TỰ HỌC GIỎI HÓA HỌC TẬP 1: HÓA HỌC VÔ CƠ (DÀNH CHO LUYỆN THI ĐẠI HỌC, CAO ĐẲNG) NXB ĐẠI HỌC SƯ PHẠM

8/10/2019 T HC GII HA HC TP 1: HA HC V C (DNH CHO LUYN THI I HC, CAO NG) NXB I HC S PH

1/165

NGUYN VN HI - NGUYN NAM TRUNG - HU NG

BNG C SM - QUC PHNG - L NG KHNG

T hc Gii Ho HC

NH XUT BN I HC S PHM

WWW.DAYKEMQUYNHON.COM

WWW.FACEBOOK.COM/DAYKEM.QUYNHON

W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


2/165

M s: 02.02.19/32 H 2013



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


3/165

Li ni u

Cc em hc sinh v qu bn c yu mn!

Cun sch T hc gii Ho hc - Tp 1: Ho hc V c c vit nhm gip

cc em hc sinh t n luyn thi i hc, Cao ng.Ni dung ch o: Lp li, tnh nhanh, tc hnh n luyn Ho hc.

im mi ca cun sch:

1.Lp li: i km vi mi bi tpgc (bi tp hay, cha kin thc trng tm) l

1 - 2 bi tp tng ng v a chiu. y l im mi nhtca cun sch, cha

ng trit lv hc tp, gip hc sinh t hc. Mun t hc, trc ht cc em cn

bit cch hc (xem v d mu bit lm), sau t lm, lp li cho thnh tho

ri mi n sng to. Lp li s ln nht nh s gip cc em nh lu v khc

su kin thc.

2.Tnh nhanh: p dng cc phng php tnh nhanh dng cho thi trc nghim.3.C nhiu hnh nhminh ho gip tng hp dn ca cun sch.

Trong qu trnh bin son, chng ti tp hp, sng tc cc cu hi, bi tp vi

mc kin thc tng ng yu cu ca thi i hc, Cao ng nhm gip

cc em c s chun b tt nht cho k thi.

Cun sch c s dng hnh nh minh ho c su tm t Internet v tng t

cc cu ni ni ting.

Chng ti hi vng cun sch s l ti liu tham kho b ch cho cc em hc sinh

n thi i hc, Cao ng v cc thy c gio trong qu trnh dy hc.

Rt mong nhn c cc gp , trao i, bnh lun ca bn c trn Facebook

mang tn cun sch: Tu hoc gioi Hoa.

Cm n bn c la chn cun sch ny.

CC TC GI



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


4/165

1

Mc lc

...................................................................................................................

1. ..........................................................

. ....................................................................

. ...............................................................................

. ......................................... 101

. ......................................................................................... 11

. ................................................................................ 1

. .................................................................................. 11

. ...................................................................

. .................................................

10. .........................................................................

11. ........................................................... 0

1. ........................................................................

1. ..............................................................................

1. ................................................................................. 0

1. , , ................



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


5/165

Nh xut bn i hc S phm

5

Chun 1

CCCCCCCC PHPHPHPHNG PHNG PHNG PHNG PHPPPP GIGIGIGII NHANHI NHANHI NHANHI NHANH

HcHcHcHc

HaHaHaHa

KhngKhngKhngKhng

KhKhKhKh

Ch eCh eCh eCh e

KhKhKhKhngngngng

ChuChuChuChu

KhKhKhKh

1 PHNG PHP BO TON KHI LNG

A. L THUYT TRNG TM1. Ni dung

Khi lng ca tt c cc nguyn t lun c bo ton trong cc phnng ho hc.

2. H qu+ Tng khi lng cc cht phn ng = Tng khi lng cc cht sn phm.+ Khi lng hn hp = Tng khi lng cc nguyn t thnh phn.

+ Khi lng mui khi c cn = Tng khi lng cc ion to mui.

B. V D CU HI

V d 1: t chy hon ton 6,4 gam hn hp Mg, Al v Zn trong kh O 2(d)thu c 9,6 gam hn hp oxit. Th tch kh O2 (ktc) tham gia phnng lA.4,48 lt. B. 2,24 lt. C. 3,36 lt. D.5,60 lt.

Bo ton khi lng:

2X O Ym + m = m

2Om = 9,6 6,4 = 3,2 (gam).

2O

n =3,2

= 0,1 (mol)32

2O

V = 2,24 lt B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


6/165

T hc gii Ho hc Tp 1: Ho hc V c

6

Cu 1: t chy hon ton 6,4 gam hn hp X gm Mg, Al v Zn trong O2(d)thu c 9,6 gam hn hp oxitY. Ho tan hon ton Y trong dung dchHNO3, to thnh dung dch Z cha m gam mui. Gi tr ca m lA.18,8. B. 25,0. C. 31,2. D.37,4.

V d 2: Ho tan hon ton 4,7 gam hn hp X gm Mg, Al v Fe bng mt

lng va dung dch H2SO4 long, thu c 4,48 lt kh H2 (ktc) vdung dch cha m gam mui. Gi tr ca m lA.24,3. B.23,9. C.16,8. D.14,5.

Nhn xt:

2 4H SOn =

2Hn = 0,2 (mol).

Bo ton khi lng:2 4 2X H SO H

m + m = m + m 4,7 + 0,2.98 = m + 0,2.2

m = 23,9 (gam) B.

Cu 2: Cho 5,2 gam hn hp X gm Al, Mg v Zn tc dng vi mt lng va

dung dch H2SO4 10%, thu c dung dch Y v 3,36 lt kh H2 (ktc).Khi lng dung dch Y lA.152,2 gam. B.146,7 gam. C. 152,0 gam. D.151,9 gam.

V d 3: Cho 3,64 gam hn hp X gm Li, Na v Ba vo 400 ml dung dch HCl0,1M, to thnh 1,12 lt kh H2(ktc) v dung dch Y. C cn Y thu c mgam cht rn khan. Gi tr ca m lA.5,06. B.6,08. C.5,40. D.5,34.

2H

1,12n = = 0,05

22,4

2HOH

n = 2n = 0,1 OH (Y)n = 0,1 Hn + = 0,06 (mol).

Bo ton khi lng: X Cl OHm = m + m + m = 3,64 + 0,04.35,5 + 0,06.17 = 6,08.

B.

Cu 3: Cho 2,65 gam hn hp X gm hai kim loi kim (thuc hai chu k k tip)vo 200 ml dung dch gm HCl 0,5M, thu c dung dch Y. C cn Y thuc 7,05 gam cht rn khan. Hai kim loi trong X lA.Li v Na. B.Na v K. C.K v Rb. D.K v Cs.

V d 4: Cho 200 ml dung dch KOH 1M vo 200 ml dung dch H3PO40,5M, thu

c dung dch X. C cn X thu c m gam mui khan. Gi tr ca m lA.17,4. B.21,0. C. 19,2. D. 15,6.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


7/165


7

KOHn = 0,2.1 = 0,2 (mol); 3 4H POn = 0,2.0,5 = 0,1 (mol).

Xt s phn ng: H3PO4 + KOH Mui + H2OBo ton khi lng:

3 4 2H PO KOH H Om + m = m + m

Nhn xt:OH (KOH)

n = 0,2;3 43 4

H POH (H PO )n = 3n = 0,3+

2H O OHn = n = 0,1 (mol).

0,1.98 + 0,2.56 = m + 0,2.18 m = 17,4 (gam) A.

Cu 4: Cho 200 ml dung dch H3PO4a mol/vo 150 ml dung dch KOH 1Mthu c dung dch Y c cha 15,5 gam hn hp mui. Gi tr ca a lA.0,3. B.0,5. C.0,7. D.0,8.

C. HNG DN GII

Cu 1: Bo ton khi lng:2X O Y

m + m = m 2O

m = 9,6 6,4 = 3,2 (gam).

2On =3,2

= 0,1 (mol)32 On = 0,2 mol 2 OO (Y)n = n = 0,2 mol.

Y + HCl: O2 3

2NO nn 23NO O

n = 2n = 0,4 (mol).

3

X NOm = m + m = 6,4 + 0,4.62 = 31,2 (gam) C.

Cu 2: Nhn xt:2 4H SO

n =2H

n = 0,15 2 4dd H SO

m = 0,15.98.10010

= 147 (gam).

Bo ton khi lng:2 4 2X H SO dd Y H

m + m = m + m

dd Y5,2 + 147 = m + 0,3 dd Ym = 151,9 (gam) D.

+ : Khngtr khi lng kh H2 dd Ym = 152,2 gam Chn A!

Cu 3: Bo ton khi lng: 7,05 X Cl OH= m + m + m 7,05 = 2,65 + 0,1.35,5 +

OHm OHm = 0,85 gam OHn = 0,05 mol.

Nhn thy: X Cl OHn = n + n = 0,15 (mol) X2,65

M = = 17,670,15

Hai kim loi l Li v Na A.

Cu 4: Y cha hn hp mui c cha mui axit KOH phn ng ht.

OH (KOH)n = 0,15;

3 43 4 H POH (H PO )

n = 3n = 0,6a+ 2H O OH

n n 0,15 (mol).= =

Xt s phn ng: H3PO4 + KOH Mui + H2O.

Bo ton khi lng: 0,2a.98 + 0,15.56 = 15,5 + 0,15.18 a = 0,5. B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


8/165


8

2 PHNG PHP BO TON NGUYN T


Trong cc phn ng ho hc, cc ngun tlun c bo ton.2. H qu

Tng s mol nguyn t ca mt nguyn t c trong cc cht trc phnng v sau phn ng lun bng nhau.C : Cn xc nh ng v y cc cht c cha nguyn t ang xt trc v sau phn ng.

B. V D CU HI

V d 1: Ho tan hon ton 0,1 mol hn hp gm Al v Al4C3vo dung dchNaOH (d), thu c V lt hn hp kh X (ktc) v dung dch Y. Sc khCO2(d) vo Y, thu c 19,5 gam kt ta. Gi tr ca V l

A. 5,60. B. 5,04. C. 6,16. D. 4,48.

Xt s phn ng: 2 2+ CO + H O+ KOH4 3 2 3Al, Al C NaAlO Al(OH)

Bo ton nguyn t Al:4 3 3Al Al C Al(OH)

n + 4n = n = 0,25 (mol).

Mt khc:4 3Al Al C

n + n = 0,1 4 3Al Al C

n = n = 0,05 (mol).

2 4 4 3X H CH Al Al C

3n = n + n n + 3n

2= = 0,225 V = 5,04 lt B.

Cu 1: t 2,8 gam Fe trong khng kh, thu c hn hp cht rn X. Cho Xtc dng vi dung dch HNO3long (d), thu c kh NO (sn phm kh

duy nht) v dung dch cha m gam mui. Gi tr ca m lA.18,15. B.24,20. C.10,89. D.12,10.

V d 2: Mt loi phn supephotphat n c cha 46,8% Ca(H2PO4)2, cn li gmcc cht khng cha photpho. dinh dng ca loi phn ln ny lA.32,2%. B.42,1%. C.28,4%. D. 35,8%.

dinh dng ca phn ln = % khi lng P2O5.Bo ton nguyn t P theo s :

Ca(H2PO4)2 P2O5

Khi lng mol: 234 gam 142 gam% khi lng: 46,8% x%



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


9/165


9

x =46,8.142

28,4234

= C.

Cu 2:Mt loi phn kali c thnh phn chnh l KCl (cn li l cc tp chtkhng cha kali) c sn xut t qung xinvinit c dinh dng 47%.Phn trm khi lng ca KCl trong loi phn kali l

A.70,2%. B.65,5%. C.74,5%. D.76,6%.

V d 3: Hp th hon ton 2,24 lt kh CO2 (ktc) vo 1,5 lt dung dchBa(OH)2a mol/, thu c 9,85 gam kt ta. Gi tr ca a lA.0,03. B.0,04. C.0,05. D.0,10.

32CO BaCO

2,24 9,85n = = 0,1 (mol); n = = 0,05 (mol);

22,4 197

Bo ton nguyn t C:2

22 3 3CO BaCO Ba(HCO )

n = n n+ 23Ba(HCO )

n = 0,025 (mol).

Bo ton nguyn t Ba:2

Ba(OH)n =23 3

BaCO Ba(HCO )n n+ = 0,075 (mol).

a = 0,05M C.

Cu 3: Hp th hon ton V lt kh CO2(ktc) vo 1,25 lt dung dch Ba(OH)20,1M (va ), thu c 9,85 gam kt ta. Gi tr ca V lA.5,60. B.3,36. C.4,48. D.2,24.

V d 4:Cho 7,8 gam Mg tc dng va vi V lt dung dch HNO 32M, thuc dung dch X v 1,12 lt (ktc) hn hp kh Y gm N2O v N2. T khica Y so vi H2l 18. Gi tr ca V l

A.0,2. B.0,4. C.0,5. D.0,3.

Mg

7,8n = = 0,325 (mol);

24 Y1,12

n = = 0,05 (mol).22,4

YM = 18.2 = 36 2

1

1

2N O

N

n 28 36= =

n 44 36

2 2N O Nn = n = 0,025 (mol).

