Phương pháp và kỹ thuật giải nhanh các dạng bài tập trắc nghiệm hóa học đại cương hóa vô cơ

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HOC

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HA

Pi6l

Npi

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Di^n t h o q i : Bien tap - C h e b a n : (04) 39 714896

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A. N O I D U N G PHlJCfNG P H A P

Fax: (04) 39 714899

Chiu
Gidm

doc

Tong

bien

Bien
Che

track

nhi^m

Trinh

tham gia phdn iing bang tong khoi lU0ig cdc chat tao thdnh "
D i e u nay giup ta giai bai toan h6a hoc mot each ddn gian, nhanh chong
X6i phan tfng: A + B -> C + D

PHAM THI TRAM

tap :

T a l u o n c o : niA + me = mc + moCl)

T H U HtTClNG

* LUu y: D i e u quan trong nhat khi ap dung phu'dng phap nay do la viec phai
xac djnh dung lifdng chat (kho'i liTdng) tham gia phan u'ng va tao thanh (c6

KHANG V I E T

:

bay

A p dung djnh luat bao toan khoi lifdng ( B T K L ) : " Tong khoi lUctng cdc chat

ban:

PHUNG QUOC BAO

:

tap:
ban

xuat

1, NOi d u n g

chii y den cac chat ket tua, bay hdi, dac biet la khoi li/dng dung djch).
2. Cac d a n g bai toan thifoTng gap

KHANG V I E T

bia:
Doi

tdc lien

ket xuat

H$ qua 1: B i e t tong khoi lifdng chat ban dau «-> khoi lu'dng chat san pham
Phifdng phap g i a i : m(dau) = m(sau) (khong phu thuoc hieu sua't phan u'ng)

ban:

qua

C o n g ty T N H H MTV D W H K H A N G V I E T
Tong

phdt

2: Trong phan u'ng c6 n chat tham gia, ne'u biet k h o i liTdng cua

(n - 1) chat thi ta de dang tinh khoi lu'dng cua chat con l a i .
//e ^Ma i ; B a i toan: K i m loai + axit -> muoi + khi

hdnh:

'^mue'i ~ "^Icimloai

C O N G T Y TNHHtVlTV
D!CH Vy VAN HOA KHANG V I E T

- Bie't k h o i luTdng k i m l o a i , khoi lu'dng anion tao muoi (tinh qua san pham khi)
-> khoi lUdng m u o i

- K h o i lu'dng anion tao muoi thudng dUdc tinh theo so mol khi thoat ra:
• V d i axit H C l va H 2 S O 4 loang

Email: khangvietbookstore®yahoo.com.vn

+ 2HC1 ^

Website: www.nhasa

+

H 2 nen S04^'

<^

'

H2

(xem them phiTdng phap bao toan electron hoac phUdng phdp bao
THUAT G I A I N H A N H C A C

T R A C N G H I E M H O A DAI C U O N G - V 6

D A N G BAI

CO

M a so : 1 L - 148 D H 2 0 1 2 . In 2.000 cuon, kho 1 6 x 2 4 c m ,
Tai Chi n h a n h con g ty T N H H M T V NXB GIAO T H O N G V A N TAI

So xuat M n : 4 9 6 - 201 2/CXB/5 - 7 7 / D H Q G H N ngay 03/05/201 2.
Q u y e t d i n h xuat h&n so: 148LK - T N / Q D - N X B D H Q G H N
ILAJ

H2SO4 ^

{ , i>

H 2 n e n 2 C r <^ H 2

• V d i axit H2SO4 dac, nong va H N O 3 : Sur dung phiTdng phap ion - electron

SACH L I E N K E T

In xong va n o p

^

- Biet khoi lUdng m u o i va kho'i lUdng anion tao muo'i -> khoi lu'dng k i m loai

Dia chl: 71 Dinh Tien Hoang - P.Da Kao - Q.1 - TP.HCM
Di§n thoai: 08. 39115694 - 39105797 - 39111969 - 39111968
F a x : 0 8 : 3911 0880

PHa(3NG PHAP & KY

'^aniontaomud'i

c h i e u quy III n a m 2 0 1 2 .

