Olimpic Hoa

8/8/2019 Olimpic Hoa

1/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Tuyn chn mt s bi t sch TUYN TP 10 NM THI OLYMPIC 30/4 HA

HC 10- NXB GIO DC

PHN I: HALOGEN

Cu 4: ( 1996 trang 7)

Xt phn ng tng hp hiro ioua:H2(kh) + I2(rn) 2HI(kh) H = +53kJ (a)

H2(kh) + I2(rn) 2HI(kh) (b)

1.Phn ng (a) l to nhit hay thu nhit?

2.Xc nh hiu ng nhit ca phn ng tng hp hiro ioua (b) da vo nng lng lin kt nu bit nng

lng lin kt ca H H, H I v I I ln lt bng 436, 295 v 150 kJ.mol-1. Gii thch s khc bit ca hai

kt qu cho (a) v (b).

3.Vit biu thc tnh hng s cn bng K ca phn ng (a) theo phng trnh ho hc ca phn ng.

4.Thc hin phn ng tng hp hiro ioua theo (b) trong mt bnh kn, dung tch 2 lit nhit T, c hng s

cn bng K = 36.

a, Nu nng ban u ca H2 v I2 bng nhau v bng 0,02M th nng ca cc cht ti thi im cn bng

l bao nhiu?

b, cn bng trn, ngi ta thm vo bnh 0,06gam hiro th cn bng cng b ph v v hnh thnh cn bng

mi. Tnh khi lng hiro ioua cn bng cui?

Gii:

1. Theo quy c H > 0 th phn ng thu nhit.

2. H2(kh) + I2(rn) 2HI(kh) (b)

Nn: H = (436 + 150) - 2. 295 = - 4kJ

Gi tr nh bt thng l do cha xt nng lng cn cung cp chuyn I2 (rn) theo phn ng (a) thnh I2(kh)

theo phn ng (b).

3. V I2 l cht rn nn:[ ]

[ ]

2

2

HIK

H=

4. H2(kh) + I2(rn) 2HI(kh)

Trc phn ng: 0,02M 0,02M 0

Phn ng: x x 2x

Cn li: 0,02 x 0,02 x 2x

Vy :( )

( ) ( )( )

22

36 2 6 0,02 0,0150,02 . 0,02

x x x x

x x= = =


2/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Kt lun: cn bng: [HI] = 0,03M, [H2] = [I2] = 0,005M

S mol H2 thm:

0,06 : 2 = 0,03 (mol) nng tng thm: 0,03: 2 = 0,015M

H2(kh) + I2(rn) 2HI(kh)

Ban u: 0,02M 0,005M 0,03M

Phn ng: a a 2a

Cn bng: 0,02 a 0,005 a 0,03 + 2a

( )

( ) ( )

20,03 2

360,02 0,005

aK

a a

+= =

a = 2,91.10-3 v 2,89.10-2.

V a < 0,005 nn ch nhn a = 2,91.10-3

Khi lng HI cn bng cui:

(0,03 + 2. 0,0029). 2. 128 = 9,165(gam)

Cu 6 (nm 1997 trang 17)

iu ch clo bng cch cho 100g MnO2 (cha 13% tp cht tr) tc dng vi lng d dung dch HCl m c.

Cho ton b kh clo thu c vo m500ml dung dch c cha NaBr v NaI. Sau phn ng, c cn dung dch, thu

c cht rn A (mui khan) c khi lng m gam.

a, Xc nh thnh phn cht rn A nu m = 117gam

b, Xc nh thnh phn cht rn A trong trng hp m = 137,6 gam. Bit rng trong trng hp ny, A gm hai

mui khan. T l s mol NaI v NaBr phn ng vi Cl2 l 3: 2. Tnh nng mol ca NaBr v NaI trong dung

dch u.Cc phn ng u hon ton.

Cho Mn = 55, Br = 80, I = 127, Cl = 35,5, Na = 23

Gii:

MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

1 mol 1 mol 1 mol

2

100 131( )

87MnOmoln

= =

Cl2 + 2NaI 2NaCl + I2

1,5a mol 3a mol 3a mol

Cl2 + 2NaBr 2NaCl + Br2

a mol 2a mol 2a mol

a, Gi s Cl2 phn ng ht mNaCl = 2.58,5 = 117(g)

Cl2 phn ng ht, NaI v NaBr phn ng ht mA = mNaCl = 117g (tha)

A ch cha NaCl

to


3/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Cl2 phn ng ht, NaI v NaBr d mA > 117 (g) (loi)

Cl2 d, NaI v NaBr ht mA < 117(g) (loi)

Vy A ch cha NaCl

b, m = 137,6g > 117g Cl2 phn ng ht

NaI, NaBr d, nNaI : nNaBr = 3 : 2 NaI phn ng ht, NaBr cn d.

nNaI : nNaBr = 3 : 2 gi 3a v 2a ln lt l s mol NaI v NaBr phn ng Cl2 ta c

21,5 2,5 1 0,4

Cl a a a an = + = = =

mA = mNaCl + mNaBr = 5a. 58,5 + mNaBr= 137,6

mNaBr = 20,6(g) 20,6

0,2( )103NaBr

moln = =

2.0,4 0,2 3.0,42 ; 2,4

0,5 0,5 NaBr NaI M C MC

+= = = =

Cu 1: 1998 trang 24

Cho kh Cl2 vo 100 ml dung dch NaI 0,2M (dung dch A). Sau , un si ui ht I2. Thm nc c

tr li 100 ml (dung dch B).

a, Bit th tch kh Cl2 dng l 0,1344 lt (ktc). Tnh nng mol/l ca mi mui trong dung dch B?

b, Thm t t vo dung dch B mt dung dch AgNO3 0,05M. Tnh th tch dung dch AgNO3 dng, nu kt

ta thu c c khi lng bng:

(1) Trng hp 1: 1,41 gam kt ta.

(2) Trng hp 2: 3,315 gam kt ta.Chp nhn rng AgI kt ta trc. Sau khi AgI kt ta ht, th mi n AgCl kt ta.

c, Trong trng hp khi lng kt ta l 3,315 gam, tnh nng mol/l ca cc ion trong dung dch thu c

sau phn ng vi AgNO3.

Gii:

20,006( )

0,134422,4Cl

moln ==

Cl2 + 2NaI 2NaCl + I2

0,006 mol 0,012 mol 0,012 mol

nNaI ban u= 0,2.0,1 = 0,02 (mol)

Vy ht Cl2 d NaI. Dung dch B cha 0,020 0,012 = 0,008 mol NaI d v 0,012 mol NaCl.