Bo ton electron: e Mgn = 2n =2N O

8n +2N

10n +4 3NH NO

8n 4 3NH NO

n = 0,025.

Bo ton nguyn t N:2 4 32

23 23

HNO N N O NH NOMg(NO )n = 2n + 2n 2n n = 0,8.+ +

V = 0,4 lt B.

+ : Khng tnh mui NH4NO3.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


10/165


10

Cu 4: Cho 7,35 gam hn hp X gm Al v Mg tc dng va vi dung dchcha a mol HNO3, thu c 49,65 gam mui v 1,12 lt hn hp kh Y(ktc) gm N2v N2O. T khi ca Y so vi H2l 18. Gi tr ca a lA.0,80. B.0,65. C.0,70. D.1,00.

V d 5:Nung nng 13,4 gam hn hp X g m Al v Fe2O3(khng c khng

kh) n phn ng hon ton, thu c cht rn Y. Chia Y thnh hai phnbng nhau. Phn mt tc dng va vi 50 ml dung dch NaOH 2M. ho tan hon ton phn hai cn va dung dch cha a mol HCl. Gitr ca a lA. 0,5. B. 0,4. C. 0,6. D. 0,8.

Nhn xt: Y + NaOH, tt c Al v Al2O3u chuyn thnh NaAlO2.Bo ton nguyn t Al:

2Al NaAlO NaOHn = n = n = 0,05.2 = 0,1 (mol).

Al (X)n = 2.0,1 = 0,2 (mol) 2 3Fe O13,4 0,2.27

n = = 0,05 (mol)160

Fe2O3 + 2Al0t Al2O3 + 2Fe

Mol: 0,05 0,1 0,05 0,1 Phn hai gm: Al (0,05 mol); Fe (0,05 mol) v Al2O3(0,025 mol).Khi cho Y + HCl:

2 3HCl Al Fe Al On = 3n + 2n + 6n = 0,4 (mol).

B.

Cu 5: Nung nng 26 gam hn hp X gm Al v Cr2O3(khng c khng kh)n phn hon ton, thu c cht rn Y. Chia Y thnh hai phn bngnhau. Phn mt tc dng va vi 100 ml dung dch KOH 2M (long).

ho tan hon ton phn hai cn va dung dch cha a mol HCl.Gi tr ca a lA. 1,0. B. 1,2. C. 1,5. D. 0,8.

C. HNG DN GII

Cu 1:Xt s phn ng: Fe 2+ OX (Fe, FexOy) 3+ HNO Fe(NO3)3

Bo ton nguyn t Fe:3 3Fe(NO ) Fe

n = n = 0,05 (mol).

3 3Fe(NO )

m = 0,05.242 = 12,1 (gam).

D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


11/165


11

Cu 2: dinh dng ca phn kali = % khi lng K2O.Bo ton nguyn t K theo s :

2KCl K2O

Khi lng mol: 149 gam 94 gam% khi lng: x% 47%

x = 47.149 74,594

= C.

Cu 3:2 3Ba(OH) BaCO

9,85n = 1,25.0,1 = 0,125 (mol); n = = 0,05 (mol).

197

Bo ton nguyn t Ba:2Ba(OH)

n =23 3BaCO Ba(HCO )

n n+ = 0,125

3 2Ba(HCO )

n = 0,125 0,05 = 0,075 (mol).

Bo ton nguyn t C:2

22 3 3CO BaCO Ba(HCO )

n = n n = 0,05 + 2.0,075 = 0,2 (mol).+

V = 0,2.22,4 = 4,48 (lt). C.

Cu 4: Y gm: N2(0,025 mol) v N2O (0,025 mol).Khi cho X + HNO3:Bo ton electron: en = 2N10n + 2N O8n + 4 3NH NO8n = 0,45 + 8x.

Khi lng mui = 49,65 =4 33

X NH NONOm + m + m = 7,35 + 62(0,45 + 8x) + 80x

x = 0,025.Bo ton nguyn t N:

2 4 3

23 23

HNO N N O NH NONOn = n + 2n 2n n = 0,8 (mol). + +

A.

+ : Khng tnh n sn phm NH4NO3!

Cu 5: Nhn xt: Y + KOH, tt c Al v Al2O3u chuyn thnh KAlO2.Bo ton nguyn t Al:

2Al KAlO KOHn = n = n = 0,1.2 = 0,2 (mol).

Al (X)n = 2.0,2 = 0,4 (mol) 2 3Cr O26 0,4.27

n = = 0,1 (mol)152

Cr2O3 + 2Al0t Al2O3 + 2Cr

Mol: 0,1 0,2 0,1 0,2 Phn hai gm: Al (0,1 mol); Cr (0,1 mol) v Al2O3(0,05 mol).Khi cho Y + HCl:

2 3

HCl Al Fe Al On = 3n + 2n + 6n = 0,8 (mol).

D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


12/165


12

3 PHNG PHP QUY I


H gm (X) s tng ng vi h gm (Y) tothnh nhng hp cht c v khi lng v s electron trao i.

2. H quBo ton khi lng v bo ton electron: X Ym = m v e (X) e (Y)n = n .

3. Cc trng hp thng gp

Ban u (X) V d Quy i (Y) S mol e nhng

Kim loi + oxitFe, FeO, Fe2O3, Fe3O4 Fe v O Fe O3n 2n

Na, Na2O, Ba, BaO Na, Ba v O Na Ba On + 2n 2n

Kim loi +mui sunfua

Fe, FeS, FeS2, S Fe v S Fe S3n + n.n (*)

FeS, FeS2, CuS, Cu2S Fe, Cu v S Fe Cu S3n + 2n + n.n (*)

(*) n = 4 nu S +4S + 4e; n = 6 nu S

+6S + 6e.

B. V D CU HI

V d 1: Nung m gam bt st trong oxi, thu c 6,8 gam hn hp cht rn X.Ho tan ht X trong dung dch HNO3d, thot ra 1,12 lt NO (sn phmkh duy nht, ktc). Gi tr ca m l:A. 5,6. B. 2,8. C. 4,2. D. 7,0.

X gm Fe v cc oxit st X cha cc nguyn t Fe v O.Quy i X thnh: Fe (a mol) v O (b mol) 56a + 16b = 6,8.

Khi cho X + HNO3: Qu trnh oxi ho: Fe +3

Fe + 3e

Qu trnh kh: O + 2e 2

O

;+5

N + 3e NOBo ton electron: Fe O NO3n = 2n + 3n 3a = 2b + 3.0,05 a = 0,1; b = 0,075.

m = 0,1.56 = 5,6 (gam) A.

Cu 1: Nung 4,2 gam bt Fe trong oxi, thu c m gam hn hp X gm Fe,FeO, Fe3O4 v Fe2O3. Ho tan ht X trong dung dch HNO3 d, thot ra

1,12 lt NO (sn phm kh duy nht, ktc). Gi tr ca m l:A. 4,6. B. 5,0. C. 4,8. D. 5,4.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


13/165


13

V d 2: Cho m gam FexOyphn ng va vi 0,3 mol H2SO4, thu c dungdch ch cha mt mui duy nht v 1,68 lt kh SO2(ktc, sn phm khduy nht ca S+6). Gi tr ca m lA. 12,0. B. 10,8. C. 11,6. D. 8,0.

Quy i FexOythnh Fe v O.

Bo ton S: 2 4 2 4 3 2H SO Fe (SO ) SOn = 3n + n 2 4 3Fe (SO )n = 0,075 Fen = 0,15 mol.Bo ton electron:

2Fe O SO3n = 2n + 2n On = 0,15 mol .

Fe On : n = 0,15 : 0,15 = 1 : 1 FeO m = 0,15.72 = 10,8 B.

Cu 2: Cho m gam hp cht FeSxphn ng va vi 0,35 mol H2SO4, thuc dung dch ch cha mt mui duy nht v 8,4 lt kh SO2 (ktc, snphm kh duy nht ca S+6). Gi tr ca m lA. 4,4. B. 6,0. C. 2,2. D. 8,0.

V d 3: Cho 8,8 gam hn hp X gm Cu v FexOy tc dng vi axit HNO3(long, d), thu c 1,12 lt kh NO(sn phm kh duy nht, ktc) vdung dch Y. Cho dung dch NH3d vo Y, thu c 10,7 gam kt ta.Cng thc v khi lng ca oxit st trong X ln lt lA. FeO v 7,2 gam. B. Fe2O3v 4,8 gam.C. Fe3O4v 5,8 gam. D. Fe2O3v 5,6 gam.

Y + NH3 (d): Fe3+ Fe(OH)3 ;

Cu2+ Cu(OH)2 [Cu(NH3)4](OH)2().

3Fe (X) Fe(OH)

n = n = 0,1 mol .

Quy i X: Cu (a mol); Fe (0,1 mol) v O (b mol) 64a + 5,6 + 16b = 8,8.Khi cho X + HNO3:Bo ton electron: Cu Fe O NO2n + 3n = 2n + 3n 2a + 0,3 = 2b + 0,15.

a = 0,025; b = 0,1 Fe On : n = 0,1 : 0,1 = 1 : 1 Oxit st l FeO.

FeOm = 0,1.72 = 7,2 (gam) A.

Cu 3: Cho 13,2 gam hn hp X gm Cu v FexOytc dng va vi 0,3 molH2SO4 (c, nng), thu c 1,12 lt kh SO2 (sn phm kh duy nht ktc). Cng thc v khi lng ca oxit st trong X ln lt l

A. FeO v 7,2 gam. B. Fe2O3v 7,2 gam.C. Fe3O4v 11,6 gam. D. FeO v 10,8 gam.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


14/165


14

V d 4: Cho 6,8 gam hn hp X g m Cu, CuO, Fe v Fe 2O3 tc dng ht viHNO3 (d), thu c 1,12 lt kh NO(sn phm kh duy nht, ktc) vdung dch Y. Cho dung dch NH3d vo Y, thu c 5,35 gam kt ta. Smol HNO3 phn ng lA. 0,2 mol. B. 0,3 mol. C. 0,4 mol. D. 0,5 mol.

Quy i X: Zn (a mol); Fe (b mol) v O (c mol) 65a + 56b + 16c = 6,8.Khi cho X + HNO3:Bo ton electron: Zn Fe O NO2n + 3n = 2n + 3n 2a + 3b = 2c + 0,15.

Khi cho Y + NH3(d):Fe3+ Fe(OH)3 ; Cu2+ Cu(OH)2 [Cu(NH3)4](OH)2().

3Fe (X) Fe(OH)n = n = 0,05 mol a = 0,05 mol. T b = c = 0,05 mol.

Bo ton nguyn t N:2 33 3 3

HNO NOCu(NO ) Fe(NO )n = 2n + 3n n = 0,3 (mol).+

B.

Cu 4a: Ho tan hon ton 7,6 gam hn hp X gm Cu2S, CuS v FeS2 trongdung dch HNO3(c nng, d) thu c a mol kh ch c NO2(sn phmkh duy nht) v dung dch Y. Cho dung dch BaCl2(d) vo Y, thu c11,65 gam kt ta; cn khi cho dung dch NH3(d) vo Y, thu c 5,35gam kt ta. Gi tr ca a lA.0,55. B.0,60. C.0,75. D.0,70.

Cu 4b: Nung 6,4 gam hn hp X gm Al, Zn, Mg trong kh oxi, thu c 7,2 gamhn hp Y. Ho tan Y vo dung dch HNO3(long, d), thu c 2,24 lt khNO (sn phm kh duy nht, ktc). S mol HNO3 phn ng l

A. 0,2 mol. B. 0,3 mol. C. 0,5 mol. D. 0,4 mol.

V d 5: Hn hp X g m Cu, Mg, CuO v MgO. Ho tan hon ton 6,8 gam Xbng dung dch HNO3, thu c 1,12 lt NO (sn phm kh duy nht, ktc) v dung dch Y cha 14,1 gam Cu(NO3)2v m gam Mg(NO3)2. Gi trca m l:A.14,8. B.7,4. C.3,7. D.11,1.

3 2Cu(NO )

14,1n = = 0,075 (mol)

188 0Cu (X)n = ,075 (mol).

Quy i X thnh: Cu (0,06 mol); Mg (a mol) v O (b mol). 64.0,075 + 24a + 16b = 6,8 3a + 2b = 0,25.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


15/165


15

Bo ton electron: Cu Mg O NO2n + 2n = 2n + 3n 0,15 + 2a = 2b + 0,15.

a = b = 0,05 3 2Mg(NO )

m = m = 0,04.148 = 7,4(gam) B.

Cu 5: Hn hp X gm Na, Ba, Na2O v BaO. Ho tan hon ton 8,4 gam Xvo nc, thu c 1,12 lt kh H2 (ktc) v dung dch Y c cha 2 gam

NaOH. Hp th hon ton 2,24 lt kh CO2(ktc) vo Y, thu c m gamkt ta. Gi tr ca m l

A. 21,92. B. 9,85. C. 19,70. D. 15,76.

C. HNG DN GII

Cu 1: Fe NO4,2 1,12

n = = 0,075 (mol); n = = 0,05 (mol).56 22,4

Quy i X thnh: Fe (0,075 mol) v O (a mol) m = 4,2 + 16a.

