. B. 11,22%. C. 22,11%. D. 12,12%.Cu 10: Khi cho a gam dung d ch H2SO4 nng C% tc dng vi lng d hn hp Na, Mg thlng kh H2 thot ra bng 4,5% lng dung dch axit dng. C% ca dung dch H2SO4 l :

A. 15%. B. 45%. C. 30%. D. 25%.Cu 11: Hn hp A gm 3 kh NH3, N2, H2. Dn hn hp A vo bnh c nhit cao. Sau phn ngphn hy NH3 (coi nh hon ton) thu c hn hp kh B c th tch tng 25% so vi A. Dn B iqua ng ng CuO nung nng sau loi n c th ch cn mt cht kh c th tch gim 75% so viB. Phn trm th tch ca cc kh NH 3, N2, H2 trong A ln lt l :

A. 25%; 20%; 55%. B. 25%; 18,75%; 56,25%.C. 20%; 25%; 55%. D. 30,5%; 18,75%; 50,75%.

Cu 12: Hn hp X gm N2 v H2 c XM 12,4 . Dn X i qua bnh ng bt Fe ri nung nng bitrng hiu sut tng hp NH 3 t 40% th thu c hn hp Y. YM c gi tr l :

A. 15,12. B. 18,23. C. 14,76. D. 13,48.Cu 13: Mt hn hp N2, H2 c ly vo bnh phn ng c nhit c gi khng i. Sau thigian phn ng, p sut ca cc kh trong b nh gim 5% so vi p sut lc u. Bit rng % s molca N2 phn ng l 10%. Phn trm th tch ca cc kh N 2, H2 trong hn hp u ln lt l :

A. 75%; 25%. B. 25%; 75%. C. 20%; 80%. D. 30%; 70%.Cu 14: Hn hp X gm N2 v c H2 c t khi hi so vi H2 bng 3,6. Sau khi tin hnh phn ngtng hp c hn hp Y c t khi hi so vi H2 bng 4. Hiu sut phn ng tng hp l :

A. 10%. B. 15%. C. 20%. D. 25%.Cu 15: Phn supephotphat kp thc t sn xut c thng ch c 40% P2O5. Vy % khi lngCa(H2PO4)2 trong phn bn l

A. 78,56%. B. 56,94%. C. 65,92%. D. 75,83%.Cu 16: Hn hp A gm mt Anken v hiro c t khi so vi H2 bng 6,4. Cho A i qua nikennung nng c hn hp B c t khi so vi H 2 bng 8 (gi thit hiu sut phn ng xy ra l 100%).Cng thc phn t ca anken l :

A. C2H4. B. C3H6. C. C4H8. D. C5H10.Cu 17: A l hn hp gm mt s hirocacbon th kh, B l khng kh. Trn A vi B cng nhit p sut theo t l th tch (1:15) c hn hp kh D. Cho D vo bnh kn dung tch khng i V.Nhit v p sut trong bnh l toC v p1 atm. Sau khi t chy A trong bnh ch c N2, CO2 v hinc vi

2 2CO H OV : V 7 : 4 a bnh v toC. p sut trong bnh sau khi t l p2. Gi tr ca p1 l :A. 2 1

47p p .48

B. p2 = p1. C. 2 116p p .17 D. 2 13p p .5

Cu 18: Cho hn hp A gm 3 hirocacbon kh X, Y, Z thuc ba d y ng ng v hn hp kh Bgm O3 v O2. Trn A vi B theo t l th tch V A:VB = 1,5:3,2 ri t chy. Hn hp sau phn ngthu c ch gm CO2 v hi nc c t l th tch l

2 2CO H OV : V 1,3 :1,2 . Bit t khi ca B so vi

H2 l 19. T khi ca A so vi hiro l :A. 24 B. 12. C. 36. D. 18.



Cu 19: t chy hon ton a gam hn hp X hai hirocacbon A, B thu c 132.a41

gam CO2 v

245a gam H O41

. Nu thm vo hn hp X mt na lng A c trong hn hp X ri t chy ho n

ton th thu c 2165a gam CO

41 v 2

60,75a gam H O41

. Bit A, B khng lm mt mu nc Br2.

a. Cng thc phn t ca A l :A. C2H2. B. C2H6. C. C6H12. D. C6H14.

b. Cng thc phn t ca B lA. C2H2. B. C6H6. C. C4H4. D. C8H8.

c. Phn trm s mol ca A, B trong hn hp X l :A. 60%; 40%. B. 25%; 75%. C. 50%; 50%. D. 30%; 70%.