CNaCl = 0,012 / 0,1 = 0,12M

CNaI = 0,008/0,1 = 0,08M

b, bit ch c AgI kt ta hay c hai AgI v AgCl kt ta, ta dng 2 mc so snh.

Mc 1: AgI kt ta ht, AgCl cha kt ta.


4/26

Tuyn tp thi OLYMPIC 30/4 HA HC 100,008 mol NaI 0,008 mol AgI

m1 = mAgI = 0,008.235 = 1,88 gam

Mc 2: AgI v AgCl u kt ta ht

0,012mol NaCl 0,012 mol AgCl

m2 = 1,88 + 0,012.143,5 = 3,602 gam

m = 1,41 gam

1,41 < m1 = 1,88 gam vy ch c AgI kt ta.

1,410,006( )

235AgImoln = =

Vy3

0,006( )AgNO moln =

3ddAgNO

0,0060,12( )

0,05litV = =

m = 3,315 gamm1 = 1,88 < 3,315 < m2 = 3,602

Vy AgI kt ta ht v AgCl kt ta mt phn

mAgCl = 3,315 1,88 = 1,435 gam

nAgCl = 1,435/143,5 = 0,01 mol

S mol AgNO3

0,008 + 0,01 = 0,018 mol

3ddAgNO 0,018 0,36( )0,05 litV = =

c, Trong trng hp th nh, dung dch ch cn cha NO3-, Na+, Cl- d

33

. d . d

0,018

0,012 0,008 0, 02

0,012 0,01 0, 002

AgNONO

NaCl b NaI b Na Na bd

Cl du Cl bd Cl

n mol n

n n n n mol

n n n mol

+ +

= =

= = + = + == = =

Th tch dung dch =3

0,100 0,36 0, 46ddB ddAgNOV lit V + = + =

3

3

0,018 0,03910,46

0,0020,0043

0,460,0434

NO

Cl

Na NO Cl

M

C M

C C C M

C

+

= =

= =

= + =

Cu 2: 1999 trang 32

1. 18oC lng AgCl c th ha tan trong 1 lt nc l 1,5 mg. Tnh tch s tan ca AgCl.


5/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Tnh nng bo ha ca Ag+ (mol/lt) khi ngi ta thm dung dch NaCl 58,5 mg/lt vo dung dch AgCl

18oC.

2. Ngi ta khuy iot nhit thng trong bnh cha ng thi nc v CS2 ngui, v nhn thy rng t l

gia nng (gam/lt) ca iot tan trong nc v tan trong CS2 l khng i v bng 17.10-4.

Ngi ta cho 50ml CS2 vo 1 lt dung dch iot (0,1 g/l) trong nc ri khuy mnh. Tnh nng (g/l) ca iot

trong nc.

Gii:

1. p dng nh lut bo ton khi lng

T = [Ag+][Cl-]

Trong 1 lt dung dch:31,5 .10

143,5 Ag Cl + = =

Vy2

3 101,5 .10 1,1.10143,5

T

=

=Khi thm 1 lng dung dch NaCl.Gi S2 l nng Ag+ mi: [Ag+] = S2 [Ag+] = [Cl-] = S2

Gi l nng ca NaCl.

Trong dung dch s ion Cl-: /1 lt

Vy [Ag+] = S2

[Cl-] = + S2

18o

C nhit khng i. T khng i.S2(S2 + ) = 1,1.10-10 S22 + S2 1,1.10-10 = 0

Ch chn nghim ng dng:2 10

2

4,4.102

S + +=

= 0,0585/58,5 = 10-3

Vy 73 3 7

2

10 10 2.10 102

S + += =

S2 gim 100 ln so vi S1

2. Theo gi thuyt ta c:2

2

417.10H OI

CSI

C

C

=

uoc 30,1 /

1000n

IC g cm=

Gi x l s mol iot t nc i vo CS2

Vy: 30,1

/1000

nuocI

x g cmC

= v 2 ( / )

50CS

I

x g mlC =


6/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10

Suy ra: 40,1

: 17.101000 50

x x = x = 0,0967

Nng iot trong nc l: 0,1 x = 0,0033 (g/l)

Cu 4: 2000 trang 38

a, Hai cc ng dung dch axit clohiric c ln hai a cn A v B. Cn trng thi cn bng. Cho a gam

CaCO3 vo cc A v b gam M2CO3 (M l kim loi kim)vo cc B. Sau khi hai mui phn ng ht v tan

hon ton, cn tr li v tr cn bng.

1. Thit lp bieetr thc tnh khi lng nguyn t M theo a v b.

2. Xc nh M khi a = 5 v b = 4,8.

b, Cho 20gam hn hp gm kim loi M v Al vo dung dch hn hp H2SO4 v HCl, trong s mol HCl gp 3

ln s mol H2SO4 th thu c 11,2 lt kh H2(ktc) v vn cn d 3,4 gam kim loi. Lc ly phn dung dch ri

em c cn th thu c mt lng mui khan.

1. Tnh tng khi lng mui khan thu c bit M c ha tr 2 trong cc mui ny.2. Xc nh kim loi M nu bit s mol tham gia phn ng ca hai kim loi bng nhau.

Gii

a, CaCO3 + 2HCl CaCl2 + CO2 + H2O (1)

M2CO3 + 2HCl 2MCl + CO2 + H2O (2)

(1) khi lng cc A tng =( )100 44

0,56100

aa

=

(2) khi lng cc B tng = ( )2 60 44 0,562 60M b aM+ =+Ta c a = 5, b = 4,8 M 22,8 M l Natri

b, M + 2H+ M2+ + H2

2Al + 6H+ 2Al3+ + 3H2

2

11,22. 2. 122,4HH

n mol n + = = =

2 4 2 4 2 42 2 3 H SO H SO H SOHClH n n n nn + = + = +

2 4

10,2

5 H SOmoln = =

0,6HCl

moln =

1. mmui = (20 3,4) + 0,2.96 + 0,6.35,5 = 57,1gam

Gi x l s mol M tham gia phn ng

2. ta c h x.M + 27x = 20 3,4 = 16,6

Hn + = 2x + 3x = 1


7/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10 M = 56 (Fe)

Cu 3: 2001 trang 44

2. 14,224 iot v 0,112g hiro c cha trong bnh kn th tch 1,12 lt nhit 400oC. Tc ban u ca

phn ng l Vo = 9.10-5 mol.l-1.pht-1, sau mt thi gian (ti thi im t) nng mol ca HI l 0,04 mol/lt v

khi phn ng: H2 + I2 2HI

t cn bng th [HI] = 0,06 mol/lt

a, Tnh hng s tc ca phn ng thun v nghch.

b, Tc to thnh HI ti thi im t l bao nhiu?

c, Vit n v cc i lng tnh c.