Khi cho X + HNO3: Qu trnh oxi ho: Fe +3

Fe + 3e

Qu trnh kh: O + 2e 2

O

;+5

N + 3e NOBo ton electron: Fe O NO3n = 2n + 3n 0,225 = 2a + 0,15 a = 0,0375.

m = 4,2 + 0,0375.16 = 4,8 (gam). C.

Cu 2: Quy i FeSxthnh : Fe (a mol) v S (b mol).Bo ton nguyn t S:

x 2 4 2 4 3 2S (FeS ) H SO Fe (SO ) SOn + n = 3n + n

b + 0,35 = 3.0,5a + 0,375 1,5a b = 0,025.Bo ton electron:

2Fe S SO S

3n + 4n = 2(n n ) 3a + 6b = 0,75

a = 0,05; b = 0,1 Fe Sn : n = 0,05 : 0,1 = 1 : 2 FeS2.

m = 0,05.120 = 6,0 (gam) B.Cu 3:

Quy i X: Cu (a mol); Fe (b mol); O (c mol) 64a + 56b + 16c = 13,2.Khi cho X + H2SO4:Bo ton electron:

2Cu Fe O SO2n + 3n = 2n + 2n 2a + 3b = 2c + 0,1.

Bo ton nguyn t S:2 4 2 4 24 3

H SO Fe (SO ) SOCuSOn = n + 3n + n 0,3 = a + 1,5b + 0,05.

a = 0,025; b = 0,15; c = 0,20 Fe On : n = 0,15 : 0,20 = 3 : 4 Fe3O4.

3 4Fe Om = 0,05.232 = 11,6 (gam). C.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


16/165


16

Cu 4a:Quy i X thnh hn hp cc n cht: Cu, Fe v S.Y gm: Fe3+, Cu2+, 24SO

, H+v3

NO .

Y + BaCl2: Ba2+ + 24SO BaSO4

4S (X) BaSOn = n = 0,05 mol.

Y + NH3: Fe3+ Fe(OH)3 ;

Cu2+

Cu(OH)2 [Cu(NH3)4](OH)2

3Fe (X) Fe(OH)n = n = 0,05 (mol)

64Cu (X)

7,6 0,05.32 0,05.56n = = 0,05 (mol).

e (X) Cu Fe Sn = 2n + 3n + 6n 0,55= 2

NO e (X)n = n 0,55 (mol).=

A.

Cu 4b:2O Y X

m = m m = 7,2 6,4 = 0,8 (gam) O0,8

n = = 0,05 (mol).16

Quy i Y thnh cc n cht: X (Al, Zn, Mg) v O (0,05 mol).Khi cho Y + HNO3:

Bo ton electron: e (X) O NOn = 2n + 3n = 0,1 + 0,3 = 0,4 (mol).Bo ton nguyn t N:

3 3HNO NO e (X) NONO

n = n + n n + n = = 0,5 (mol).

C.

Cu 5: NaOH2

n = = 0,05 (mol)40

0Na (X)n = ,05 (mol).

Quy i X thnh: Na (0,05 mol); Ba (a mol) v O (b mol). 23.0,05 + 137a + 16b = 8,4 137a + 16b = 7,25.Khi cho X + H2O:

Bo ton electron: 2Na Ba O H1n + 2n = 2n + 2n 0,05 + 2a = 2b + 0,1. a = 0,05; b = 0,025

2NaOH Ba(OH)OH (Y)n = n + 2n = 0,15 (mol).

Khi cho Y + CO2:

2

OH

CO

n 0,15= = 1,5 > 1

n 0,1

to ra hai mui: cacbonat v hirocacbonat.

Tnh nhanh: 223

COCO OHn = n n = 0,15 0,1 = 0,05 (mol).

Do: 2 2+33

BaCOCO Ban = n = 0,05 n = 0,05 m = 0,05.197 = 9,85 (gam).

B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


17/165


17

HcHcHcHc

TpTpTpTp

HmHmHmHm

NayNayNayNay

TngTngTngTng

LaiLaiLaiLai

NgyNgyNgyNgy

MaiMaiMaiMai

4 PHNG PHP TNG GIM KH I LNG

A. L THUYT TRNG TM1. Ni dungKhi nguyn t (nhm nguyn t) ca cht phn ng c hoc bi nguyn t (nhm nguyn t) mi s to ra s tng gim khilng mol (M) ca cht.

2. H qu

S mol ca nguyn t (nhm nguyn t) ban u =mM

.

3. Cc trng hp thng gp

Cht u S thay th (cng hp) M (gam)

Mui cacbonat2

3CO + HCl 2Cl

2

3CO 2 4+ H SO 2

4SO

+ 11

+ 36

Oxit kim loi2O + HCl 2Cl

2O 2 4+ H SO SO 24

+ 55

+ 80

Kh oxit kim loiCO + O CO2;H2 + O H2O

+ 16+ 16

Trao i ion NaCl AgCl + 85

iu ch kim loiNaCl NaMO M

35,5 16



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


18/165


18

B. V D CU HI

V d 1: t chy hon ton 7,1 gam hn hp X gm Mg, Al v Cu trong khO2 (d) thu c 11,1 gam hn hp oxit Y. Ho tan Y trong dung dchH2SO4 (long, d) ri c cn dung dch sau phn ng thu c m gammui khan. Gi tr ca m l

A.23,8. B. 31,1. C.21,1. D.32,0.

Bo ton khi lng:2X O Y

m + m = m 2O

m = 11,1 7,1 = 4 (gam).

2O

n =4

= 0,125 (mol)32

2O O (Y)n = n 0,25 (mol). =

Khi cho Y + H2SO4: 1 mol O2 1 mol 24SO M = 80 gam.

Ta c: 2O (Y)m

n =M

m 11,10,25 =

80 m = 31,1 (gam) B.

Cu 1: t chy hon ton 5,6 gam hn hp X gm Li, Cu v Cr trong kh O2

(d) thu c 8,8 gam hn hp oxit Y. Ho tan Y trong dung dch HCl(d) ri c cn dung dch sau phn ng thu c m gam mui khan. Gitr ca m lA.14,3. B. 25,2. C.20,1. D.19,8.

V d 2: X v Y l hai kim loi thuc hai chu k k tip trong nhm IIA, ZX< ZY.Ho tan hon ton 9,2 gam hn hp M gm XCO3v YCO3vo dung dchHCl, thu c 10,3 gam mui clorua.Phn trm s mol XCO3trong M lA.80%. B.40%. C.50%. D.60%.

Khi cho M + HCl: 1mol2

3CO

2 mol Cl

M = 11 gam. 2

3CO

m 10,3 9,2n = = = 0,1 (mol)

M 11 Mn = 0,1 (mol).

M9,2

M = = 920,1

XCO3< 92 < YCO3 X < 32 < Y Mg v Ca.

3

3

MgCO

CaCO

n 92 100 1= =

n 92 84 1

3MgCO%n = 50% C.

Cu 2: X v Y l hai kim loi thuc nhm IIA. Ho tan ht 7,6 gam hn hp Mgm hai oxit ca X v Y bng dung dch HCl, thu c 15,85 gam mui

clorua.in phn nng chy ton b lng mui clorua (hiu sut 100%)thu c m gam hp kim. Gi tr ca m lA.8,75. B.5,20. C.5,60. D.6,25.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


19/165


19

V d 3: Cho V lt hn hp kh (ktc) g m CO v H2phn ng vi mt lngd hn hp rn gm CuO v Fe2O3 nung nng. Sau khi cc phn ng honton, khi lng hn hp rn gim 3,2 gam. Gi tr ca V lA. 4,48. B. 1,12. C. 2,24. D. 5,60.

1 mol CO + O 1 mol CO2 M = 16 gam.

1 mol H2 + O 1 mol H2O M = 16 gam.1 mol hn hp (CO + H2) + O 1 mol (CO2+ H2O) M = 16 gam.

Om 3,2

n = = = 0,2 (mol)M 16

2CO H

n + n = 0,2 (mol) V = 4,48 (lt).

A.

Cu 3: Hn hp Y gm FeO, Fe2O3v CuO. Kh hon ton 6,8 gam Y bng khCO (d), thu c 5,2 gam kim loi. Mt khc, ho tan hon ton 6,8 gamY bng dung dch HCl, thu c m gam mui clorua. Gi tr ca m l:A.12,3. B. 11,6. C.13,8. D.12,8.

V d 4: Hn hp X gm CuO, FeO v Fe2O3. Ho tan hon ton 9,6 gam Xbng dung dch H2SO4, thu c 21,6 gam mui. Mt khc, nu kh ht 9,6gam X bng CO (d), dn hn hp kh thu c vo dung dch Ca(OH)2(d), to thnh mgam kt ta. Gi tr ca m lA.10. B.15. C.20. D.25.

Khi cho X tc dng vi axit H2SO4:1 mol O2 1 mol SO 2

4

M = 80 gam.

2O (X) m 21,6 9,6n = = = 0,15 (mol)M 80 Cln = 0,3 (mol).

Khi cho Xtc dng viCO:2

2 3CO CO CaCOO (X)n = n = n = n = 0,15 (mol) m = 0,15.100 = 15 (gam).

B.

Cu 4: Hn hp X gm ZnO v FexOy. Cho 11,25 gam Xtc dng ht vi H2SO4(long), thu c dung dch Y cha 23,25 gam mui. Cho dung dch NaOHd vo Y, lc kt ta v nung trong khng kh n khi lng khng i thuc 8 gam cht rn. Cng thc v khi lng ca oxit st trong X l

A. FeO v 7,2 gam. B.FeO v 9,0 gam.C. Fe3O4v 5,8 gam. D. Fe2O3v 8,0 gam.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


20/165


20

C. HNG DN GII

Cu 1: Bo ton khi lng:2X O Y

m + m = m 2O

m = 8,8 5,6 = 3,2 (gam).

2O

n =3,2

= 0,1 (mol)32

2O O (Y)n = n = 0,2 (mol).

Khi cho Y tc dng vi axit HCl:

1 mol O2 2 mol Cl M = 55 (gam).

Ta c:2O (Y)

mn =

M

m 8,80,2 =

55 m = 19,8 (gam) D.

Cu 2: Khi cho M + HCl: 1 mol O2 2 mol Cl M = 55 gam.

2Om 15,85 7,6

n = = = 0,15 (mol)M 55

Cl

n = 0,3 (mol).

Bo ton khi lng: 15,85 = m + 0,3.35,5 m = 5,2 (gam) B.

Cu 3: Khi cho Y + CO: 1 mol CO+ O

1 mol CO2

M = 16 gam. O

m 6,8 5,2n = = = 0,1 (mol)

M 16 2O (Y)n = 0,1 (mol).

Khi cho Y + HCl: 0,1 mol O2 0,2 mol ClBo ton khi lng:

Clm = 5,2 + m = 5,2 + 0,2.35,5 = 12,3 (gam)

A.

Cu 4: Khi cho X tc dng vi axit H2SO4long:1 mol O2 1 mol 24SO

M = 80 gam.

2O (X)

m 23,25 11,25

n = = = 0,15 (mol)M 80 O (X)n = 0,15 (mol).Khi cho Y + NaOH d v em nung kt ta:

Fe2+, Fe3+ + OH

Fe(OH)2, Fe(OH)3 20+ O , t Fe2O3

Zn2+ + 2OH

ZnOH)2 + 2OH

ZnO 22

(tan)

2 3Fe (X) Fe O

n = 2n = 0,1 (mol) Zn (X)11,25 0,1.56 0,15.16

n = = 0,05 (mol)65

x yO (Fe O ) O (X) O (ZnO)

n = n n = 0,15 0,05 = 0,1 (mol).

Trong FexOy th Fe On : n = 0,1 : 0,1 = 1 : 1 FeO FeOm = 7,2 (gam).

A.+ :Tnh ngay Fe On : n = 0,1 : 0,15 = 2 : 3 Fe2O3 Chn D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


21/165


21

5 PHNG PHP BO TON ELECTRON


S electron trao i trong cc phn ng oxi ho kh lun c bo ton.2. H qu

Tng s mol electron cc cht kh nhng = Tng s mol electron cc chtoxi ho nhn.

3. Cch p dngXc nh ng v y cc cht kh v cht oxi ho cng nh s bini trng thi oxi ho ca chng.Vit cc qu trnh oxi ho (nhng electron) v qu trnh kh (nhnelectron) xc nh s mol electron trao i.

B. V D CU HI

V d 1: Ho tan hon ton 4,8 gam Mg trong dung dch HNO3 long, d thuc 0,56 lt kh N2O (ktc) v dung dch X. C cn X thu c m gammui khan. Gi tr ca m lA.29,6. B.45,6. C.33,6. D.31,6.