Cu 20: Trn a gam hn hp X gm 2 hirocacbon C 6H14 v C6H6 theo t l s mol (1:1) vi m gam

mt hirocacbon D ri t chy hon ton th thu c 2275a gam CO82

v 94,5a82

gam H2O.

a. D thuc loi hirocacbon no ?A. CnH2n+2. B. CmH2m2. C. CnH2n. D. CnHn.

b. Gi tr m l :A. 2,75 gam. B. 3,75 gam. C. 5 gam. D. 3,5 gam.

Cu 21: Oxi ha C2H5OH bng CuO nung nng, thu c hn hp cht lng gm CH 3CHO,C2H5OH d v H2O c M = 40 vC. Hiu sut phn ng oxi ha l :

A. 25%. B. 35%. C. 45%. D. 55%.Cu 22: Cho dung dch axit axetic c nng x% tc dng va vi dung dch NaOH 10% th thuc dung dch mui c nng 10,25%. Gi tr ca x l :

A. 20%. B. 16%. C. 15%. D.13%.4.3. Bi tp dnh cho hc sinh lp 12Cu 23: Ha tan ht mt lng hiroxit kim loi M(OH) n bng mt lng va dung dch HCl10% thu c dung dch A, cho dung dch A phn ng va vi dung dch AgNO 3 20% thu cdung dch mui B c nng 8,965%. Cng thc ca M(OH) n l :

A. Ca(OH)2. B. KOH. C. NaOH. D. Mg(OH)2.Cu 24: Ho tan mt mui cacbonat kim loi M ha tr n bng mt l ng va dung dch H2SO49,8% ta thu c dung dch mui sunfat 14,18%. Kim loi M l :

A. Cu. B. Fe. C. Al. D. Zn.Cu 25: Ly m gam A gm Na, Al chia lm 2 phn bng nhau :Phn 1 cho vo nc d n khi phn ng xy ra hon ton, thy thot ra V1 lt kh H2 (ktc); phn 2cho vo dung dch Ba(OH)2 d n khi phn ng xy ra hon ton, thy thot ra V2 lt kh H2 (ktc).Thnh phn % v khi lng ca Al trong hn hp l (bit V2=7,75V1) :

A. 92,15%. B. 80,16%. C. 95,05%. D. 72,34%.Cu 26: Ho tan ht hn hp X gm Fe v Mg bng mt lng va dung dch HCl 20%, thuc dung dch D. Nng ca FeCl 2 trong dung dch D l 15,757%. Nng phn trm ca MgCl 2trong dung dch D l

A. 11,787%. B. 84,243%. C. 88,213%. D. 15,757%.



Cu 27: Nu ho tan a gam hn hp A cha Fe, FeO, Fe2O3 bng dung dch HCl d th lng khthot ra bng 1% khi lng hn hp ban u. Nu kh a gam hn hp A bng H 2 nng d th thuc mt lng nc bng 21,15% khi lng hn hp ban u. Phn trm khi l ng Fe, FeO,Fe2O3 trong hn hp A ln lt l :

A. 36%; 36%; 28%. B. 28%; 38%; 34%.C. 28%; 30%; 42%. D. 28%; 36%; 36%.

Cu 28: Hn hp CaCO3, CaSO4 c ho tan bng axit H2SO4 va . Sau phn ng un nng chobay hi nc v lc c mt lng cht rn bng 121,43% lng hn hp ban u. Phn trm khilng CaCO3, CaSO4 trong hn hp u ln lt l :

A. 55,92%; 44,08% B. 59,52%; 40,48%C. 52,59%; 47,41% D. 49,52%; 50,48%

Cu 29: Mt loi cha 80% CaCO 3 phn cn li l tp cht tr. Nung ti phn ng hon ton(ti khi lng khng i) thu c cht rn R. Vy % khi lng CaO trong R l

A. 62,5%. B. 69,14%. C. 70,22%. D. 73,06%.Cu 30: Nung m gam X cha 80% khi lng gam CaCO3 (phn cn li l tp cht tr) mt thigian thu c cht rn Y cha 45,65 % CaO. Tnh hiu sut phn hy CaCO 3.