Gii

1. Tnh hng s tc ca phn ng thun v phn ng nghch:

22

14,224 0,0560,056 0,05 /254 1,12

Ibd

mol I mol l n

= = = =

2 2

0,112 0,0560,056 0,05 /2 1,12H bd

H mol l n = = = =

Phn ng: H2 + I2 2HI

v1 = k1 [I2][H2] 1

1

2 2

v

I Hk

=

a,5 1 -1

3 1 -11 1 1

9.10 . . . t36.10 . . t

0, 05. . .0,05. .

mol l phl mol ph

mol l mol l

k

= =

Mc khc:

22 4 2 2

12

2 2 2 2 2

6 .10 . .

0,060,05 . .2

HIK mol l

K I Hmol l

K

= =

=

3 1 -1

2

36.10 . . . t9

9 l mol phK k

= =

k2 = 4.10-3.l.mol-1.pht-1.

b, Tc to thnh HI ti thi im t: v HI = vt vn = v1 v2

v1 = k1[I2][H2] = 36.10-3 l.mol-1.pht-1.2

0,060,05

2

mol2.l-2

v1 = 144. 10-7 mol. l-1. pht-1

v2 = k2 [HI]2 = 4.10-3 l. mol-1. pht-1. 42. 10-4 . mol2. l-2

v2 = 64 . 10-7 mol. l-1. pht-1

VHI = (144.10-7 - 64.10-7) mol. l-1. pht-1

(1)

(2)


8/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10VHI = 0,8. 10-5 mol. l-1. pht-1

Chuyn : L THUYT V PHN NG HA HC

Cu 1: trang 112

Trong bnh kn dung tch khng i cha 35,2x (g) oxi v 160x (g) SO2. Kh SO2 136,5oC c xc tc V2O5.

un nng bnh mt thi gian, a v nhit ban u, p sut bnh l P. Bit p sut bnh ban u l 4,5 atm

v hiu sut phn ng l H%.

a, Lp biu thc tnh p sut sau phn ng P v t khi hi d ca hn hp kh sau phn ng so vi khng kh,

theo H.

b, Tm khong xc nh P, d?

c, Tnh dung tch bnh trong trng hp x = 0,25?

Hng dn gii:

2

35,21,1 ( )

32O bdaux

x moln = =

2

160 2,5 ( )64SO bdau

xx moln = =

2SO2 + O2 2SO3

Ban u: 2,5x 1,1x 0

Phn ng: 2,2xH 1,1xH 2,2xH

Sau phn ng: (2,5x 2,2xH) (1,1x 1,1xH) 2,2xH

n2 = 2,5x - 2,2xG + 1,1x - 1,1xH + 2,2xH = x(3,6 - 1,1H) (mol)

Trng hp bi ton ng V, ng T.

( )

( )1 2

2 1

3,6 1,1 4,5

' 1,25 3,6 1,1' 3,6

x Hn n PP

P H P n n x

= = = = b, Khi H = 0 P = 4,5 (atm)

H = 1 P = 3,125 (atm)

Vy trong thi gian phn ng th 3,125 < P < 4,5

T khi hi so vi khng kh:

160 35,2 195,2(3,6 1,1 ) 3,6 1,1

sau truocsau

sau truoc

m m x xM

n n x H H

+= = = =

xt, to


9/26


( )/195,2 6,731

29 3,6 1,129 3,6 1,1sau

hhsau kk

M

HHd = = =

Khi H = 0 d = 1,87

H = 1 d = 2,69

Vy 1,87 < d < 2,69

C, p dng cng thc: PV = nRT

Pu = 4,5atm

Nu = 3,6x = 3,6.0,25 = 0,9(mol)

( )

22,40,9. 273 136,5273 6,72( )

4,5nRT

V lP

+= = =

Cu 11: trang 126

Tnh nng lng mng tinh th ion ca mui BaCl2 t cc d kin:Nhit to thnh tiu chun ca BaCl2 tinh th: - 205,6 kcal/mol

Nng lng lin kt Cl2: + 57 kcal/mol

Nhit thng hoa Ba: + 46 kcal/mol

Nng lng ion ha th nht ca Ba: + 119,8 kcal/mol

Nng lng ion ha th hai ca Ba: + 230,0 kcal/mol

Gii:

Nng lng mng tinh th ion ca BaCl2 tc l hiu ng nhit ca qu trnh sau, (trong nng lng tnh theo

n v kcal/mol):

20( ) ( ) 2( )

2 ; ?k k r

Cl BaCl H Ba + + =

Qu trnh to thnh mui BaCl2 tinh th qua nhng bc sau,

Phn li phn t Cl2: Cl2(k) 2Cl- ; H1 = +57,0

Clo nhn electron: 2Cl + 2e 2Cl- ; H2 = 2.(-87)

Ba rn thang hoa: Ba(r) Ba(k); H3 = +46,0

Ba mt electron: Ba(k) 1e Ba+(k); H4 = +119,8

Ba+(k) 1e Ba2+(k); H5 = +230,0

To mng li:2

0( ) ( ) 2( )2 ; ?

k k rCl BaCl H Ba+ + =

Qu trnh chung: Ba(r) + 2Cl-(k) BaCl2(r); H = -205,6

Theo nh lut Hess: H = H1 + H2 + H3 + H4 + H5 + H0

H0 = H (H1 + H2 + H3 + H4 + H5 )

= -205 57 (-174) - 46 119,8 230

= - 484,4 kcal/mol


10/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Cu 6: chuyn phn ng oxi ha kh trang 147

1. Vit cc phn ng ha hc trong cc trng hp sau:

a, Ozon oxi ha I- trong mi trng trung tnh.

b, Sc kh CO2 qua nc Javen.

c, Cho nc clo vo dung dch KI.

d, H2O2 kh MnO4- trong mi trng axit.

e, Sc kh flo qua dung dch NaOH long lnh.

Gii:

a, O3 + 2I- + H2O O2 + I2 + 2OH-

b, CO2 + NaClO + H2O NaHCO3 + HclO

c, Cl2 + KI 2KCl + I2

d, 5H2O2 + 2MnO-4 + 6H+ 5O2 + 2Mn2+ + 8H2O

e, 2F2 + 2NaOH 2NaF + H2O + OF2Cu 9: trang 150

Th tch kh clo cn phn ng vi kim loi M bng 1,5 ln lng kh sinh ra khi cho cng lng kim loi tc

dng hon ton vi axit clohiric d trong cng iu kin. Khi lng mui clo sinh ra trong phn ng vi clo

gp 1,2886 ln lng sinh ra trong phn ng vi axit axit clohiric.

a, Xc nh kim loi M.

b, Phn ng gia HCl v mui M (VI) xy ra theo chiu no khi nng cc cht u trng thi chun v khi

tng nng H+ ln hai ln.