2Mg N O

4,8 0,56n = = 0,2 (mol); n = = 0,025 (mol).

24 22,4

Qu trnh oxi ho: Zn +2

Zn + 2e e Znn = n = 0,4 (mol).

Qu trnh kh:

+5

2N + 8e N2O 2e N On = 8n = 0,2 (mol). Nh vy, s mol electron trao i cha bng nhau xy ra c qu trnh:

2N+5 + 8e NH4NO3Mol: 0,2 0,025 m =

3 2 4 3Mg(NO ) NH NOm + m = 0,2.148 + 0,025.80 = 31,6 (gam) D.

+ : Khng vit qu trnh to thnh NH4NO3 Chn A.

Cu 1: Cho 6,5 gam Zn tc dng vi dung dch HNO3(d). Sau khi phn ngxy ra hon ton thu c 0,28 lt kh X (ktc) v dung dch Y. C cn Y

thu c 19,9 gam mui khan. Kh X lA.NO. B.NO2. C.N2O. D.N2.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


22/165


22

V d 2: Cho hn hp kh X gm Cl2v O2tc dng va vi hn hp btgm 10,8 gam Al v 2,4 gam Mg, thu c 40,9 gam hn hp cht rn Y.Phn trm th tch ca kh Cl2trong X lA.80%. B.40%. C.50%. D.60%.

Al Mg

10,8 2,4n = = 0,4 (mol); n = = 0,1 (mol).

27 24

Qu trnh oxi ho:

Al +3

Al + 3e; Mg +2

Zn + 2eMol: 0,4 1,2 Mol: 0,1 0,2Qu trnh kh:

O2 + 4e 2O2 Cl2 + 2e 2ClMol: x 4x Mol: y 2yBo ton electron: 4x + 2y= 1,2 + 0,2 = 1,4 (mol).Bo ton khi lng: 32x + 71y + 10,8 + 2,4 = 40,9 32x + 71y = 27,7.

x = 0,2; y = 0,3 2Cl 0,3%V = .100% = 60%0,2+0,3 D.

Cu 2: Hn hp kh X gm Cl2v O2c t khi so vi hiro bng 25,75. Cho 0,2mol X tc dng va vi m gam kim loi R, thu c 22,3 gam cht rn.Kim loi R lA.Mg. B.Ca. C.Fe. D.Cu.

V d 3: Hn hp X gm Mg, Al v Zn. Ho tan ht 5,2 gam X trong dung dchHCl, thu c 3,36 lt H2(ktc). Mt khc, t chy hon ton 5,2 gam X

trong oxi, thu c m gam oxit. Gi tr ca m lA.7,6. B.6,4. C.8,4. D.6,0.

Khi cho X + HCl, ion H+ nhn electron tr thnh kh H2:2H+ + 2e H2

Mol: 0,3 0,15 2e H

n = 2n = 0,3 (mol).

Khi cho X + O2, O2 nhn electron tr thnh O2:

O2 + 4e 2

2O

Mol: 0,075 0,3

Bo ton khi lng: 2X Om = m + m = 5,2 + 0,075.32 = 7,6 (gam). A.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


23/165


23

Cu 3a: Hn hp Y gm Al, Zn v Cu. t chy hon ton 12,4 gam X gmtrong kh oxi, thu c 17,2 gam oxit. Mt khc, ho tan ht 12,4 gam Xtrong dung dch HNO3(long, d), thu c 1,12 lt kh N2O (ktc). S molHNO3 tham gia phn ng lA.0,85 mol. B.0,60 mol. C.0,70 mol. D.0,75 mol.

Cu 3b: Ho tan hon ton 3,9 gam hn hp X gm Al v Mg Zn bng lngva 500 ml dung dch HNO31M. Sau khi cc phn ng kt thc, thuc 0,56 lt kh N2O (ktc) duy nht v dung dch Y cha m gam mui.Gi tr ca m lA.31,8. B.30,7. C.34,9. D.28,7.

C. HNG DN GII

Cu 1: Zn X6,5 0,28

n = = 0,1 (mol); n = = 0,0125 (mol).65 22,4

3 2Zn(NO )m = 0,1.189 = 18,9 (gam) 4 3NH NOm = 19,9 18,9 = 1 (gam).

4 3NH NOn = 0,0125 mol.

Qu trnh oxi ho: Zn +2

Zn + 2e e Znn = n = 0,2 (mol).

Qu trnh kh:+5

N + ne X e Xn = n.n = 0,0125n (mol).+5

2N + 8e NH4NO3Mol: 0,1 0,0125Bo ton electron: Zn2n = 4 3X NH NOn.n + 8n 0,2= 0,0125n + 0,1 n = 8.

Kh X l N2O ( +52N + 8e N2O). C.

Cu 2: XM = 25,75.2 = 51,5 22

Cl

O

n 32 51,5 1= =

n 71 51,5 1

2 2Cl On = n = 0,1 (mol).

Bo ton khi lng:2 2Cl O

m + m m = 22,3+ m = 12 (gam).

Qu trnh oxi ho:

R +n

R + ne

Mol: 12R

12nR



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


24/165


24

Qu trnh kh:O2 + 4e 2O2 Cl2 + 2e 2Cl

Mol: 0,1 0,4 Mol: 0,1 0,2

Bo ton electron:12n

= 0,4 + 0,2 = 0,6R

R = 20n.

n = 2 v R = 40 (Ca).

B.

Cu 3a: Khi cho X + O2:

2Om = 17,2 12,4 = 4,8 (gam)

2O

4,8n = = 0,15 (mol).

32

Qu trnh kh: O2 + 4e 2

2O

2e (X) O

n = 4n = 0,6 (mol).

Khi cho X + HNO3:

Qu trnh kh:+5

2N + 8e N2O 2e N O

n = 8n = 0,4 (mol).

Nh vy, s mol electron trao i cha bng nhau xy ra c qu trnh:2N+5 + 8e NH4NO3Mol: 0,2 0,025Bo ton nguyn t N:

2 23 4 3 4 33HNO N O NH NO e (X) N O NH NONO

n = n + 2n + 2n n + 2n + 2n = 0,75 (mol). =

D.

Cu 3b:2 3N O HNO

0,56n = = 0,025 (mol); n = 0,5.1 = 0,5 (mol).

22,4

Khi cho X + HNO3:

Qu trnh kh:+5

2N + 8e N2O 2e N O

n = 8n = 0,2 (mol).

v c th xy ra qu trnh:+5

2N + 8e NH4NO3 (a mol).

4 32e (X) N O NH NOn = 8n + 8n = 0,2 + 8a (mol).

Bo ton nguyn t N:

23 4 3HNO e (X) N O NH NOn = n + 2n + 2n = 0,25 + 10a = 0,5 a = 0,025 (mol).

Bo ton khi lng:m =

4 33X NH NONO

m + m + m = 3,9 + 0,4.62 + 0,025.80 = 30,7 (gam).

B.+ : Khng xt s to mui NH4NO3 Chn D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


25/165


25

6 PHNG PHP TRUNG HO IN


Dung dch cc cht in li lun lun v in.2. Biu thc

Tng s mol in tch dng = Tng s mol in tch m:t (+) t ()n = n

Trong : S mol in tch = s mol ion in tch ion.

B. V D CU HI

V d 1: Dung dch X cha cc ion: Fe2+(0,05 mol ), Al3+ (0,1 mol), Cl (x mol),

v SO 24

(y mol ). C cn dung dch X thu c 22,2 gam cht rn khan Y.

Gi tr ca x v y ln lt l

A. 0,2 v 0,1. B. 0,1 v 0,2. C. 0,2 v 0,3. D. 0,1 v 0,3. + p dng nh lut trung ho in:Tng s mol in tch dng = Tng s mol in tch m 2+ 3+ 2

4Fe Al Cl SO2n + 3n = n + 2n 0,1 + 0,3 = x + 2y x + 2y = 0,4.

+ Bo ton khi lng: 2+ 3+ 24

Y Fe Al Cl SOm = m + m + m + m

56.0,05 + 27.0,1 + 35,5x + 96y = 22,2. T x = 0,2; y = 0,1. A.+ : Khng p dng biu thc nh lut trung ho in tch.

Cu 1: Dung dch X gm: Na+(0,05 mol); Ca2+(0,10 mol); HCO3

(0,10 mol) v

ion Y (0,15 mol). un nng c cn dung dch X thu c m gam cht rnkhan. Ion Y v gi tr ca m lA. OH v 12,00. B.

3NO v 17,45. C. 23CO

v 20,25. D.3

NO v 20,55.

V d 2: Cho dung dch X gm: Na+(0,1 mol), Fe2+(0,05 mol), NO3

(0,1 mol) v

Cl. Cho lng d dung dch AgNO3vo X, to thnh m gam kt ta. Gitr ca m lA.14,35. B.28,70. C.3,95. D.19,75.

p dng nh lut trung ho in vi X: + 2+

3Na Fe NO Cln + 2n = n + n



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


26/165


26

0,1 + 0,1 = 0,1 +Cl

n Cl = 0,1 mol.

Cc phn ng ion rt gn:Ag+ + Cl AgCl

Mol: 0,1 0,1Ag+ + Fe2+ Ag + Fe3+ (*)

Mol: 0,05 0,05 m = 0,1.143,5 + 0,05.108 = 19,75 (gam) D+ : Khng vit phn ng (*) Chn A.

Cu 2: Cho dung dch X (long) gm cc ion: H+(0,1 mol), Fe2+(0,1 mol) v Cl.Cho lng d dung dch AgNO3vo X, sau khi cc phn ng hon tonto thnh m gam kt ta. Gi tr ca m lA.43,05. B.53,85. C.45,75. D.28,70.

V d 3: Dung dch X c cha: Fe2+(0,05 mol), Na+(0,15 mol), Cl(0,05 mol) v

SO

2

4

. Cho dung dch Ba(OH)2

d vo X, thu c kt ta Y. Nung Y nhit cao ngoi khng kh n khi lng khng i thu c m gamcht rn. Gi tr ca m lA.30,5. B.23,3. C.31,3. D.27,3.

p dng nh lut trung ho in: t (+) t ()n = n

42+ + 2Fe Na Cl SO

2n + n = n 2n + 0,1 + 0,15 = 0,05 + 24SO

2n 24SO

n = 0,1 (mol).

Cc s phn ng:

SO 24

2++ Ba BaSO4

0+ t BaSO4

Fe2+ + 2OH Fe(OH)2 2 2+ O + H O Fe(OH)30+ t 1

2Fe2O3

m = 233.0,1 + 160.0,025 = 27,3 (gam) D.+ : Nhit phn Fe(OH)2ngoi khng kh thnh FeO Chn A.

Cu 3:Dung dch Y cha cc ion: Mg2+, NH4

+, SO 24

, Cl . Chia Y thnh hai

phn bng nhau.+ Phn mt cho tc dng vi dung dch NaOH d, un nng, thu c 2,9gam kt ta v 2,24 lt kh (ktc).+ Phn hai tc dng vi dung dch BaCl2d, thu c 11,65 gam kt ta.

C cn Y thu c m gam cht rn khan. Gi tr ca m lA.32,30. B.22,70. C. 11,35. D.16,35.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


27/165


27

V d 4: Mt cc nc cha: Ca2+(a mol), Mg2+(b mol) v HCO3

. Cho ti thiu

V lt dung dch Ca(OH)2x mol/vo cc lm gim tng nng ion kimloi trong cc xung mc nh nht. Biu thc tnh V theo a, b, x l

A. 2a bVx+

= . B. a bVx+

= . C. a 2bVx

+= . D. a bV

2x+

= .

Cc phn ng ho hc:Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O

Mol: a aMg(HCO3)2+ Ca(OH)2 MgCO3 + CaCO3 + 2H2O

Mol: b b

Vx = a + b a b

Vx+

= B.

Cu 4:Cho dung dch X gm: Na+ (0,15 mol), Ca2+ (0,05 mol); Cl (0,05 mol);

3HCO

(x mol) v 3NO

(0,05 mol). loi b ht Ca2+trong X cn mt

lng va V lt dung dch Ca(OH)20,01M. Gi tr ca V lA. 2,0. B. 5,0. C. 2,5. D. 4,5.

C. HNG DN GII

Cu 1: Gi in tch ion Y l n.+ p dng nh lut trung ho in: + 2+ YNa Ca Cln + 2n = n + n.n 0,05 + 0,2 = 0,1 + 0,15n n = 1 Loi C.

n y c 3 phng n tho mn l A, B v D. Vy chn Y l ion no?

Loi A v ion OHc phn ng vi ion HCO 3 :

OH + HCO3

CO 23

+ H2O.

Ion Y l3

NO . Khi un nng dung dch X:

Ca2+ + 2HCO3

CaCO3 + CO2 + H2O. (*)

Mol: 0,05 0,1 0,05 m = 0,05.23 + 0,05.40 + 0,15.62 + 0,05.100 = 17,45 (gam). B.+ : Khng vit phn ng (*) Chn D.