A. 50%. B. 75%. C. 80%. D. 70%.



CHUYN 3 : PHNG PHP BO TON NGUYN T

I. Phng php bo ton nguyn t1. Ni dung nh lut bo ton nguyn t :- Trong phn ng ha hc, cc nguyn t lun c bo ton.2. Nguyn tc p dng :- Trong phn ng ha hc, tng s mol ca nguy n t trc v sau phn ng lun bng nhau.3. Cc v d minh ha :

Dng 1 : Xc nh cng thc phn t ca mt hp cht v c hoc mt hp chthu c

Phng php gii- Bc 1 : T gi thit ta c th xc nh c thnh phn nguyn t trong hp cht, ring i

vi nguyn t oxi c nhng trng hp ta khng th xc nh chnh x c trong hp cht cn tm coxi hay khng, trong nhng trng hp nh vy ta gi s l hp cht c oxi.

- Bc 2 : t cng thc phn t ca hp cht : CxHyOzNt . Lp s chuyn ha :CxHyOzNt + O2 CO2 + H2O + N2

- Bc 3 : p dng nh lut bo ton nguyn t tm s nguyn t C, H, O, N trong hpcht, suy ra cng thc ca hp cht CxHyOzNt

x y z t 2

x y z t 2

x y z t 2

x y z t 2 2 2

C(C H O N ) C(CO )

H(C H O N ) H(H O)

N(C H O N ) N(N )

O(C H O N ) O(O ) O(CO ) O(H O)

n nx

n n yn n z

tn n n n

Lu : t c cng thc phn t ca hp cht th iu quan trng nht l ta phi xc nhc thnh phn nguyn t ca hp cht v cc hp cht khc nhau s c thnh phn nguyn tkhc nhau.

Cc v d minh ha i vi dng 1 Cc v d dnh cho hc sinh lp 10V d 1: Hp cht A (khng cha clo) chy c trong kh clo to thnh nit v hiro clorua. Bit tl th tch clo tham gia phn ng v th tch nit to thnh l 3 : 1. Cng thc phn t ca kh A l :

A. N3H. B. NH3.C. N2H4. D. Khng xc nh c.

Hng dn gii Theo gi thit ta suy ra thnh phn nguyn t trong A gm N v H t cng thc phn t ca A l : NxHy S phn ng :

NxHy + Cl2 N2 + HClTh tch: a 3 1 b



p dng nh lut bo ton nguyn t i vi Cl, N, H ta c :3.2 b.1

x 1ax 1.2

y 3ay b.1

. Vy cng thc phn t ca A l NH3.

p n B.

V d 1 : t chy hon ton 3,4 gam hp cht A thu c 2,24 lt kh SO2 (kc) v 1,8 gam H2O.Xc nh cng thc phn t ca hp cht A?

Hng dn gii Theo gi thit ta suy ra thnh phn nguyn t trong A gm S, H v c th c hoc khng c oxi t cng thc phn t ca A l : SxHyOz p dng nh lut bo ton khi lng ta c :

2 2 2 2 2A O SO H O O Om m m m m 4,8 gam n 0,15 mol

S phn ng :SxHyOz + O2 SO2 + H2O

mol: a 0,15 0,1 0,1 p dng nh lut bo ton nguyn t i vi S, H, O ta c :

a.x 0,1.1 a.x 0,1x 1

ay 0,1.2 a.y 0,2y 2

az 2.0,15 0,1.2 0,1.1 az 0

.

Vy cng thc phn t ca A l H2S. Cc v d dnh cho hc sinh lp 11V d 2: t chy 1 lt hi hirocacbon vi mt th tch khng kh (lng d). Hn hp kh thu csau khi hi H2O ngng t c th tch l 18,5 lt, cho qua dung d ch KOH d cn 16,5 lt, cho hn hpkh i qua ng ng photpho d th cn li 16 lt. Xc nh CTPT ca hp cht tr n bit cc th tchkh o cng iu kin nhit , p sut v O2 chim 1/5 khng kh, cn li l N2. A. C2H6. B. C2H4. C. C3H8. D. C2H2.

Hng dn gii Theo gi thit, ta c :

2COV 2 lt ;

2OV (d) = 0,5 lt ;

2NV 16 lt

2OV (ban u) = 4 lt

S phn ng :CxHy + O2 CO2 + H2O + O2 d + N2

lt: 1 4 2 a 0,5 p dng nh lut bo ton nguyn t i vi cc nguyn t C, H, O ta c :

1.x 2.1 x 21.y a.2 y 64.2 2.2 a 0.5.2 a 3

Cng thc ca hirocacbon l C2H6

p n A.