Bit 6 30

/1,33

M MVE+ + = v

2

0

/21,36

Cl Cl VE =

Hng dn gii:

M +2

nCl2 MCln

M + mHCl MClm +2

mH2

a, T 2

n

= 1,5 2

m

v m, n = 1, 2, 3 n = 3, m = 2

v M + 106,5 = 1,2886.(M + 71)

M = 52 g/mol, M l Crom

b, 14H+ + 6Cl- + Cr2O72- Cl2 + 2Cr3+ + 7H2O

Eo = 1,33 1,36 = -0,03V: phn ng xy ra theo chiu nghch.

14

0,059 10,03 lg 0,105( )6 2

E V = = : phn ng xy ra theo chiu thun.


11/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Cu 11: trang 152

1. Ag kim loai c kh nng tc dng c vi dung dch HI 1N tp thnh kh H2 khng?

Cho TAgI = 8,3.10-17

E0(Ag+/Ag) = +0,799V

2. Trn 250ml dung dch AgNO3 0,01M vi 150ml dung dch HCl 0,1M. Tnh nng cc ion ti thi im cn

bng TAgCl = 10-10.

Hng dn gii:

[I-] = 1ion g/l [Ag+] = 8,3.10-17 ion g/l

0

/

0,059 lg Ag Ag

AgE

n AgE +

+

= + = 0,799 + 0,059.lg8,3.10-17 = -0,15V

Nu c phn ng xy ra, xt phn ng:

2Ag + 2H+ 2Ag+ + H2

E = +0,15V

Vy Ag c th y H2 ra khi HI trong iu kin cho.

2.3

30,25.0,01 2,5.10AgNO moln= =

20,15.0,1 1,5.10HCl

n mol = =

33

3

2,5.10 6,25.100,4

Ag M NO

+ = = =

2 21,5.10 3,75.100,4

Cl M H + = = =

Nu phn ng ht:

Ag+ + Cl- AgCl

6,25.10-3 6,25.10-3

Cn bng: AgCl Ag+ + Cl-

Ban u: 3,125.10-2M

Phn ng: x xCn bng: x 3,125.10-2 + x

TAgCl = 10-10 x(3,125.10-2 + x) = 10-10

x qu nh:10

9

2

103,2.10

3,125.10Mx

==

[Ag+] = 3,2.10-9M; [NO3-] = 6,25.10-3M

[Cl-] = 3,125.10-2M; [H+] = 3,75.10-2M

Cu 12: trang 154


12/26

Tuyn tp thi OLYMPIC 30/4 HA HC 101. MnO4- c th oxi ha ion no trong s cc ion Cl-, Br-, I- cc gi tr pH ln lt bng 0, 3, 5? Trn c s

ngh mt phng pho nhn bit cc ion halogenua c trong hn hp gm Cl -, Br-, I-.

Bit1,512/4

oE VMnO Mn

= + , 1,36/ 22

oE VCl Cl

= , 1,08/ 22

oE V Br Br

= , 0,62/ 22

oE VI I

=

2. A l dung dch cha AgNO3 0,01M, NH3 0,23M; v B l dung dch cha Cl-, Br-, I- u c nng 10-2M.

Trn dung dch A vi dung dch B (gi thuyt nng ban u ca cc ion khng i) th kt ta no c to

thnh? Trn c s hy ngh phng pho nhn bit s c mt ca ion Cl- trong mt dung dch hn hp

cha Cl-, Br-, I-.

Bit ( )3 32 2 NH Ag NH Ag+ + + K = 10-7,24

TAgCl = 10-10, TAgBr = 10-13, TAgI = 10-16

Hng dn gii:

1. 8H+

+ MnO4-

+ 5e Mn2+

+ 4H2O8

4

2

0,059 lg5

oMnO H

MnE E

+

+

+=

* pH = 0, 24 / 2 2 2

, ,/2 /2 /2

1,51MnO Mn

o o oCl Cl Br Br I I

V E E E E + = >

Nh vy MnO4- oxi ha c c Cl-, Br-, I-.

* pH = 3, 24 / 2

/21, 23

MnO Mn

oCl Cl

VE E + = < nhng ln hn2 2

,/2 /2

o o Br Br I I

E E . Nh vy MnO4- ch oxi ha

c Br-, I-.

* pH = 5, 24 / 2 2

,/2 /2

1,04MnO Mn

o oCl Cl Br Br

V E E E + = < nhng ln hn2

/2o

I IE . Nh vy MnO4- ch oxi ha

c I-.

Nh vy nhn bit dung dch hn hp Cl-, Br-, I- ta c th dng dung dch KmnO4 v dung mi chit CCl4.

Lc u tin hnh phn ng pH = 5, trong lp dung mi chit s c mu tm ca iot. Thay lp dung mi, pH

= 3, s thy dung mi co mu vng ca brom. Cui cng loi lp dung mi v kh lng MnO4- d v nhn bit

ion Cl- d bng AgNO3.

2. Coi phn ng gia AgNO3 v NH3 xy ra hon ton, nh vy dung dch A s gm ( )3 2 Ag NH+

0,01M v

NH3 0,23M.

( )3 32 2 NH Ag NH Ag+ + + K = 10-7,24

Nng ban u: 0,01 0,23

Nng cn bng 0,01- x x 0,23 + 2x


13/26


( )2

7,24. 0,23 2

100,01

x x

xK

+=

= . Gn ng ta c: [Ag+] = x 10-8M

[Ag+]. [Cl-] = 10-10 TAgCl = 10-10 nhng ln hn TAgBr = 10-13, TAgI = 10-16, nn ch c ion Br- v I- kt ta. Sau

dng axit ph phc lm tng nng ca ion Ag+ v nhn c Cl- nh kt ta AgCl.