Cu 2: p dng nh lut trung ho in vi dung dch X: + 2+H Fe Cln + 2n = n 0,1 + 0,2 = Cln Cl

(0,3 mol).



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


28/165


28

Cc phn ng ion rt gn:Ag+ + Cl AgCl

Mol: 0,3 0,33Fe2+ + NO

3

+ 4H+ 3Fe3+ + NO + 2H2O (*)

Mol: 0,075 0,1Ag+ + Fe2+ Ag + Fe3+ (**)

Mol: 0,025 0,025 m = 0,3.143,5 + 0,025.108 = 45,75 (gam) C.+ : Khng vit phn ng (*) Chn B.

Cu 3: Phn mt + NaOH d:

Mg2+ + 2OH Mg(OH)2 22Mg(OH)Mg

2,9n = n = = 0,05 (mol).

58+

NH4

+ + OH NH3+ H2O +34

NHNH

2,24n = n = = 0,1 (mol).

22,4

Phn hai + BaCl2

d:Ba2+ + SO 2

4

BaSO4 24BaSOBa

11,65n = n = = 0,05 (mol).

233+

p dng nh lut trung ho in: 24 4

2+ +Mg NH SO Cl2n + n = 2n n +

0,05.2 + 0,1.1 = 0,05.2 +Cl

n .1 Cln = 0,1 (mol).

Bo ton khi lng: 22+ +4 4Mg NH Cl SO

m = m + m + m + m

= 2.(0,05.24 + 0,1.18 + 0,1.35,5 + 0,05.96) = 22,7 (gam).

B.+ : Ch tnh lng cht tan trong mt phn Chn C.

Cu 4: p dng nh lut trung ho in: + 2+3 3Na Ca Cl HCO NO

n + 2n = n + n + n

0,15 + 0,1 = 0,05 + x + 0,05 x = 0,15 (mol). Ghp Ca2+(0,05 mol) v 3HCO

(0,15 mol) thnh Ca(HCO3)2= 0,05 mol.

Khi cho X + dung dch Ca(OH)2:Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O

Mol: 0,05 0,05 V.0,01 = 0,05 V = 5 (lt). B.+ : Ch tnh kt ta ht ion Ca2+trong dung dch ban u Chn C.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


29/165


29

MunMunMunMun

HcHcHcHc

HHHHoooo

HayHayHayHay

NhNhNhNh

CyCyCyCy

ViViViVi

SchSchSchSch

7 PHNG PHP NG CHO


Vi hn hp bt k gm hai cht X v Y, khi bit c gi tr khi lngmol trung bnh ( M) ta s c t l:

X Y Y

Y XX

n M M M=

n MM M

=

2. S ho

3. Vn dngPhng php ng cho thng c p dng vi cc i lng nh:nguyn t khi, phn t khi, s nguyn t cacbon, s nguyn t hiro...Cng thc tnh ton vi cc i lng khc c dng tng t nh trn.

B. V D CU HI

V d 1: Trong t nhin, nguyn t ng c hai ng v l 6329

Cu v 6529

Cu.

Nguyn t khi trung bnh ca ng l 63,54. Thnh phn phn trm tngs nguyn t ca ng v 63

29Cu l

A.27%. B.50%. C.54%. D.73%.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


30/165


30

Gi s nguyn t 6329

Cu l a v 6529

Cu l b.

p dng cng thc ca phng php ng cho:

=63

65

a Cu 65 63,54 1,46 73= = =

b 63 63,54 0,54 27Cu % 63

29Cu = 73%.

D.

Cu 1: Trong tnhin, nguyn t brom c haing vl 7935 Br v81

35 Br. Nguyn

t khi trung bnh ca brom l 79,92. Thnh phn phn trm tng snguyn t ca ng v 7935 Br l

A.46%. B.54%. C.40%. D.60%.

V d 2: Hn hp X gm N2v H2c t khi hi so vi hiro bng 3,6. Nungnng X trong bnh kn nhit khong 4500C c bt Fe xc tc. Sau phnng thu c hn hp Y. T khi ca Y so vi hiro bng 4. Hiu sut ca

phn ng tng hp NH3lA.20%. B. 25%. C. 30%. D. 10%.

XM = 3,6.2 = 7,2.

p dng cng thc phng php ng cho: 2

2

H

N

n

n28 7,2 20,8 4

=2 7,2 5,2 1

= =

Chn s mol: N2(1 mol), H2(4 mol) Xm = 28.1 + 2.4 = 36 (gam).

Bo ton khi lng: Y Xm = m = 36 (gam) Y36

n = = 4,5 (mol).8

N2 + 3H2 2NH3

Mol: x 3x 2xMt khc: X Yn = n n = 0,5 4x 2x = 0,5 x = 0,25.

H =0,25

.100%1

= 25%.

B.

Cu 2a: Hn hp kh X gm N2v H2c t khi so vi hiro bng 3,6. Nungnng X trong bnh kn c bt Fe xc tc, thu c hn hp kh Y c s mol

gim 8% so vi ban u. Hiu sut ca phn ng tng hp NH3lA.25%. B.20%. C. 16%. D.23%.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


31/165


31

Cu 2b:Hn hp X gm SO2v O2c t khi hi so vi hiro bng 28. Nungnng X trong bnh kn cha xc tc V2O5, thu c hn hp kh Y. T khica Y so vi hiro bng 35. Hiu sut phn ng oxi ho SO2thnh SO3lA.60%. B.80%. C.50%. D.90%.

C. HNG DN GII

Cu 1: p dng cng thc ca phng php ng cho:79

81

Ag 81 79,92 1,08 54= =

79 79,92 0,92 46Ag = % 7935 Br = 54%.

B.

Cu 2a: XM = 3,6.2 = 7,2 22

H

N

n

n28 7,2 4

=2 7,2 1

=

Chn s mol: N2(1 mol), H2(4 mol).N2 + 3H2 2NH3

Mol: x 3x 2x

S mol kh gim: 4x 2x = 2x =8

.5 = 0,4100

x = 0,2 H = 20%.

B.

Cu 2b: XM = 28.2 = 56.p dng cng thc phng php ng cho:

2

2

SO

On 32 56 24 3 = = =n 64 56 8 1 Chn s mol: SO

2(3 mol) v O2(1 mol).

Xm = 64.3 + 32.1 = 224 (gam).

Bo ton khi lng: Y Xm = m = 224 (gam) Y224

n = = 3,2 (mol).70

2SO2 + O2 2 50

V O

t 2SO3

Mol: 2x x 2xMt khc: X Yn = n n = 4 3,2 = 0,8 3x 2x = 0,8 x = 0,8.

H =

0,8

.100%1 = 80%. B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


32/165


32

8 PHNG PHP TRUNG BNH


Trong mt hn hp nhiu cht, c th biu din mt i lng no cacc cht thng qua mt i lng chung, i din cho c hn hp, gi l

i lng trung bnh.2. Biu thc v vn dng

n

i ii 1

n

ii 1

X .nX

n

=

=

=

(X i lng ang xt; n s mol; i s th t cht).

+ Trong Ho V c, thng p dng biu thc trn vi nguyn t khi trung

bnh hoc phn t khi trung bnh: hhhh

mM

n=

+ Trong Ho Hu c cn p dng vi s nguyn t C trung bnh, s nguyn

t H trung bnh, s nhm chc trung bnh,

B. V D CU HI

V d 1: phn ng hon ton 3,8 gam hn hp X gm kim loi R (ch c hotr II) v oxit ca n cn va 250 ml dung dch HCl 1M. Kim loi R lA.Ba. B.Be. C.Mg. D.Ca.

Cc phng trnh phn ng:R + 2HCl RCl2 + H2RO + 2HCl RCl2 + H2O

Nhn thy: X HCl1

n = n = 0,125 (mol)2

XM = 3,8

= 30,40,125

R < 30,4 < RO R < 30,4 < R + 16

14,4 < R < 32 R = 24 (Mg) C.

Cu 1: Cho 3,2 gam hn hp gm hai kim loi ( hai chu k lin tip thucnhm IIA) tc dng ht vi dung dch HCl (d), thot ra 2,24 lt kh H2(ktc).

Hai kim loi lA.Be v Mg. B. Mg v Ca. C. Sr v Ba. D.Ca v Sr.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


33/165


33

V d 2: Cho hn hp gm mt kim loi kim X v mt kim loi kim th Y(MX< MY) tc dng va vi 0,1 mol kh Cl2, thu c 9,3 gam mui clorua.Kim loi X lA.Li. B.Na. C.Rb. D.K.

Gi cng thc chung ca hai kim loi l M, ho tr l n (1 < n < 2).

2M + nCl2 2MClnBo ton khi lng:

2M Clm + m = 9,3 Mm 2,2 (gam).=

Ta c:2e Cl

n = 2n = 0,2 (mol). Bo ton electron: e (M) en = n = 0,2 mol.

Mn.n = 0,2 mol 2,2.n

= 0,2M

M = 11n.

11 < M < 22. Vy MX < M < 22 X l Li. A.

Cu 2: Cho hn hp gm mt kim loi kim X v mt kim loi kim th Ytc dng va vi 5,6 lt kh Cl2(ktc), thu c 24,85 gam mui clorua.Kim loi X, Y lA.kali v bari. B.liti v beri.C.natri v magie. D.kali v canxi.

V d 3: Cho 4,75 gam hn hp X gm mui cacbonat v hirocacbonat ca kimloi kim M tc dng ht vi dung dch Ca(OH)2(d), to thnh 5 gam kt ta.Kim loi M lA.Na. B.K. C.Rb. D.Li.

Cc phn ng ha hc:

M2CO3 + Ca(OH)2 2MOH + CaCO3

MHCO3 + Ca(OH)2 MOH + CaCO3 + H2O (*)

Nhn thy :3X CaCO

5n = n = = 0,05 (mol).

100

XM =4,75

= 950,05

MHCO3 < 95 < M2CO3 M + 61 < 95 < 2M + 60

17,5 < M < 34 M l Na. A.

+ :Vit sn phm phng trnh (*) to thnh M2CO3,, CaCO3v H2O.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


34/165


34

Cu 3: Cho 5,0 gam hn hp X gmkim loi M (nhm IIA) v Zn tc dng vikh Cl2 d, to thnh 12,1 gam mui clorua. Mt khc, cho 1,6 gam M tcdng vi dung dch HCl d th th tch kh H2sinh ra cha n 1,12 lt (ktc).Kim loi X lA.Ba. B.Ca. C.Sr. D.Mg.

C. HNG DN GII

Cu 1:2H

2,24n = = 0,1 (mol).

22,4

Gi cng thc chung ca hai kim loi l M. Phn ng ho hc:

M + 2HCl MCl2 + H2

Nhn thy:2M H

n = n = 0,1 (mol) M =3,2

= 320,1

Mg v Ca.

B.

Cu 2: Gi cng thc chung ca hai kim loi l M, ho tr n (1 < n < 2).2M + nCl2 2MClnBo ton khi lng:

2M Clm + m = 24,85 Mm 7,1 (gam).=

Ta c:2

e Cln = 2n = 0,5 (mol) Bo ton electron: e (M) en = n = 0,5 mol.

Mn.n = 0,5 7,1.n

= 0,5M

M = 14,2n 14,2 < M < 28,4.

Mt kim loi < M < 28,4 Loi A v D. Mt kim loi > M > 14,2 Loi B. C.

Cu 3: Bo ton khi lng:

2X Clm + m = 12,1

2Clm 12,1 5 = 7,1 (gam)=

2Cln = 0,1 (mol).

Cc phng trnh phn ng:

X + Cl2 XCl2

Zn + Cl2 ZnCl2

Nhn thy:2M Zn Cl

n + n = n = 0,1 (mol) XM = 5

= 500,1

M ZnM < 50 < M

M l Mg hoc Ca.

Mt khc: M X 1,6M > M > = 320,05 M l Ca B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


35/165


35

9 PHNG PHP GII THEO PHNG TRNH ION


Trong dung dch cha cht in li, cc iontrc tip tham gia vo cc phnng ho hc, cn xt phn ng dng ion rt gn.

2. Vn dng+ P :

1 Dung dch hn hp axit tc dng vi dung dch hn hp baz;2 Hp th kh cacbonic vo dung dch cha hn hp baz;3 Trn hai dung dch cha nhiu cht in li vi nhau.

+ P :1 Dung dch cha axit HNO3v HCl (hoc H2SO4long);2 Dung dch cha mui nitrat v axit HCl (hoc H2SO4long).

B. V D CU HI

V d 1: Trn 200 ml dung dch X gm H2SO40,05M v HCl 0,1M vi 200 mldung dch Y gm NaOH 0,2M v Ba(OH)2 0,1M, thu c dung dch Z.Dung dch Z c pH lA.13. B.7. C.1. D.12.

24

2 4HCl H SOH

Cl SO

n = n + 2n = 0,02 + 2.0,01 = 0,04 (mol)Trong X:

n = 0,02 (mol); n = 0,01 (mol).+

2

+ 2+

NaOH Ba(OH)OH

Na Ba

n = n + 2n = 0,04 + 2.0,02 = 0,08 (mol)Trong Y: n = 0,04 (mol); n = 0,02 (mol).