V d 3: Cho 0,5 lt hn hp gm hirocacbon v kh cacbonic vo 2,5 lt oxi (ly d) ri t. Thtch ca hn hp thu c sau khi t l 3,4 lt. Cho hn hp qua thit b lm lnh, th tch hn hpkh cn li 1,8 lt v cho li qua dung dch KOH ch cn 0,5 lt kh. Th tch cc kh c o trongcng iu kin. Tn gi ca hirocacbon l :

A. propan. B. xiclobutan. C. propen. D. xiclopropan.Hng dn gii

Theo gi thit, ta c :2H O

V 1,6 lt ;2CO

V 1,3 lt ;2O

V (d) = 0,5 lt

S phn ng :(CxHy + CO2) + O2 CO2 + H2O + O2 d

lt: a b 2,5 1,3 1,6 0,5 p dng nh lut bo ton nguyn t i vi cc nguyn t C, H, O ta c :

a.x b.1 1,3 x 3a.y 1,6.2 y 8b.2 2,5.2 1,3.2 1,6.1 0,5.2 a 0,4a b 0,5 b 0,1

Cng thc ca hirocacbon l C3H8

p n A. Cc v d dnh cho hc sinh lp 12V d 4: t chy hon ton 10 ml mt este cn dng ht 45 ml O2, thu c V 2CO : V 2H O= 4 : 3.Ngng t sn phm chy thy th tch gim 30 ml. Cc th tch o c ng iu kin. Cng thc caeste l :

A. C8H6O4. B. C4H6O2. C. C4H8O2 D. C4H6O4.Hng dn gii

Theo gi thit suy ra : V 2H O= 30 ml ; V 2CO = 40 ml S phn ng :

CxHyOz + O2 CO2 + H2Oml : 10 45 40 30

p dng nh lut bo ton nguyn t cho cc nguyn t C, H, O ta c :10.x 40.1 x 410.y 30.2 y 610.z 45.2 40.2 30.1 z 2

Vy este c cng thc l : C4H6O2

p n B.



Dng 2: Tnh s mol ; th tch kh ; khi lng ca cc cht ban u hoc cccht sn phmPhng php gii

- Bc 1 : Da vo gi thit lp s chuyn ha gia cc cht .- Bc 2 : Da vo yu cu ca bi xc nh xem cn p dng n h lut bo ton nguyn

t i vi mt hay nhiu nguyn t m c th l nhng nguyn t no.- Bc 3 : p dng nh lut bo ton nguyn t : Tng s mol nguyn t trc v sau phn

ng lun bng nhau t m ra kt qu m bi yu cu.

Cc v d minh ha i vi dng 2 Cc v d dnh cho hc sinh lp 10V d 5: Cho V lt hn hp kh gm H2S v SO2 tc dng vi dung dch brom d. Thm dung dchBaCl2 d vo hn hp trn th thu c 2,33 gam kt ta. Gi tr ca V l : A. 0,112 lt. B. 2,24 lt. C. 1,12 lt. D. 0,224 lt.

Hng dn gii Phng trnh phn ng :

H2S + 4Br2 + 4H2O 8HBr + H2SO4SO2 + Br2 + 2H2O 2HBr + H2SO4H2SO4 + BaCl2 BaSO4 + 2HCl

2 2 2 4 4(H S,SO ) H SO BaSOn n n 2,33 0,01233 mol 2 2(H S,SO )V 0,01.22,4 = 0,224 lt.

p n D.V d 6: Cho hn hp Fe v FeS tc dng vi dung dch HCl (d), thu c 2,464 lt hn hp kh iu kin tiu chun. Dn hn hp kh ny qua dung dch Pb(NO3)2 d, sinh ra 23,9 gam kt ta en.Phn trm khi lng ca Fe v FeS trong hn hp ln lt l :

A. 5,98% v 94,02%. B. 94,02% v 5,98%.C. 25% v 75%. D. 75% v 25%.

Hng dn gii Phng trnh phn ng :

Fe + HCl FeCl2 +H2 (1)FeS + HCl FeCl2 + H2S (2)

p nh lut bo ton nguyn t i vi lu hunh ta c :

2FeS H S PbS23,9

n n n 0,1239

mol.

2H

2, 464n 0,1 0,01

22, 4 mol

2Fe Hn n 0,01 mol

Vy phn trm v khi lng ca Fe v FeS trong hn hp ban u l :

%Fe = 0,01.56 .100% 5,98%0,01.56 0,1.88

; %FeS = (100 5,98)% = 94,02%p n A.