Cu 13: trang 155Vit phng trnh di dng ion thu gn phn ng xy ra khi cho dung dch KI tc dng vi dung dch KmnO4

(trong mi trng axit) trong cc trng hp sau:

1. Sau phn ng cn d ion ioua (c gii thch).

2. Sau phn ng cn d ion pemanganat (c gii thch).

Bit gin th kh ca I v Mn trong mi trng axit nh sau:

1,70 1,14 1,45 0,544 6 3 3 H IO IO HIO I I

+ + + +

40,56 2,26 0,95 1,51 1,182 3 2

4 2MnO MnO MnO Mn Mn Mn + + + + + +

Hng dn gii:

Da vo gin th kh ca I- ta suy ra HIO khng bn v

3 3

0 0/ / HIO I IO HIO

EE > nn HIO s d phn thnh 3IO v 3I

Ta vit li gin th kh ca I nh sau:1,70 1,20 0,54

4 6 3 3 H IO IO I I + +

Da vo th kh ca Mn ta suy ra 24MnO v Mn3+ khng bn v chng c th kh bn phi ln hn th kh

bn tri nn chng s b d phn thnh hai tiu phn bn cnh nh HIO.

i vi qu trnh Mn2+ Mn ta cng khng xt v Mn kim loi khng th tn ti trong dung dch nc khi c

mt H+ do th kh ca Mn2+/Mn qu m.

Do ta c th vit li gin th kh ca Mn nh sau:

4

1,70 1,23 22MnO MnO Mn

+ + +

Ta c phng trnh ion thu gn trong cc trng hp nh sau:

1. Trng hp sau phn ng c I- d:

4 6 H IO

hoc 3IO

khng th cng tn ti vi I- v:

4 6 3 3

0 0

/ /1,7 0,54

H IO IO I I E V E V = > = v

3 3

0 0

/1, 2 0, 54

IO I I E V E V = > =

Nn 4 6 H IO hoc 3IO

u c th oxi ha I thnh 3I .

+1,20

+1,7+1,23

+1,51


14/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Nh vy I ch b oxi ha thnh 3I

.

Khi I d th 4MnO v 2MnO khng th tn ti v

4 2

0

/MnO MnOE v 2

2

0

/Mn O MnE + u ln hn

3

0

/I IE nn 4MnO

v

2MnO u c th oxi ha I thnh 3I . Nh vy 4MnO

b kh hon ton thnh 2Mn + . Do phng trnh phn

ng xy ra khi I d di dng ion thu gn nh sau:

2

4 3 22 15 16 5 2 8MnO I H I Mn H O + ++ + + +

Trng hp sau phn ng c d 4MnO :

2Mn + khng th tn ti khi 4MnO d v 2

4 2 2

0 0

/ /MnO MnO MnO MnE E +> nn 4MnO

s oxi ha 2Mn + thnh 2MnO .

Khi 2MnO d th 3I v I cng khng th tn ti v:

4 2 3 3 3

0 0 0

/ / /,

MnO MnO I I IO I E E E > nn 4MnO

oxi ha l 3I v I .

Nh vy sn phm sinh ra khi Ib oxi ha l 3IO v mt lng nh 4 6 H IO

v4 2 4 6

0 0

/1,7

MnO MnO H IO E E V = = .

Do phng trnh ny xy ra khi 4MnO

d nh sau:

4 2 3 22 2 2MnO I H MnO IO H O + + + + +

4 2 2 4 68 3 8 2 8 3MnO I H H O MnO H IO + + + + +

Cu 7: trang 170

nh gi kh nng ha tan ca HgS trong:

a, Axit nitric

b, Nc cng toan

Bit3

0 02/ 0,96 NO N O E E V = = ; 2

0 1

/ 0 0,17S H S E E V = = ;51,810HgST

=

Hng dn gii:

a, Trong axit nitric:

Cc qu trnh xy ra: 3 3 HNO H NO+ +

2 23 HgS Hg S + + 51,810t HgS T T= =

23 H S HS + + 2

1 12,9210ak =

23 HS H H S + +

1

1 710ak =

23 2 2 H S e S H + +

102

1 0,0591 10

E

k

=

023

1 0,0593 2 2

2

3 2

2 4 3 2 10

3 2 8 3 3 2 4

E

NO H e NO H k

HgS NO H Hg S NO H O

+

+ +

+ + + =

+ + + + +

2 1

3 3 3 3 2

1 2. . . .t a ak T k k k k =


15/26


2 1 1 2lg 3lg 3lg 3lg 3lg 2 lga ak T k k k k = +

0 0

1 22 33( 51,8) 3( 12,92) 3( 7) 3 20,059 0,059

E E = +

155,4 38,76 21 17,29 97,63 15,3= + + + =

15,3

10k

=V k rt nh nn xem nh HgS khng tan trong HNO3.

b, Trong nc cng toan:

Cc qu trnh xy ra:

2

3 23 2 8 3 3 2 4 HgS NO H Hg S NO H O + ++ + + + + 15,310k =

2 2

43 4 Hg Cl HgCl + + 14,924 10 =

2

3 4 23 2 8 12 3 3 2 4 HgS NO H Cl HgCl S NO H O + + + + + + +

4

3 4' . lg ' lg 3lg 15,3 3.14,92 29,46k k k k = = + = + =

29,4610k = rt ln. Vy HgS tan mnh trong nc cng toan.

Cu 8: trang 171

Thm 1 ml dung dch H2S 0,01M vo 1ml dung dch hn hp:

Fe3+ 0,01M v H+ 0,1M.

C xut hin kt ta khng? Bit:

2

7,02

1( ) 10H SK= ;

2

10,9

2( ) 10H SK= ; 3 20

/

0,77 Fe Fe

E V+ + = ;2

0

/ 0,14S H SE V= ;17,410

FeS

T =

Hng dn gii:

Nng cc cht sau khi trn: [H2S] = 5.10-3 mol/l

[Fe3+] = 5.10-3 mol/l

[H+] = 5.10-2 mol/l

3 21 Fe e Fe+ ++ (1)0

130,059

1 10 10nE

K = =

22 2S H e H S ++ + (2) 4,7452 10K =

T hp (1) v (2)

3 2

22 2 2 Fe H S Fe S H + + ++ + + (3) 2 1 21,2553 1 2. 10 K K K

= =

5.10-3 2,5.10-3 5.10-3 5.10-3(M)

V K3 rt ln nn phn ng (3) xy ra hon ton:

2 H S H HS + + (4) K4

2 HS H S + + (5) K5

T hp (4) v (5)


16/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10H2S 2H+ + S2- (6) K6 = K4. K5 = 10-19,92

Cn bng (2,5.10-3 x) (5,5.10-2 + 2x) x

( )

( )

22

19,92

6 3

5,5.10 210

2,5.10

x xK

x

+= =

x = [S2-

] = 5,2.10-20

.Ta c: [Fe2+].[S2-] = 2,6.10-23 < TFeS

Vy FeS cha kt ta.