Phng trnh ion: H+ + OH H2O OHd 0,04 mol.

[OH] = 10,04

= 0,1 = 100,4

[H+] = 314

110 = 10[OH ]

pH = 13.

A.

Cu 1: Trn 100 ml dung dch X gm HCl 0,10M v H2SO40,1M vi 400 mldung dch Ba(OH)2a mol/, thu c m gam kt ta v 500 ml dung dch

c pH = 13. Gi tr m v a ln lt lA. 2,33 v 0,10. B.2,33 v 0,05. C. 4,66 v 0,02. D.4,66 v 0,10.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


36/165


36

V d 2: Cho gam hn hp X g m Na v Ba (t l mol 1 : 1) tc dng vi nc(d), thu c dung dch Y v 3,36 lt H2 (ktc). Cho 200 ml dung dchAl2(SO4)30,2M vo Y, thu c a gam kt ta. Gi tr ca a lA.29,54. B.27,98. C.24,86. D.23,30.

Gi s mol trong X: Na (a mol) v Ba (a mol).

Theo bi: 2Hn = 0,15 mol 0,5a + a = 0,15 a = 0,1 (mol).

Y gm: OH(0,3 mol), Na+(0,1 mol) v Ba2+(0,1 mol).Phng trnh ion:

Ba2+ + 24SO BaSO4

Mol: 0,1 0,1 0,1

Al3+ + 3OH Al(OH)3

Mol: 0,08 0,24 0,08

Al(OH)3 + OH AlO2

+ 2H2O (*)

Mol: 0,06 0,06a = 0,1.233 + 0,02.78 = 24,86 (gam) C.+ :Khng vit phn ng (*) Chn A.

Cu 2: Trn 200 ml dung dch X gm Ba(OH)20,1M v NaOH 0,3M vi 100 mldung dch Y gm Al2(SO4)30,1M v H2SO40,1M, thu c a gam kt ta.Gi tr ca a lA. 5,24. B.4,66. C.5,82. D.6,22.

V d 3: Cho 6,4 gam bt Cu tc dng vi 100 ml dung dch X hn hp gmHNO31M v H2SO40,5M. Sau khi cc phn ng xy ra hon ton, sinh ra Vlt kh NO (sn phm kh duy nht, ktc). Gi tr ca V lA.2,24. B.0,84. C.1,68. D.1,12.

3

23 4

2 4HNO H SOH

NO SO

n = n + 2n = 0,1 + 2.0,05 = 0,2 (mol)Trong X:

n = 0,1 (mol); n = 0,05 (mol).+

Phng trnh ion rt gn:

3Cu + 8H+ + 2NO3

3Cu2+ + 2NO + 4H2O

Mol: 0,075 0,2 0,05 0,05 V = 0,05.22,4 = 1,1 (lt) D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


37/165


37

Cu 3a: Cho 4,8 gam Cu vo 100 ml dung dch X gm NaNO31M v H2SO41M. Sau khi cc phn ng xy ra hon ton thu c kh NO (sn phmkh duy nht l NO) v dung dch Y. C cn cn thn Y thu c m gamcht rn khan. Gi tr ca m lA.7,9. B.11,0. C.22,9. D.19,8.

Cu 3b: Cho 4,8 gam bt Cu trong 100 ml dung dch X gm HNO 3 1M vH2SO41,5M thu c dung dch Y. Cho bt Fe d vo Y, thy c m gambt Fe phn ng. Bit cc phn ng hon ton, kh NO l sn phm khduy nht ca N+5. Gi tr ca m lA. 5,6. B. 7,0. C. 8,4. D. 4,2.

C. HNG DN GII

Cu 1:24

2 4HCl H SOH

Cl SO

n = n + 2n = 0,01 + 2.0,01 = 0,03 (mol)Trong X:n = 0,01 (mol); n = 0,01 (mol).

+

Dung dch sau phn ng c mi trng baz OH d:OH + H+ H2O

Mol: 0,03 0,03 OH d (0,8a 0,03) mol.

Theo bi: pH = 13 [ OH ] = 101= 0,1

OH

n = 0,5.0,1 = 0,05 (mol) 0,8a 0,03 = 0,05 a = 0,1.

Ba2+ + 24SO BaSO4

Mol: 0,01 0,01 0,01 m = 0,01.233 = 2,33 (gam) A.

Cu 2: Trong X: Ba2+(0,02 mol), Na+(0,02 mol), OH (0,1 mol).

Trong Y: Al3+(0,02 mol); 24SO (0,04 mol), H+ (0,02 mol).

Cc phn ng ion rt gn:

H+ + OH H2O

Mol: 0,02 0,02

Al3+

+ 3OH

Al(OH)3

Mol: 0,02 0,06 0,02



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


38/165


38

Al(OH)3 + OH AlO2

+ 2H2O (*)

Mol: 0,02 0,02

Ba2+ + SO 24

BaSO4

Mol: 0,02 0,02 0,02 a = 0,02.233 = 4,66 (gam).

B.+ :Khng vit phn ng (*) Chn D.

Cu 3a:2

3 4

H Na

NO SO

n = 0,2 (mol); n = 0,1 (mol)Trong X:

n = 0,1 (mol); n = 0,1 (mol).

+ +

Phng trnh ion rt gn:

3Cu + 8H+ + 2NO3

3Cu2+ + 2NO + 4H2O

Mol: 0,075 0,2 0,05 0,05

Y gm cc ion: Cu2+ (0,075 mol), Na+ (0,1 mol), NO3

(0,05 mol) v

SO 24

(0,1 mol).

m = 0,075.64 + 0,1.23 + 0,05.62 + 0,1.96 = 19,8 (gam). D.

+ : Khng tnh khi lng ion Na+v SO 24

.

Cu 3b: 32

3 4

2 4HNO H SOH

NO SO

n = n + 2n = 0,1 + 2.0,15 = 0,4 (mol)Trong X:

n = 0,1 (mol); n = 0,15 (mol).+

3Cu + 8H+ + 2NO3

3Cu2+ + 2NO + 4H2O

Mol: 0,075 0,2 0,05 0,05Y + Fe (d):

Fe + 4H+ + NO3

Fe3+ + NO + 2H2O

Mol: 0,05 0,2 0,05 0,05Fe + 2Fe3+ 3Fe2+ (*)

Mol: 0,025 0,05Fe + Cu2+ Fe2+ + Cu

Mol: 0,05 0,05 m = 0,125.56 = 7,0 (gam).

B.+ :Khng vit phn ng (*) Chn A.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


39/165


39

Chun 2

CC nguyn tCC nguyn tCC nguyn tCC nguyn t halogenhalogenhalogenhalogen

ThiThiThiThi

GianGianGianGian

KhngKhngKhngKhng

ChChChCh

iiii

NuNuNuNu

BnBnBnBn

CnCnCnCn

HamHamHamHam

ChiChiChiChi

DNG 1: CC N CHT HALOGEN

A. L THUYT TRNG TM1. Tnh oxi ho mnh

L tnh cht c trng ca cc halogen. Trong dy halogen, tnh oxi hogim theo th t: F2> Cl2> Br2> I2.

a. (to mui vi s oxi ho cao nht ca kim loi)

2Fe + 3Cl2ot 2FeCl3 Cu + Cl2

ot CuCl2b.

F2 + H2 2HF Cl2 + H2 as 2HClc.

Cl2 + H2O HCl + HClO

2F2 + 2H2O 4HF + O2 d.

Cl2 + 2NaOH NaCl + NaClO + H2O (N G)

3Cl2 + 6KOHo80 C 5KCl + KClO3 + 3H2O

Cl2 + Ca(OH)2 ( )030 C CaOCl2 + H2O (C )

e.

Cl2 + 2NaBr 2NaCl + Br2Br2 + 2NaI 2NaBr + I2



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


40/165


40

f. Cl2 + 2FeCl2 2FeCl34Cl2 + H2S + 4H2O 8HCl + H2SO43Cl2 + 2NH3 N2 + 6HClBr2 + SO2 + 2H2O 2HBr + H2SO4

2. Tnh kh

F2khng th hin tnh kh.Cl2th hin tnh cht t oxi ho, t kh khi tc dng vi dung dch kim.Br2v I2th hin tnh kh khi tc dng vi cht oxi ho mnh hn:Br2 + 5Cl2 + 6H2O 10HCl + 2HBrO3

3. iu ch clo+ :

Cho cc cht oxi ho mnh tc dng vi axit HCl c:

MnO2 + 4HClot MnCl2 + Cl2 + 2H2O

2KMnO4+ 16HCl 2KCl + 2MnCl2 + 5Cl2 + 8H2O

KClO3

+ 6HCl

KCl + 3Cl2

+ 3H2

OK2Cr2O7 + 14HClot 2KCl + 2CrCl3 + 3Cl2 + 7H2O

+ :in phn dung dch NaCl, c mng ngn in cc:

2NaCl + 2H2O pddmn 2NaOH + Cl2 + H2

B. V D CU HI

V d 1 (B12): Cho phng trnh ho hc (vi a, b, c, d l cc h s):aFeSO4 + bCl2 cFe2(SO4)3 + dFeCl3

T l a : c lA. 4 : 1. B. 3 : 2. C. 2 : 1. D. 3 : 1.

Cc h s a v c lin quan n cc mui sunfat Bo ton nguyn t S viv tri v v phi ca phn ng:

4 4SO (tr) SO (p)n = n a = 3c a : c = 3 : 1 D.

Cu 1: Cho phng trnh ho hc (vi a, b, c, d l cc h s):aFeSO4 + bCl2 cFe2(SO4)3 + dFeCl3

T l a : b lA. 3 : 1. B. 3 : 2. C. 2 : 1. D. 1 : 1.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


41/165


41

V d 2 (B13): Cho gi tr m in ca cc nguyn t: F (3,98); O (3,44);C (2,55); H (2,20); Na (0,93). Hp cht no sau y l hp cht ion?A. NaF. B. CO2. C. CH4. D. H2O.

H L T 0,0 n < 0,4 cng ho tr khng c cc

T 0,4 n < 1,7 cng ho tr c cc 1,7 ion

A.

Cu 2: Cho gi tr m in ca cc nguyn t: F (3,98); O (3,44); N (3,04);C (2,55); H (2,20). Phn t cha lin kt cng ho tr khng c cc lA. NH3. B. CH4. C. CO2. D. H2O.

V d 3: X v Y l hai nguyn t trong cng mt nhm A v thuc hai chu klin tip (ZX< ZY). Tng s ht proton trong nguyn t X v Y l 26. Nhn

xt no sau y v X, Y l sai?A. n cht X, Y u l cht kh iu kin thng.B. m in ca X ln hn m in ca Y.C. Lc axit ca HX ln hn lc axit ca HY.D. n cht X tc dng vi hiro mnh lit hn Y.

X YZ + ZZ = = 132

X thuc chu k 2, Y thuc chu k 3 ZY ZX= 8.

ZY+ ZX= 26 ZX= 9 (Flo) v ZY= 17 (Clo).A ng: iu kin thng, F2v Cl2u tn ti th kh.

B ng: m in ca F ln hn ca Cl.C sai: Axit HF l axit yu, axit HCl l axit mnh.D ng: F2tc dng vi hiro ngay trong bng ti, phn ng gy n v tonhit rt mnh. Cl2tc dng vi hiro khi chiu sng hoc t nng. C.

Cu 3: X v Y l hai nguyn t trong cng mt nhm A v thuc hai chu klin tip (ZX< ZY). Tng s ht proton trong nguyn t X v Y l 52. Nhnxt no sau y v X, Y l ng?A. n cht X, Y u l cht kh iu kin thng.B. Bn knh nguyn t X ln hn bn knh nguyn t Y.

C. Lc axit ca HX ln hn lc axit ca HY.D. n cht X c tnh oxi ho mnh hn n cht Y.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


42/165


42

V d 4:Cho s chuyn ho gia clo v hp cht cha clo:

NaCl pddmn X0+ KOH, 80 C Y

0xt, t Z + T AgCl

Cc cht Y v T ln lt lA. KClO3v KCl. B. KClO3v AgNO3.C. KCl v HNO3. D. KCl v AgNO3.

2NaCl + 2H2O pddmn 2NaOH + Cl2 + H2

3Cl2 + 6KOH080 C 5KCl + KClO3 + 3H2O

2KClO30xt, t 2KCl + 3O2

KCl + AgNO3 AgCl + KNO3 B.

Cu 4: Cho s chuyn ho gia clo v hp cht cha clo:KClO3 + HCl X + Fe Y + NaOH Z 2 4+ H SO () T

Cc cht Y v T ln lt lA. FeCl3v HCl. B. FeCl3v Cl2. C. FeCl2v NaOH. D. FeCl2v NaCl.

V d 5:Trong tnhin, nguyn t clo c haing vl 3517

Clv 3717

Cl. Nguyn

tkhi trung bnhca clo l 35,5. Phn trm khi lng ca 3717

Cl trong

NaCl lA. 60,68%. B. 15,17%. C. 15,81%. D. 30,34%.