V d 7: Cho hn hp gm 0,3 mol Fe, 0,15 mol Fe 2O3 v 0,1 mol Fe3O4 tc dng ht vi dung dchH2SO4 long thu c dung dch A. Cho dung dch A tc dng vi dung dch NaOH d , lc kt taem nung trong khng kh n khi lng khng i thu c m gam cht rn C. Tnh m

A. 70. B. 72. C. 65. D. 75.Hng dn gii

Tng s mol nguyn t Fe trong hn hp Fe, Fe 2O3, Fe3O4 l : 0,3 + 0,15.2 + 0,1.3 = 0,9 mol. Sau tt c cc phn ng thu c cht rn C l Fe2O3 s mol Fe2O3 thu c l 0,45 mol. Khi lng Fe2O3 thu c l: 0,45.160 = 72 gam.p n B. Cc v d dnh cho hc sinh lp 11V d 8: Ha tan ht 0,02 mol Al v 0,03 mol Cu vo dung d ch HNO3, c cn dung dch sau phnng v nung n khi lng khng i th thu c m gam cht rn. Gi tr ca m l :


S phn ng :2Al 2Al(NO3)3 Al2O3 (1)0,02 0,02 0,01Cu Cu(NO3)2 CuO (2)0,03 0,03 0,03

p dng nh lut bo ton nguyn t cho s (1), (2) ta thy :2 3Al O

n 0,01 mol ; CuOn 0,03 mol Vy khi lng cht rn thu c l : 0,01.102 + 0,03.80 = 3,42 gam.p n A.V d 9: Ho tan hon ton hn hp gm x mol FeS2 v y mol Cu2S vo axit HNO3 (va ), thuc dung dch X (ch cha hai mui sunfat) v kh duy nht NO. T l x : y l :

A. 1 : 3. B. 3 : 1. C. 1 : 2. D. 2 : 1.Hng dn gii

S phn ng :2FeS2 + Cu2S Fe2(SO4)3 + 2CuSO4 (1)

p dng nh lut bo ton nguyn t cho s (1), ta thy :2

2

FeS

Cu S

n 2n 1

x : y = 2 : 1

p n D.V d 10: t chy hon ton m gam hn hp X gm CH4, C3H6 v C4H10 thu c 4,4 gam CO2 v2,52 gam H2O. m c gi tr l :

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

X C H4, 4 2, 52m m m .12 .2 1, 2 0, 28 1, 4844 18

gam.

p n A.



V d 11: Tch nc hon ton t hn hp X gm 2 ru A v B ta c hn hp Y gm cc olefin.Nu t chy hon ton X th thu c 1,76 gam CO2. Vy khi t chy hon ton Y th tng khilng nc v CO2 to ra l :


Theo nh lut bo ton nguyn t, ta c: 2C(X) C(Y) CO 0,04n n n mol

M khi 2OY 2COn = 2H On = 0,04 mol

2 2CO H O 1,76 (0, 04.18) 2, 48m gam.p n B.V d 12: Chia hn hp 2 anehit no n chc thnh 2 phn bng nhau :

- t chy hon ton phn 1 thu c 0,54 gam H2O.- Phn 2 cng H2 (Ni, to ) thu c hn hp A.

Nu t chy hon ton A th th tch kh CO2 thu c (ktc) l :A. 0,112 lt. B. 0,672 lt. C. 1,68 lt. D. 2,24 lt.

Hng dn gii t chy hn hp l anehit no, n chc

2 2CO H On n 0,03 mol Theo nh lut bo ton nguyn t ta c :

1 2C(A) C(P ) C(P )n n n 0,03 mol 2COV 0,672 lt.

p n B.V d 13: Cho hn hp kh X gm CH3CHO v H2 i qua ng s ng bt Ni nung nng. Sau khiphn ng xy ra hon ton thu c hn hp kh Y gm hai cht. t chy ht ht Y t hu c 11,7gam H2O v 7,84 lt CO2 (ktc). Phn trm theo th tch ca H 2 trong X l :

A. 63,16%. B. 46,15%. C. 53,85%. D. 35,00%.Hng dn gii

Theo nh lut bo ton nguyn t ta thy : Thnh phn nguyn t trong hn hp X v Y l nhnhau nn ta suy ra t chy hn hp Y cng nh t chy hn hp X. Phng trnh phn ng :

2CH3CHO + 5O2 4CO2 + 4H2O (1)mol : 0,175 0,35 0,35

2H2 + O2 2H2O (2)mol : 0,3 (0,65 0,35) = 0,3

Theo cc phn ng v gi thit suy ra phn trm th tch ca hiro trong hn hp l :