Cu 2: trang 192

1. Hy cho bit s bin thin tnh axit ca dy HXO4 (X l halogen). Gii thch?

2. Mt hn hp X gm 3 mui halogen ca kim loi Natri nng 6,23g ha tan hon ton trong nc c dung

dch A. Sc kh clo d vo dung dch A ri c cn hon ton dung dch sau phn ng c 3,0525g mui khan

B. Ly mt na lng mui ny ha tan vo nc ri cho phn ng vi dung dch AgNO3 d th thu c

3,22875g kt ta. Tm cng thc ca cc mui v tnh % theo khi lng mi mui trong X.

Hng dn gii:

Tnh axit ca dy HXO4 gim dn khi X: Cl I

Gii thch:

Cu to ca HXO4.

O O

H O X O hoc H O X = O

O OV Cl I m in gim lm cho phn cc ca lin kt O H gim.

2. Gi s lng mui khan B thu c sau khi cho clo d vo dung dch A ch c NaCl

3,05250,0522

58,5NaCln mol = =

NaCl + AgNO3 AgCl + NaNO3 (1)

Theo (1) 3,22875

.2 0,045 0,0522

143,5 NaCl AgCl n n mol mol = = = 106,11 l iot. Vy cng thc ca mui th 2 l NaI.Do c hai trng hp:

* Trng hp 1: NaF, NaCl v NaI

Gi a, b ln lt l s mol ca NaCl v NaI

Ta c:58,5 150 5,81 0,01027

0,045 0,03472

a b a

a b b

+ = = + = =

mNaCl = 58,5.0,01027 = 0,6008(g)

mNaI = 150. 0,03472 = 5,208 (g)

Vy:0,6008

% .100% 9,64%6,23

NaCl = =

0,6008% .100% 9,64%

6,23NaCl = =

% 6,77%NaF =

% 83,59%NaI =

Trng hp 2: NaF, NaBr v NaI

Ta c:103 ' 150 ' 5,81 ' 0,02

' ' 0,045 ' 0,025

a b a

a b b

+ = = + = =

mNaBr = 103.0,02 = 2,06(g)

mNaI = 150.0,025 = 3,75 (g)

Vy2,06

% .100% 33,07%6,23

NaBr = = ;3,75

% .100% 60,19%6,23

NaI = = ; % 6,74%NaF =

Cu 8: X l mt loi mui kp ngm nc c cha kim loi kim clorua v magie clorua. xc nh cng thc

ca X, ngi ta lm cc th nghim sau:* Ly 5,55g X ha tan vo nuoc ri em dung dch thu c tc dng vi lng d dung dch AgNO3 to

thnh 8,61gam kt ta.

* Nung 5,55g X n khi lng khng i th khi lng gim 38,92%. Cht rn thu c cho tc dng vi mt

lng d dung dch NaOH to kt ta. Lc ly kt ta, ra sch ri nung n khi lng khng i thu c

0,8gam cht rn.

Hy xc nh cng thc ca X.

Hng dn gii:


18/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10* Th nghim 1:

Ag+ + Cl- AgCl;8,61

0,06143,5AgCl Cl

n n mol = = =

Th nghim 2:

Khi nung, xy ra s loi nc c mui khan, nn khi lng nc ngm trong mui bng

38,92%5,55=2,16gam, ng vi 2,16/18 = 0,12mol H2O.

Khi tc dng vi dung dch NaOH: Mg2+ + 2OH- Mg(OH)2

Nung Mg(OH)2 MgO + H2O

nMgO = 0,8/40 = 0,02mol = 2Mgn + ng vi 0,02 mol MgCl2 ban u.

Cn li 0,02 mol Cl- s kt hp vi ion kim loi M+ cho 0,02 mol MCl c khi lng bng:

5,55 (2,16 + 0,02.95) = 1,49 gam.

Tnh c:

1,4935,5 39

0,02M

= = vC. Vy M l KaliCng thc ca mui l: 0,02 mol KCl, 0,02 mol MgCl2, 0,12 mol H2O hay KCl.MgCl2.6H2O.

Cu 10: trang 206

Cho hn hp A gm 3 mui MgCl2, NaBr, KI. Cho 93,4 gam hn hp A tc dng vi 700 ml dung dch AgNO3

2M. Sau khi phn ng kt thc thu c dung dch D v kt ta B. Lc kt ta B, cho 22,4 gam bt Fe vo dung

dch D. Sau khi phn ng kt xong thu c cht rn F v dung dch E. Cho F vo dung dch HCl d to ra 4,48

lt H2 (kc). Cho dung dch NaOH d vo dung dch E thu c kt ta, nung kt ta trong khng kh cho n

khi lng khng i thu c 24 gam cht rn.1. Tnh khi lng kt ta B.

2. Ha tan hn hp A trn vo nc to ra dung dch X. Dn V lt Cl2 sc vo dung dch X, c cn dung dch

sau phn ng thu c 66,2 gam cht rn. Tnh V(kc)?

Hng dn gii:

Gi a, b, c ln lt l s mol ca MgCl2, NaBr, KI.

Phng trnh phn ng:

Cl- + Ag+ AgCl (1)

Cl- + Ag+ AgBr (2)

I- + Ag-+ AgI (3)

Fe + 2Ag+(d) Fe2+ + 2Ag (4)

Fe(d) + 2H+ Fe2+ + H2 (5)

Fe2+ + 2OH- Fe(OH)2 (6)

2Fe(OH)2 +1

2O2 + H2O 2Fe(OH)3 (7)

to


19/26

Tuyn tp thi OLYMPIC 30/4 HA HC 102Fe(OH)3 Fe2O3 + 3H2O (8)

Mg2+ + 2OH- Mg(OH)2 (9)

Mg(OH)2 MgO + H2O (10)

Theo (5) nFe(d) = 24,48

0,222,4H

n mol = =

( )0,2 2 0,4

Ag dun mol + = =

Theo (1) (2) (3)

(0,7 2) 0,4 2 1Ag

n a b c mol + = = + + = (I)

mrn = 2 3 160 0,1 40 24 Fe O MgOm m a= + = + =

a = 0,2 (II)

mA = 95.0,2 + 103b + 166c = 93,4 (III)

2. Phng trnh phn ng: Cl2 + 2I-

2Cl-

+ I2 (1)Cl2 + 2Br- 2Cl- + Br2 (2)

Khi phn ng (1) xy ra hon ton khi lng mui gim:

0,2(127 35,5) = 18,3 gam

Khi c hai phn ng (1) v (2) xay ra hon ton khi lng mui gim:

0,2(127 35,5) + 0,4(80 35,5) = 36,1 gam

Theo bi ta co khi lng mui gim:

93,4 66,2 = 27,2 gam

18,3 < 27,2 < 36,1 chng t phn ng (1) xy ra hon ton v c mt phn phn ng (2).

t s mol Br2 phn ng bng x th khi lng mui gim:

18,3 + x(80 35,5) = 27,2

Suy ra x = 0,2 mol

Vy2

1(0,2 0,2) 0,2

2Cln mol = + =

2

22,4.0,2 4,48Cl litV = =

Cu 11: trang 208

Hn hp A: KClO3, Ca(ClO3)2, Ca(ClO)2, CaCl2, KCl nng 83,68 gam. Nhit phn hon ton A thu c cht

rn B gm CaCl2, KCl v mt th tch oxi va oxi ha SO2 thnh SO3 iu ch 191,1 gam dung dch

H2SO4 80%. 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 gp 22/3 ln lng KCl trong A.