T l s nguyn t 3517 Cl v

3717 Cl c tnh theo cng thc ng cho:

35

37

Cl 37 35, 5 1, 5 3 75= = = =35 35, 5 0, 5 1 25Cl % s nguyn t

3717 Cl = 25%.

Xt 1 mol Cl: 3717 100Cl

25m = .37 = 9,25 37

17 Cl

9,25%m = .100% = 15,81 %.

58,5

C.

+ :35,5

%Cl = .100% = 60,68 %58,5

100

3717 Cl

25%m = .60,68% = 15,17% .

Cu 5a:Trong tnhin, nguyn t clo c haing vl 3517

Clv 3717

Cl. Nguyn

tkhi trung bnhca clo l 35,5. Phn trm khi lng ca 3517

Cl trong

KClO3l

A. 21,43%. B. 28,98%. C. 25,15%. D. 21,73%.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


43/165


43

Cu 5b: Trong t nhin, nguyn t brom c hai ng v l 7935

Brv 8135

Br.

Nguyn t khi trung bnh ca brom l 79,9. Phn trm khi lng ca79

35Br trong HBrO3l

A. 34,09%. B. 33,71%. C. 27,89%. D. 40,29%.

V d 6:Nguyn t X l phi kim thuc chu k 3, c cng thc hp cht vihiro l HX. Nguyn t X to vi kim loi M hp cht c cng thc MX,trong M chim 52,35% v khi lng. Kim loi M lA. Li. B. Ag. C. Na. D. K.

X to hp cht vi hiro l HX X thuc nhm 8 1 = 7 X l halogen.Theo bi, X thuc chu k 3 X l Cl (Clo).

Ta c:%M 52,35

=%Cl 47,65

35 5 47 65

M 52,35=

, , M = 39 (K) D.

Cu 6: Nguyn t Y l phi kim thuc chu k 2, c cng thc hp cht vi hirol HY. Nguyn t Y to vi kim loi M hp cht c cng thc MY 2, trong M chim 51,28% v khi lng. Kim loi M lA. Zn. B. Ca. C. Mg. D. Fe.

V d 7:Pht biu no sau y lsai?A.Halogen l nhng cht oxi ho mnh.B.Tnh oxi ho ca halogen gim t flo n iot.C.Cc halogen u c th c s oxi ho : 1, +1, +3, +5, +7.D.Cc halogen c tnh cht ho hc tng t nhau.

Nguyn t flo c m in ln nht flo ch th hin s oxi ho 1trong cc hp cht C sai. C.

Cu 7a: Pht biu no sau y l sai?A. Bn knh nguyn t ca clo ln hn flo.B. m in ca brom ln hn iot.C.Tnh kh ca ion Br ln hn ion Cl.D. Lc axit ca HF mnh hn HCl.

Cu 7b (B13): Cho cc pht biu sau:

(a) Flo ch th hin tnh oxi ho trong cc phn ng ho hc.(b) Axit flohiric l axit yu.(c) Dung dch NaF long c dng lm thuc chng su rng.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


44/165


44

(d) Cc halogen u c th c s oxi ho: 1, +1, +3, +5, +7.(e) Tnh kh ca cc ion halogenua tng dn theo th t: F, Cl, Br, I.Trong cc pht biu trn, s pht biu ng lA. 3. B. 5. C. 2. D. 4.

V d 8:Dn 1,344 lt kh Cl2(ktc) vo 200 ml dung dch KOH a mol/ 800C.

Sau khi phn ng hon ton, thu c 3,725 gam KCl. Gi tr ca a lA.0,3. B.0,6. C.0,4. D.0,2.

2Cl KCl

1,344 3,725n = = 0,06 (mol); n = = 0,05 (mol).

22,4 74,5


Mol: 0,03 0,06 0,05

a =0,06

0,2= 0,3 (mol/) A.

+ : Tnh theo s mol Cl2:2

KOH Cln = 2n = 0,12 a = 0,6 Chn B.

Cu 8:Hp th hon ton 1,344 lt kh Cl2(ktc) vo 200 ml dung dch KOH 1M 800C. C cn dung dch sau phn ng thu c m gam cht rn khan.Gi tr ca m lA.9,90. B.14,38. C.16,62. D.11,93.

V d 9:Cho cc phn ng:(a) O3 + KI + H2O (b) F2 + H2O

(c) MnO2 + HCl () (d) Cl2 + Br2+ H2O

S phn ng to ra n cht l

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

O3 + 2KI + H2O O2 + 2KOH + I2 (*)

2F2 + 2H2O 4HF + O2 (*)

MnO2 + 4HClot MnCl2 + Cl2 + 2H2O (*)

5Cl2 + Br2 + 6H2O 10HCl + 2HBrO3 C.

Cu 9:Cho cc phn ng xy ra trong dung dch iu kin thng:(a) Cl2 + KBr (b)NaCl ()+ H2SO4()

(c) KMnO4() + HCl () (d) Br2+ SO2 + H2O S phn ng to ra cht kh lA. 1. B. 2. C. 3. D. 4.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


45/165


45

V d 10:Cho hn hp kh X gm O2v Cl2tc dng va vi hn hp btgm 1,2 gam Mg v 1,3 gam Zn, thu c 5,27 gam hn hp cht rn Y.Phn trm th tch ca kh O2trong X lA.80%. B.40%. C.50%. D.60%.

65Mg Zn

1,2 1,3n = = 0,05 (mol); n = = 0,02 (mol).

24

Qu trnh oxi ho: Mg

+2

Mg + 2e; Zn

+2

Zn + 2e

Qu trnh kh: O2 + 4e 22

O ; Cl2 + 2e 21

ClBo ton electron:

2 2Mg Zn O Cl2n + 2n = 4n + 2n

2 2O Cl4n + 2n = 0,14.

X Mg Zn Ym + m + m = m 2 2O Cl32n + 71n = 2,77 2 2O Cln = 0,02; n 0,03.=

2O

%V = 40% B.

Cu 10a:t chy hon ton 3,6 gam kim loi R (ho tr II) trong hn hp kh

Cl2v O2. Sau phn ng thu c 11,5 gam cht rn v tng s mol kh tham gia phn ng l 0,125 mol. Kim loi R lA.Be. B.Cu. C.Ca. D.Mg.

Cu 10b:Hn hp kh X gm Cl2v O2. Cho X phn ng va ht vi hn hp gm2,4 gam Mg v 2,7 gam Al to ra 17,35 gam cht rn. Phn trm v th tchca kh Cl2trong X lA.85%. B.55%. C.75%. D.65%.

V d 11: Nguyn t X c electron mc nng lng 3p v c 1 electron lpngoi cng. Nguyn t Y c electron mc nng lng cao nht l 3p.Nguyn t X v Y c s electron hn km nhau l 2. Cng thc v bn chtlin kt trong hp cht gia X v Y ln lt lA.XY2, lin kt ion. B.XY, lin kt ion.C.XY2, lin kt cng ho tr. D.XY, lin kt cng ho tr.

Nguyn t X c electron mc nng lng 3p v c 1 electron lp ngoicng Lp ngoi cng ca X l 4s1.Y c mc nng lng cao nht l 3p Y c t electron hn X.

Lp electron ngoi cng ca Y l 3s23p5v ca X l 4s1. X to ra ion X+v Y to ra ion Y To hp cht XY B.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


46/165


46

Cu 11: Nguyn t X c electron mc nng lng 3p v c 2 electron lpngoi cng. Nguyn t Y c electron mc nng lng cao nht l 3p.Nguyn t X v Y c s electron hn km nhau l 3. Cng thc v bn chtlin kt trong hp cht gia X v Y ln lt lA.XY2, lin kt ion. B.X2Y, lin kt ion.C.XY2, lin kt cng ho tr. D.XY, lin kt cng ho tr.

V d 12: Nguyn t X c tng s ht electron trong cc phn lp p l 11. Sht mang in ca nguyn t X nhiu hn s ht mang in ca nguynt Y l 10 ht. Cng thc v bn cht lin kt hp cht gia X v Y lA. YX, lin kt cng ho tr. B. XY2, lin kt ion.C. YX2, lin kt ion. D. XY2, lin cng ho tr.

X c tng s electron p l 11 Cu hnh electron ca X: 1s22s22p63s23p5 X c Z = 17 v thuc nhm VIIA (X l phi kim).Theo bi: X Y2Z 2Z = 10 X YZ Z 5= YZ 12= Cu hnh electron ca

Y l 1s22s22p63s2 Y thuc nhm IIA (Y l kim loi).Cng thc hp cht gia X v Y l YX2, lin kt ion C.

Cu 12: Nguyn t X c tng s ht electron trong cc phn lp p l 7. S htmang in ca nguyn t X nhiu hn s ht mang in ca nguyn t Y l8 ht. Cng thc v bn cht lin kt hp cht gia X v Y lA. XY, lin kt cng ho tr. B. XY3, lin kt ion.C.X2Y, lin kt ion. D. XY2, lin cng ho tr.

V d 13: Dy g m cc dung dch u th hin tnh kh khi phn ng vi khCl2lA. NaBr, NaI, NaOH. B. NaOH, FeCl2, SO2.C. NH3, Ca(OH)2, NaBr. D. SO2, Br2, FeCl2.

Cl2ng vai tr t oxi ho, t kh khi tc dng vi cc dung dch NaOHv Ca(OH)2 Loi A, B, C D.Cl2 + 2NaOH NaCl + NaClO + H2O

Cl2 + Ca(OH)2 ( )030 C CaOCl2 + H2O

Cl2 + SO2 + 2H2O 2HCl + H2SO4

Cl2 + 2FeCl2 2FeCl35Cl2 + Br2 + 6H2O 10HCl + 2HBrO3



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


47/165


47

Cu 13: Cho dy gm cc cht: F2, Cl2, Br2, HCl, H2O. S cht trong dy va ctnh oxi ho, va c tnh kh lA. 4. B. 3. C. 2. D. 1.

V d 14: Cho cc phn ng ho hc:(a) Cl2 + 2NaOH NaCl + NaClO + H2O

(b) 3Cl2 + 6KOH o80 C 5KCl + KClO3 + 3H2O(c) KClO3 + 6HCl KCl + 3Cl2 + 3H2O(d) Br2 + 5Cl2 + 6H2O 10HCl + 2HBrO3(e) CaOCl2 + 2HCl CaCl2 + Cl2 + H2OS phn ng trong nguyn t clo va ng vai tr cht oxi ho, vang vai tr cht kh lA. 3. B. 2. C. 4. D. 5.

Cc phn ng clo va ng vai tr cht oxi ho, va ng vai tr cht kh Cl va tng, va gim s oxi ho Phn ng (a), (b), (c), (e). C.

Cu 14: Cho cc phn ng ho hc:(a) Br2 + 5Cl2 + 6H2O 10HCl + 2HBrO3(b) 3Br2 + 6KOH 5KBr + KBrO3 + 3H2O(c) Br2 + 2FeBr2 2FeBr3(d) Br2 + SO2 + 2H2O 2HBr + H2SO4(e) Br2 + 2NaI 2NaBr + I2S phn ng trong Br2ch ng vai tr cht oxi ho l

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

V d 15: Cho cc phn ng ho hc sau:2FeBr2 + Br2 2FeBr3 (1)2NaBr + Cl2 2NaCl + Br2 (2)Pht biu ng l:A.Cl c tnh kh mnh hn Br. B.Br2c tnh oxi ho mnh hn Cl2.C.Br c tnh kh mnh hn Fe2+. D.Cl2 c tnh oxi ho mnh hn Fe3+.

Nhn xt: (1) Tnh oxi ho Br2> Fe3+; tnh kh Fe2+> Br.

(2) Tnh oxi ho Cl2> Br2; tnh kh Br> Cl. Tnh oxi ho Cl2> Br2> Fe3+; tnh kh Fe2+> Br> Cl D.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


48/165


48

Cu 15: Cho cc phn ng ho hc sau:2FeCl2 + Cl2 2FeCl3 (1)2FeCl3 + 2HI 2FeCl2 + I2 + 2HCl (2)Pht biu khng ng l:A.Fe2+ c tnh kh mnh hn I. B.Cl2c tnh oxi ho mnh hn I2.C.Fe3+ c tnh oxi ho mnh hn I2. D.Cl2 c tnh oxi ho mnh hn Fe3+.

V d 16: Cho 1,12 gam kim loi M tc dng ht vi kh clo (d), thu c 3,25gam mui clorua. Kim loi M lA.Cu. B.Mg. C.Ca. D.Fe.

Bo ton khi lng:2Cl

m = 3,25 1,12 = 2,13 2Cl

n =2,1371

= 0,03 (mol).

Phng trnh ho hc: 2M + nCl20t 2MCln

2

M Cl

2 0,6n = n =

n n M =

1,12n

0,06 =

56n

3 n = 3 v M = 56 (Fe).

D.