% H2 =0,3

.100% 63,16%0,175 0,3

.p n A.V d 14: t chy hon ton 0,1 mol mt axit cacboxylic n chc cn va V lt O2 (ktc), thuc 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 O nn c th t l ROOH. p dng nh lut bo ton nguyn t i vi oxi ta c :



2 2 2O (ROOH) O (O ) O (CO ) O (H O)n n n n 0,1.2 + 2O (O )n = 0,3.2 + 0,2.1

2O (O )n = 0,6 mol 2On 0,3 mol 2OV 6,72 lt.p n C.V d 15: Hn hp X gm ru no n chc A v axit no n chc B. Chia thnh 2 phn bng nhau.

- Phn 1: B t chy hon ton thy to ra 2,24 lt CO2 (ktc)- Phn 2: c este ha hon ton v va thu c este E.

Khi t chy este E th lng nc sinh ra l :A. 1,8 gam. B. 3,6 gam. C. 19,8 gam. D. 2,2 gam.

Hng dn gii Theo nh lut bo ton nguyn t, ta c :

2 1C(este) C(P ) C(P ) 0,1n n n mol

Este no, n chc 22

OH O C C(este)n n n 0,1mol

2H Om 0,1.18 1,8 gam.

p n A. Cc v d dnh cho hc sinh lp 12V d 16: in phn 200 ml dung dch CuSO 4 (dung dch X) vi in cc tr sau thi gian ngngin phn th thy khi lng X gim. Dung dch sau in phn tc dng va vi 500 ml dungdch BaCl2 0,3M to kt ta trng. Cho bit khi lng ring dung dch CuSO4 l 1,25 g/ml; sau inphn lng H2O bay hi khng ng k. Nng mol/lt v nng % dung dich CuSO 4 trc inphn l : A. 0,35M, 8%. B. 0,52, 10%. C. 0,75M, 9,6%. D. 0,49M, 12%.

Hng dn gii Khi in phn dung dch CuSO 4, ion SO42- khng b in phn nn p dng nh lut bo tonnguyn t ta c :

4 4 2CuSO BaSO BaCln n n 0,15 mol.

Nng mol ca dung dch CuSO 4 l : CM = 0,150, 2 = 0,75M.

Nng % ca dung dch CuSO 4 l : C% = 0,15.160 .100%200.1, 25 9,6%.p n C.V d 17: Kh hon ton 24 gam hn hp CuO v FexOy bng H2 d nhit cao thu c 17,6 gamhn hp 2 kim loi. Khi lng H2O to thnh l :


Thc cht phn ng kh cc oxit trn lH2 + O H2O

mO (trong oxit) = moxit m kim loi = 24 17,6 = 6,4 gam. 2O H Om 6,4 gam 2H O O

6, 4n n 0, 4

16 mol

2H Om 0, 4.18 7, 2 gam

p n C.



V d 18: Kh ht m gam Fe3O4 bng CO thu c hn hp A gm FeO v Fe. A tan va trong0,3 lt dung dch H2SO4 1M cho ra 4,48 lt kh H2 (ktc). Tnh m ?


S phm ng:Fe3O4 (FeO, Fe) 3FeSO4

24 4Fe trong FeSO SOn n 0,3 mol t s mol ca Fe3O4 l n p dng nh lut bo ton nguyn t Fe : 3 4 4Fe trongFe O Fe trongFeSOn n

3n = 0,3 n = 0,1

3 4Fe Om 23,2 gam.p n A.V d 19: Thi t t V lt hn hp kh (ktc) gm CO v H2 i qua mt ng ng 16,8 gam hn hp3 oxit : CuO, Fe3O4, Al2O3 nung nng, phn ng hon ton. Sau phn ng thu c m gam cht rnv 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 H2 + O H2O

Khi lng hn hp kh to thnh nng hn hn hp kh ban u chnh l khi lng ca nguynt oxi trong cc oxit tham gia phn ng. Do vy :

mO = 0,32 gam O 0,32n 0,02 mol16 2CO Hn n 0,02 mol .p dng nh lut bo ton khi lng ta c :

moxit = mcht rn + 0,3216,8 = m + 0,32 m = 16,48 gam.