1. Tnh khi lng kt ta C?

2. Tnh thnh phn phn trm v khi lng ca KClO3 trong A?


20/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10Hng dn gii:

Hn hp A

3

3 2

2

2

: ( )

( ) : ( )

( ) : ( )

: ( )

: ( )

KClO a mol

Ca ClO b mol

Ca ClO c mol

CaCl d mol

KCl e mol

3 23

2

ot KClO KCl O +

Mol a a 3

2a

2 23 2 3( )ot CaCl OCa ClO +

Mol b b 3b

2 22( )o

t CaCl OCa ClO +Mol c c c

* Theo nh lut bo ton khi lng:22A O KCl CaCl

m m m m= + +

83,68 = 74,5(a + e) + 111(b + c + d) + 32(3

2

a+ 3b + c) (1)

,2 2 32 2oxt tSO O SO+

Mol

33

2

ab c

+ + 3 6 2a b c+ + 2 2 43 H O H SOSO +

Mol 3 6 2a b c+ + 3 6 2a b c+ +

98(3 6 2 ) 100

80191,1

a b c+ + = (2)

80 191,13 6 2 1,56

100 98a b c

+ + = =

* Cht rn B 2 2 3: ( ): ( ) ddKCaCl b c d mol CO KCl a e mol

+ + + +

KCl + K2CO3

CaCl2 + K2CO3 2KCl + CaCO3

Mol (b + c + d) (b + c + d) 2(b + c + d) (b + c + d)

S mol K2CO3 = 0,36. 0,5 = 0,18 (mol) = b + c + d (3)

* Kt ta C: CaCO3

Khi lng kt ta CaCO3 = 100(b + c + d) = 100. 0,18 = 18 gam


21/26

Tuyn tp thi OLYMPIC 30/4 HA HC 102. Dung dch D (KCl)

nKCl = a + e + 2(b + c + d) = (a + e) + 2. 0,18

= a + e + 0,36

(ddD) ( ) (ddA) ( )

22 223 3 KCl KCl A KCl KCl A

m m n n= =

223

0,36 ea e+ + = (4)

T (1), (2), (3), (4) ta c:

383,68 74,5( ) 111( ) 32 3

2

3 6 2 1,56

0,18

220,36

3

aa e b c d b c

a b c

b c d

ea e

= + + + + + + + + + =

+ + =

+ + =

1,56

83,68 74,5( ) (111 0,18) 322

a e = + + +

( )74,5 38,74a e + =

0,52

220,36

3

a e

a e e

+ =

+ + =

3 122,5. .100% 58,56%83,68aKClO trongA = =

Cu 12: trang 210

Cho 50g dung dch X cha 1 mui halogenua kim loi ha tr II tc dng vi dung dch AgNO3 d th thu c

9,40g kt ta. Mt khc, dng 150g dung dch X phn ng vi dung dch Na2CO3 d th thu c 6,30g kt ta.

Lc kt ta em nung n khi lng khng i, kh thot ra cho vo 80g dung dch KOH 14,50%. Sau phn

ng, nng dung dch KOH gim cn 3,85%.

a, Xc nh CTPT ca mui halogen trn.

b, Tnh C% mui trong dung dch X ban u.

Hng dn gii:

a, CTPT mui MX2:

2 3 3 22 2 ( )MX AgNO AgX M NO+ + (1)

2 2 3 3 2MX Na CO MCO NaX + + (2)

3 2MCO MO CO + (3)


22/26


2 2 3 22 duCO KOH K CO H O+ + (4)

L lun:

(1) s mol AgX(1)9,4

108 X=

+(5)

(2) s mol MX2(2) = s mol MCO3(2) = s mol CO2 =

6,3

60M + (6)

(4) mKOHpu(4) =6,3

2 5660M

+

(7)

M mKOH(b) = 11,6g

mKOHsau p6,3 3,85

44 8060 100M

= + + (8)

mKOHpu(4) = mKOH(b) + mKOHsau p

6,3 6,3 3,852 56 11,6 44 80

60 60 100M M

= + + +

(9)

Gii ra M = 24 (Mg).

(6) s mol MX2(2) = 0,075 s mol MX2(1) = 0,025

(1) s mol ca AgX(1) = 2 ln s mol MX2(1)

(5) X = 80 (Br)

CT mui: MgBr2.

b, Khi lng MgBr2 (trong 50gam dung dch X) = 4,6g

C% MgBr2 = 9,2%.

Cu 19: trang 224

X l mui c cng thc NaIOx.

Ha tan X vo nc thu c dung dch A. Cho kh SO2 i t t qua dung dch A thy xut hin dung dch mu

nu, tip tc sc SO2 vo th mt mu nu v thu c dung dch B. Thm mt t dung dch axit HNO3 vo dung

dch B v sau thm lng d dung dch AgNO3, thy xut hin kt ta mu vng.

- Thm dung dch H2SO4 long v KI vo dung dch A, thy xut hin dung dch mu nu v mu nu mt ikhi thm dung dch Na2S2O3 vo.

a, Vit cc phng trnh phn ng xy ra di dng ion thu gn.

b, xc nh chnh xc cng thc ca mui X ngi ta ha tan 0,100 gam vo nc, thm lng d KI v vi

ml dung dch H2SO4 long, dung dch coa mu nu. Chun I2 sinh ra dung dch Na2S2O3 0,1M vi cht ch th

h tinh bt cho ti khi mt mu, thy tn ht 37,40ml dung dch Na2S2O3. Tm cng thc ca X.

hng dn gii:

a, Vit phng trnh phn ng xy ra di dng ion thu gn:


23/26

Tuyn tp thi OLYMPIC 30/4 HA HC 10(2x - 1)SO2 + 2IOx- + (2x - 2)H2O I2 + (2x - 1)SO42- + (4x - 4)H+ (1)

SO2 + I2 + 2H2O 2I- + SO42- + 4H+ (2)

Ag+ + I- AgI

2IOx- + (2x - 1)I- + 2xH+ xI2 + xH2O (3)

I2 + 2S2O32- 2I- + S4O62- (4)

b, Theo (4): 22 2 30,5 0,5.0,0374.0,1 0,00187( )I S On n mol = = =

x = 4

Vy cng thc mui X l NaIO4.