Cu 16: Cho 2,4 gam hn hp gm Cu v kim loi M (t l mol 1 : 1) tc dnght vi kh Cl2(d), thu c 5,95 gam mui clorua. Kim loi M lA.Mg. B.Ca. C.Al. D.Fe.

V d 17: t chy hon ton 2,2 gam hn hp gm mt kim loi kim X vmt kim loi kim th Y (MX < MY) trong kh clo d, thu c 9,3 gammui clorua. Kim loi X l

A. Li. B. Na. C. Rb. D. K.

Gi cng thc chung ca hai kim loi l M, ho tr n (1 < n < 2).

2M + nCl20t 2MCln

Bo ton khi lng:2Cl

m = 9,3 2,2 = 7,1 2Cl

n = 0,1 mol.

Bo ton electron:2M Cl

n.n = 2n = 0,2 mol M0,2

n = mol.n

M

2,2 2,2nM = = 11n.

n 0,2= Do 1 < n < 2 11 < M < 22 MX < M < 22.

X l Li A.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


49/165


49

Cu 17: t chy hon ton 4,07 gam hn hp gm mt kim loi kim X vmt kim loi kim th Y trong kh Cl2d, thy c 0,185 mol kh Cl2 phnng. Kim loi X, Y lA. Na v Be. B. Li v Be. C. Na v Mg. D. K v Ca.

C. HNG DN GII

Cu 1: Bo ton nguyn t Fe: a = 2c + d.Bo ton nguyn t S:

4 4SO (tr) SO (p)n n= a = 3c c = d a = 3d.

Bo ton nguyn t Cl: 2b = 3d 2b = a a : b = 2 : 1 C. Cn bng phn ng: 6FeSO4 + 3Cl2 2Fe2(SO4)3 + 2FeCl3 a : b = 6 : 3 = 2 : 1 C.

Cu 2:Lin kt cng ho tr khng c cc khi hiu m in ca hai nguyn tto thnh lin kt nh hn 0,4.Trong phn t CH4, hiu m in ca cacbon v hiro bng 0,35 lin kt CH khng c cc. B.

Cu 3: X YZ + Z

Z = = 262

X thuc chu k 3, Y thuc chu k 4 ZY ZX= 18.

ZY + ZX= 52 ZX= 17 (Clo) v ZY= 35 (Brom).A sai: iu kin thng, Cl2l cht kh, Br2l cht lng.B sai: Bn knh nguyn t Cl nh hn bn knh nguyn t Br.C sai: Lc axit HF


50/165


50

Cu 5a: Phn trm s nguyn t mi ng v: 3517 Cl = 75%;3717 Cl = 25%.

Xt 1 mol Cl:3517 100Cl

75m = .35 = 26,25 35

17 Cl

26,25%m = .100% = 21,43 %

122,5

A.

+ :35,5

%Cl = .100% = 28,98 %122,5

100

3517 Cl

75%m = .28,98% = 21,73%

Chn D.

Cu 5b:T l s nguyn t 79

35Br v 81

35Br c tnh theo cng thc ng cho:

79

81

Br 81 79,9 1,1 11 55= = =

7979,9 0,9 9 45Br = % s nguyn t 79

35Br = 55%.

Xt 1 mol Br: 79 Br55

m = .79 = 43,45100

79 Br43,45

%m = .100% = 33,71 %128,9

B.

+ : 79,9%Br = .100% = 61,99 %128,9

7935 100Br

55%m = .61,99% = 34,09%

Chn A.

Cu 6: Y to hp cht vi hiro l HY Y thuc nhm 8 1 = 7 Y l thucnhm VIIA. Theo bi, Y thuc chu k 2 Y l F (Flo).

Ta c:%M 51,28

=%F 48,72

M 51, 38

=2.19 48,62

M = 40 (Ca).

B.

Cu 7a: Nhn xt: Do HF c bn lin kt ln HF kh phn li axit HFl axit yu D sai. D.

Cu 7b (B13): (d) sai: Flo ch th hin s oxi ho 1 trong cc hp cht.Cc pht biu cn li u ng D.

Cu 8: 442Cl KOH

1,3n = = 0,06 (mol); n = 0,2.1 = 0,2 (mol).

22,4


Mol: 0,06 0,12 0,1 0,02



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


51/165


51

3KCl KClO KOH

m = m + m + m = 0,1.74,5 + 0,02.122,5 + 0,08.56 = 14,38.

B.+ : Khng tnh cht rn KOH m = 9,90 Chn A.

Cu 9: Cl2 + 2KBr 2KCl + Br2 ()NaCl () + H2SO4() NaHSO4 + HCl (*)2KMnO4+ 16HCl 2KCl + 2MnCl2 + 5Cl2 + 8H2O (*)Br2 + SO2 + 2H2O 2HBr + H2SO4 B

Cu 10a:22Cl O

m + m = 11,5 3,6 = 7,9 22Cl O

71n + 32n = 7,9.

Theo bi:2 2Cl O

n + n = 0,125 2 2Cl O

n = 0,025 (mol); n = 0,1 (mol).

Bo ton electron:2 2R O Cl R

2n = 4n + 2n = 0,3 n = 0,15 3,6

R = = 24 (Mg)0,15

D.

Cu 10b:22Cl O

m + m = 17,35 2,4 2,7 = 12,25 22Cl O

71n + 32n = 12,25.

Bo ton electron:2 2Cl O Mg Al

2n + 4n 2n + 3n = 0,5= 2 2Cl O

n = 0,15; n = 0,05.

2Cl

0,15%V .100% 75%

0,2= = C.

Cu 11: X c mc nng lng 3p v c 2 electron lp ngoi cng Lpngoi cng ca X l ns2(n 4).Y c mc nng lng cao nht l 3p Y c t electron hn X.

Lp electron ngoi cng ca Y l 3s23p5v ca X l 4s2. X to ra ion X2+v Y to ra ion Y To hp cht XY2 A.

Cu 12: Nguyn t X c tng s electron p l 7 X c cu hnh electron l:1s22s22p63s23p1 XZ = 13 v X l kim loi nhm IIIA.

X Y2Z 2Z = 8 X YZ Z = 4 9YZ = Y: 1s22s22p5 (Y l halogen).

To hp cht XY3, lin kt ion B.

Cu 13: F2ch c tnh oxi ho; Cl2, Br2, HCl, H2O c c tnh oxi ho v tnh kh.

Cl2

+ 2NaOH

NaCl + NaClO + H2

O (Cl2

: t oxi ho, t kh)Br2 + 2NaI 2NaBr + I2 (Br2: cht oxi ho)



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


52/165


52

Br2 + 5Cl2 + 6H2O 10HCl + 2HBrO3 (Br2: cht kh)2HCl + Fe FeCl2 + H2 (HCl: cht oxi ho)

4HCl + MnO20t MnCl2 + Cl2 + H2O (HCl: cht kh)

2H2O + 2Na 2NaOH + H2 (H2O: cht oxi ho)

2H2O + 2F2 4HF + O2 (H2O: cht kh)

B.

Cu 14: (a) Br2ng vai tr cht kh.(b) Br2va ng vai tr cht oxi ho, va ng vai tr cht kh.(c), (d), (e) tho mn. A.

Cu 15: Nhn xt: (1) Tnh oxi ho Cl2> Fe3+; tnh kh Fe2+> Cl.(2) Tnh oxi ho Fe3+> I2; tnh kh I> Fe2+.

Tnh oxi ho Cl2> Fe3+ > I2; tnh kh I> Fe2+> Cl. D.

Cu 16:2Cl

m = 5,95 2,4 = 3,55 (gam) 2Cl

n = 3,5571

= 0,05 (mol).

Gi s mol mi kim loi l a.+ Nu M ho tr II:

a + a = 0,05 a = 0,025 64 + M 2,4

M = = = 482 0,05

M = 32: Loi.

+ Nu M ho tr III:

a + 1,5a = 0,05 a = 0,02 64 + M 2,4

M = = = 602 0,04

M = 56 (Fe).

D.

Cu 17: Gi cng thc chung ca hai kim loi l M, ho tr n (1 < n < 2).

2M + nCl20t 2MCln

Bo ton electron:2M Cl

n.n = 2n = 0,37 mol M0,37

n = mol.n

M

4,07 4,07nM = = 11n.

n 0,37= Do 1 < n < 2 11 < M < 22.

Mt kim loi c nguyn t khi < M < 22 Loi C, D. Mt kim loi c nguyn t khi > M > 11 Loi B. A.



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


53/165


53

ThiThiThiThi

SSSS

DDDD

HnHnHnHn

NgayNgayNgayNgay

KhiKhiKhiKhi

BnBnBnBn

GiiGiiGiiGii

HnHnHnHn

DNG 2. AXIT HALOGENHIRIC (HX)

A. L THUYT TRNG TM1. Tnh axit

Tnh axit tng dn trong dy: HF


54/165


54

+ HI th hin tnh kh khi tc dng vi H2SO4c, dung dch FeCl3:

8HI + H2SO4()ot 4I2 + H2S + 4H2O

2HI + 2FeCl3 2FeCl2 + I2 + 2HCl3. iu ch

+ HF: CaF2 + H2SO4()o250 C 2HF + CaSO4

+ HCl: NaCl () + H2

SO4

()

ot NaHSO

4

+ HCl

Cl2 + H2 as 2HCl

B. V D CU HI

V d 1:C cc th nghim sau:(a) Cho SiO2vo dung dch HF; (b) Sc kh SO2vo nc brom;(c) Cho nc brom vo dung dch NaI; (d) Cho H2tc dng vi F2.S th nghim xy ra phn ng oxi ho kh lA.3. B.1. C.2. D.4.

Cc phn ng ho hc:SiO2 + 4HF SiF4 + 2H2O

SO2 + Br2 + 2H2O H2SO4 + 2HBr (*)

Br2 + 2NaI 2NaBr + I2 (*)

H2 + F2 2HF (*) A.

Cu 1: Pht biu no sau y l sai?A. Lin kt HF phn cc hn lin kt HCl.

B. AgI khng tan trong nc, AgF tan trong nc.C. Lc axit ca HCl mnh hn HI.D. Dung dch HF ho tan c SiO2.

V d 2: Cho dy cc hp cht: CaCO3, Fe(OH)3, FeS, BaSO4, AgNO3, CuS,NaHCO3, CuO.S hp cht trong dy tc dng c vi dung dch HCllong lA. 7. B. 8. C. 6. D. 5.

Cc phng trnh ho hc:

CaCO3 + 2HCl CaCl2 + CO2 + H2OFe(OH)3 + 3HCl FeCl3 + 3H2O



W

W

W

D

Y

K

E

M

QU

Y

N

H

O

N

U

C

O

Z

CO

M


55/165


55

AgNO3 + HCl AgCl + HNO3NaHCO3 + HCl NaCl + H2O + CO2FeS + 2HCl FeCl2 + H2SCuO + 2HCl CuCl2 + H2O A.+ : Tnh c BaSO4hoc CuS.

Cu 2: Dy gm cc cht u tc dng c vi dung dch HCllong lA.KNO3, CaCO3, Fe(OH)3. B.FeS, BaSO4, KOH.C.AgNO3, (NH4)2CO3, CuS. D.NaHCO3, FeS, CuO.

V d 3:Cho dy cc cht: Al, Al(OH)3, Zn(OH)2, NaHCO3, Na2SO4. S chttrong dy va phn ng c vi dung dch HCl, va phn ng c vidung dch NaOH lA. 5. B. 4. C. 3. D. 2.

Cc cht tho mn: Al, Al(OH)3, Zn(OH)2, NaHCO3

Al + NaOH + H2O NaAlO2 +32

H2

Al(OH)3 + NaOH NaAlO2 + 2H2OZn(OH)2 + 2NaOH Na2ZnO2 + 2H2ONaHCO3 + NaOH Na2CO3 + H2OTng t, cc em c th vit phn ng vi HCl B.

Cu 3:Cho dy cc cht rn: MnO2, FeS, CuS, Fe(NO3)2, CaCO3, Fe, CaC2. Scht trong dy tc dng vi dung dch HCl long to ra cht kh lA. 5. B. 4. C. 3. D. 2.

V d 4:Ho tan hon ton 2,43 gam hn hp g m Mg v Zn vo mt lngva dung dch HCl, sau phn ng thu c 1,12 lt H2 (ktc) v dungdch X. C cn X thu c m gam mui khan. Gi tr ca m lA. 6,08. B. 6,18. C. 6,33. D. 5,98.

Bo ton nguyn t H: 2HCl H2

2HCl Hn = 2n = 0,1 (mol)

Cl (X)n = 0,1 (mol).

Clm = 2,43 + m = 2,43 + 0,1.35,5 5,98 (gam) = D.



W

W

W

D

Y

K

E

M

QU

Y

N

Documents

TỰ HỌC GIỎI HÓA HỌC TẬP 1: HÓA HỌC VÔ CƠ (DÀNH CHO LUYỆN THI ĐẠI HỌC, CAO ĐẲNG) NXB ĐẠI HỌC SƯ PHẠM