2(CO H )V 0,02.22, 4 0, 448 lt.p n D.V d 20: Cho 4,48 lt CO (ktc) t t i qua ng s nung nng ng 8 gam mt oxit st n khiphn ng xy ra hon ton. Kh thu c sau phn ng c t khi so vi hiro bng 20. Cng thcca oxit st v phn trm th tch ca kh CO 2 trong hn hp kh sau phn ng l :

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

Phng trnh phn ng :FexOy + yCO xFe + yCO2

Kh thu c c M 40 gm 2 kh CO2 v CO d2CO

CO

n 44 1240

n 28 4



2COCO

n 3n 1

2CO%V 75% .

Mt khc:2CO ( ) CO

75n n .0, 2 0,15

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

Thc cht phn ng kh oxit st l doCO + O

(trong oxit st) CO2 nCO = nO (trong oxit st) = 0,15 mol mO = 0,15.16 = 2,4 gam. mFe = 8 2,4 = 5,6 gam nFe = 0,1 mol.

Ta c: FeO

n x 0,1 2n y 0,15 3

Oxit st c cng thc l Fe2O3.

p n B.V d 21: em 11,2 gam Fe ngoi khng kh, sau mt thi gian thu c mt hn hp gm Fe vcc oxit. Ha tan hon ton hn hp trong dung dch H 2SO4 c, nng d thu c 3,36 lt khSO2 (ktc). S mol H2SO4 tham gia phn ng l :

A. 0,4 mol. B. 0,3 mol. C. 0,5 mol. D. 0,45 mol.Hng dn gii

S phm ng :Fe (FeO, Fe3O4, Fe2O3, Fe) Fe2(SO4)3 + SO2 + H2O (1)

mol : 0,2 0,1 0,15 Cn c vo s (1) v p dng nh lut bo ton nguyn t ta c :

22 4 24 2 4 3H SO SOSO trong Fe (SO )n n n 0,1.3 0,15 0,45 mol.

p n D.

4. Bi tp p dng :4.1. Bi tp dnh cho hc sinh lp 10Cu 1: Cho hn hp gm 0,15 mol Fe, 0,075 mol Fe 2O3 v 0,05 mol Fe3O4 tc dng ht vi dungdch H2SO4 long thu c dung dch A. Cho dung dch A tc dng ht vi dung dch NaOH d, lckt ta em nung trong khng kh n khi l ng khng i thu c m gam cht rn C. Gi tr cam l :

A. 36. B. 72. C. 65. D. 75.Cu 2: Kh 16 gam Fe2O3 thu c hn hp A gm Fe, Fe 2O3, FeO, Fe3O4. Cho A tc dng ht vidung dch H2SO4 c, nng. Khi lng mui sunfat to ra trong dung dch l :

A. 48 gam. B. 50 gam. C. 32 gam. D. 40 gam.Cu 3: Ha tan hon ton hn hp gm 0,1 mol Fe 2O3 v 0,2 mol FeO vo dung dch HCl d thuc dung dch A. Cho NaOH d vo dung dch A thu c kt ta B. Lc ly kt ta B ri emnung trong khng kh n khi lng khng i c m gam cht rn, m c gi tr l :

A. 16 gam. B. 32 gam. C. 48 gam. D. 52 gam.Cu 4: Cho hn hp X gm SO2 v O2 theo t l s mol 1:1 i qua V2O5 xc tc, un nng thu chn hp Y c khi lng 19,2 gam. Ho tan Y vo nc sau thm Ba(NO3)2 d thu c 37,28gam kt ta. Hiu sut phn ng SO 2 + O2 l :

A. 40%. B. 75%. C. 80%. D. 60%.



4.2. Bi tp dnh cho hc sinh lp 11Cu 5: Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo axit HNO3 (va ), thuc dung dch X (ch cha hai mui sunfat) v kh duy nht NO. Gi tr ca a l :

A. 0,04. B. 0,075. C. 0,12. D. 0,06.Cu 6 : t chy 9,8 gam bt Fe trong khng kh thu c hn hp cht rn X gm 3 oxit ca Fe. ho tan X cn va ht 500 ml dung dch HNO 3 1,6M thu c V lt NO (sn phm duy nht). Gitr V l :