Cu 20: trang 225

1. a, Cho m gam hn hp gm NaBr v NaI phn ng vi dung dch H2SO4 c, nng thu c hn hp kh A

iu kin chun. iu kin thch hp, A phn ng va vi nhau to cht rn c mu vng v mt cht lng

khng lm chuyn mu qu tm. Cho Na d vo phn cht lng c dung dch B. Dung dch B hp th va

vi 2,24 lt CO2 iu kin tiu chun c 9,5 gam mui.

Tm m.

b, ngh mt phng php tinh ch NaCl khan c ln cc mui khan NaBr, NaI, Na2CO3.

2. a, Mt axit mnh c th y c axit yu ra khi mui, nhng mt axit yu cng c th y c axit mnh

ra khi mui. Ly v d minh ha v gii thch.

b, Ti sao H2SO4 khng phi l axit mnh hn HCl v HNO3 nhng li y c nhng axit ra khi mui?

c, C mt hn hp gm 2 kh A v B:

- Nu trn cng mt th tch th t khi hi ca hn hp so vi Heli l 7,5(d 1).

- Nu trn cng khi lng th t khi hi ca hn hp so vi oxi l ( )111

15d

- Tm khi lng mol ca A v B. Bit th tch kh c o iu kin tiu chun.

Hng dn gii:

1. a, A phn ng va vi nhau to cht rn mu vng A l hn hp SO2 v H2S.

Mt khc, NaBr c tnh kh yu hn NaI.

2NaBr + 2H2SO4 Na2SO4 + Br2 + SO2 + 2H2O

0,15mol 0,075mol

8NaI + 5H2SO4 4Na2SO4 + 4I2 + H2S + 4H2O

(0,15.8)mol 0,15mol

2H2S + SO2 3S + 2H2O

0,15mol 0,075mol 0,15mol

Cht lng l H2O:

2Na + 2H2O 2NaOH + H2


24/26

Tuyn tp thi OLYMPIC 30/4 HA HC 100,15mol 0,15mol

B l NaOH

CO2 + NaOH NaHCO3

x(mol) x x (mol)

CO2 + 2NaOH Na2CO3 + H2O

y 2y y

2

2,240,1( )

22,4COn mol = =

0,10,05

84 106 9,5

x yx y

x y

+ = = = + =

mhn hp = (0,15.103) + (0,15.8.150) = 195,45(g)

b, Cho hn hp trn vo dung dch HCl, ch Na2CO3 phn ng:

Na2CO3 + 2HCl 2NaCl + H2O + CO2Sc kh clo vo dung dch thu c:

2NaBr + Cl2 2NaCl + Br2

2NaI + Cl2 2NaCl + I2

C cn dung dch, Br2 v I2 ha hi thot ra, NaCl kt tinh li.

2. a, Mt axit mnh c th y c mt axit yu ra khi mui v axit yu l cht in li yu hoc cht khng

bn.

Na2CO3 + 2HCl 2NaCl + H2O + CO2

CO2 + H2O H2CO3 H+ + HCO3- (1)

HCO3- H+ + CO32- (2)

HCl H+ + Cl-

Khi cho HCl vo dung dch Na2CO3 lm tng nng H+ lm cho cc cn bng (1) (2) chuyn sang tri to ra

H2CO3 ri sau l CO2 v H2O

Ngc li, 1 axit yu c th y c 1 axit mnh ra khi mui

Pb(NO3)2 + H2S PbS + 2HNO3

Axit yu axit mnh

V PbS khng tan.

b, H2SO4 khng phi l axit mnh hn HCl v HNO3 nhng y c 2 axit ra khi mui v H2SO4 l axit

khng bay hi cn HCl v HNO3 l axit d bay hi.

2NaCl + H2SO4ot Na2SO4 + 2HCl

2NaNO3 + H2SO4ot Na2SO4 + 2HNO3


25/26


c, 11 17,5 7,5.4 304

Md M= = = =

22 2

11 11.32 352

32 15 15 15

Md M= = = =

1 30 602

A B

M A B

+

= = + =

2

2 2 3521 1 15

m m ABM

m m A B

A B A B

+= = = =

++ +

352.60704

2.15AB = =

60 44

704 16

A B A

AB B

+ = = = =

hoc16

44

A

B

= =

Cu 32: trang 243

Nung hn hp bt Mg v S trong bnh kn ri ngui. Ly ton b cc cht sau phn ng cho tc dng vi

lng d dung dch HCl thu c sn phm kh c t khi hi so vi khng kh l 0,9. t chy hon ton 3 lt

sn phm kh (kc) trn ri thu sn phm chy vo 100ml dung dch H2O2 5% (t khi bng 1).

a, Vit phng trnh phn ng xy ra.

b, Tnh phn trm khi lng Mg v S trong hn hp u.

c, Tnh nng % ca dung dch thu c cui cng.

Hng dn gii:a, Cc phng trnh phn ng xy ra.

Mg + S MgS (1)

a (mol)

MgS + 2HCl MgCl2 + H2S (2)

a (mol) a (mol)

MTB kh = 29.0.9 = 26,1< 2H SM

Vy trong sn phm c kh H2 do Mg dMg + 2HCl MgCl2 + H2 (3)

b(mol) b (mol)

2H2S + 3O2 2SO2 + 2H2O (4)

H2 + 1/2O2 H2O (5)

SO2 + H2O2 H2SO4 (6)

b, Gi a, b l s mol Mg tham gia phn ng (1) v (3)


26/26

Tuyn tp thi OLYMPIC 30/4 HA HC 1034 2

26,1tba b

Ma b

+= =

+

7,9a = 24b

( )

( )

24%

24 32Mga b

ma b a

+=

+ +

% 50,08%Sm =

c,2

0,74%H

V = ;2

0,033H

n mol = ;2

2,26( )SO

V lit = ;2

0,1SO

n mol

2 2

100.50,147( )

100.34H On mol = =

Dung dch sau cng cha H2SO4, H2O2.

mdd = 100 + 18.(0,033 + 0,1) + 64.0,1 = 108,794 (g)

2 4 H SOm = 0,1.98 = 9,8 (g)

2 4% 9% H SOC =

2 2% 1, 45%

H OC =

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