A. 6,16 lt. B. 10,08 lt. C. 11,76 lt. D. 6,72 lt.Cu 7: Khi t 1 lt kh X cn 6 lt O2 thu c 4 lt CO2 v 5 lt hi H2O (cc th tch kh o cngiu kin nhit , p sut). CTPT ca X l : A. C4H10O. B. C4H8O2. C. C4H10O2. D. C3H8O.Cu 8: t chy hon ton mt hirocacbon trong 0,5 lt hn hp ca n vi CO 2 bng 2,5 lt O2 thuc 3,4 lt kh. Hn hp ny sau khi ngng t ht hi nc cn 1,8 lt, tip tc cho hn hp kh cnli qua dung dch kim d th cn li 0,5 lt kh. Cc th t ch c o cng iu kin nhit , psut. CTPT ca hirocacbon l :

A. C4H10. B. C3H8. C. C4H8. D. C3H6.Cu 9: t chy hon ton 0,6 gam hp cht hu c X ri cho sn phm chy qua b nh ng dungdch Ca(OH)2 d thy c 2 gam kt ta v khi lng bnh tng thm 1,24 gam. T khi ca X so viH2 bng 15. CTPT ca X l :

A. C2H6O. B. CH2O. C. C2H4O. D. CH2O2.Cu 10: t 0,15 mol mt hp cht hu c thu c 6,72 lt CO2 (ktc) v 5,4 gam H2O. Mt khct 1 th tch hi cht cn 2,5 th tch O2. Cc th tch o cng iu kin nhit , p sut.CTPT ca hp cht l :

A. C2H6O2. B. C2H6O. C. C2H4O2. D. C2H4O.Cu 11: t chy 200 ml hi mt hp cht hu c X cha C, H, O trong 900 ml O 2, th tch hnhp kh thu c l 1,3 lt. Sau khi ngng t hi nc ch cn 700 ml. Tip theo cho qua dung dchKOH d ch cn 100 ml kh bay ra. Cc th tch kh o cng iu kin nhit , p sut. CTPT caY l :

A. C3H6O. B. C3H8O2. C. C3H8O. D. C3H6O2.Cu 12: Cho 224,00 lt metan (ktc) qua h quang c V lt hn hp A (ktc) cha 12% C 2H2 ;10% CH4 ; 78% H2 (v th tch). Gi s ch xy ra 2 phn ng :

2CH4 C2H2 + 3H2 (1)CH4 C + 2H2 (2)

Gi tr ca V l A. 407,27. B. 448,00. C. 520,18 D. 472,64.Cu 13: t chy hon ton mt th tch kh thin nhin gm metan, etan, propan bng oxi khngkh (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 nhin trn l :

A. 70,0 lt. B. 78,4 lt. C. 84,0 lt. D. 56,0 lt.Cu 14: X l mt ancol no, mch h. t chy hon ton 0,05 mol X cn 5,6 gam oxi, thu c hinc v 6,6 gam CO2. Cng thc ca X l :

A. C2H4(OH)2. B. C3H7OH. C. C3H6(OH)2. D. C3H5(OH)3.Cu 15: t chy hon ton 0,1 mol mt anehit n chc X cn dng va 12,32 lt O2 (ktc) thuc 17,6 gam CO2. X l anehit no sau :

A. HCC CH2 CHO. B. CH3 CH2 CH2 CHO.C. CH2=CH CH2 CHO. D. CH2=C=CH CHO.



Cu 16: 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 chy hon 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.

Cu 17: t chy hon ton 0,1 mol mt axit cacboxylic n chc, cn va V lt O2 (ktc), thuc 0,3 mol CO2 v 0,2 mol H2O. Gi tr ca V l :

A. 8,96. B. 11,2. C. 6,72. D. 4,48.4.2. Bi tp dnh cho hc sinh lp 12Cu 18: t chy hon ton mt hp cht hu c X (C, H, N) bng lng khng kh va (gm1/5 th tch O2, cn li l N2) c kh CO2 , H2O v N2. Cho ton b sn phm chy qua b nh ngdung dch Ba(OH)2 d thy c 39,4 gam kt ta, khi lng dung dch gim i 24,3 gam. Kh thotra khi bnh c th tch 34,72 lt (ktc). Bit

2OXd < 2. CTPT ca X l :

A. C2H7N. B. C2H8N. C. C2H7N2. D. C2H4N2.Cu 19: t chy hon ton 1,605 gam hp cht hu c A thu c 4,62 gam CO2 ; 1,215 gam H2Ov 168 ml N2 (ktc). T khi hi ca A so vi khng kh khng vt qu 4. Cng thc phn t ca Al :

A. C5H5N. B. C6H9

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Phương pháp giải nhanh bài toán hóa học - Trắc nghiệm Vô cơ + Hữu cơ + 40 Đề thi - Nguyễn Minh Tuấn