TAP

nguyento)

toan

,

qua 3: B a i toan khuT hon hdp oxit k i m loai bdi cdc cha't k h i ( H 2 , CO)
Sd do: Oxit k i m loai + (CO, H 2 ) ^

r ^ n + hon hdp k h i ( C O 2 , H 2 O , H 2 , CO)

Ban chat la cac p h i n i^ng: C O + [O] -> C O 2
H2 +
_ _ _ ^ n [ 0 ]

=

n(C02) =

n(H20) ^

[O] ^

H2O

m^n = moxu - m[oj

"''-''^

"


Phuong ph^p vJi ky thugt giSi nhanh B T T N H6a J j i cuong - vO cO - D 5 XuSn Hung

3. D a n h gia phUrfng phap bao toan kho'i iif(/ng

Phifdng phap bao toan kho'i li/'dng cho phep giai nhanh di/dc nhieu bai toan
khi bie't quan he ve khoi liTdng cua cic chat triTdc va sau phan ufng.
Dac biet, khi chiTa bie't ro phan iJng xay ra hoan toan hay khong hoan toan
thi viec sijr dung phu'cJng phap nay cang giup dcfn gian hoa bai toan hdn.
PhU'dng phap bao toan khoi lu'dng thu'dng du'cJc suT dung trong cac bai toan
nhieu chat.

4. C a c btfdfc giai

- Lap sd do bien doi cac chat tru'dc va sau phan iJng.
- Tu" gia thie't cua bai toan tim X w d c ~ S s a u ('^hong can bie't phan uTng la
hoan toan hay khong hoan toan)
- Van dung djnh luat bao toan khoi lu'dng de lap phUdng trinh toan hoc, ket
hdp diJ kien khac de lap he phi/dng trinh toan.
- Giai he phuTdng trinh.
B. B A I T A P M I N H H Q A

CSu 1: Dot chay hoan toan 17,4 gam hon hdp Mg va Al trong khi oxi (duT) thu
diifdc 30,2 gam hon hdp oxit. The tich khi oxi (dktc) da tham gia phan iJng la:
A. 17,92 lit
B. 4,48 lit
C. 11,20 lit

D. 8,96 lit
(Trich de thi tuyen sink Dai hoc khoi A nam 2011)
Hi^dng dSn

Ap dung DLBTKL ta c6: m^^ = moxu ^ "02 ^^"^'^

= 30,2 - 17,4 = 12,8 (g)

CSu 2: Dun nong m gam h6n hdp Cu va Fe c6 ti le kho'i lifdng tiTdng iJng 7 : 3
vdi mot liTdng dung djch HNO3. Khi cac phan tfng ket thuc, thu diTdc 0,75m
gam chat r^n, dung djch X va 5,6 lit hon hdp khi (dktc) gom NO va NO2
(khong C O san pham khu" khac cua N"^^). Bie't lu'dng HNO3 da phan iJng la 44,1
gam. Gia tn cua m la
A. 44,8.
B. 40,5.
C. 33,6.
D. 50,4.
(Trich de thi tuyen xinh Dai hoc khoi A nam 2011}
giai

- Khoi lu'dng Fe - 0,3m gam va khoi liTdng Cu = 0,7m gam
Sau phan iJng con 0,75m gam => Fe chi phan (Sng 0,25m gam; Fe dif vay sau
phan uTng ch? thu diTdc muo'i Fe^*
Ta c6: n^.^^ = 0,7 ; n^^ + n^^^ = 0,25, so' mol ciia Fe(N03)2 = ^'^^"^
A

I).
''
CSu 3: Cho hdi nu'dc di qua than nong do, thu diTdc 15,68 lit hon hdp khi X (dktc)
gom CO, CO2 va H2. Cho toan bp X tac dung het vdi CuO (du') nung nong, thu

diTdc h6n hdp chat ran Y. Hoa tan toan bo Y bang dung djch HNO3 (loang,
du') diTdc 8,96 lit NO (san pham khuT duy nha't, d dktc). Phan tram the tich khi
CO trong X la:
A. 18,42%
B. 28,57%
C. 14,28%
D. 57,15%
(Trich de thi tuyen sinh Dai hoc khoi B nam 2011)
Hufdng d i n giai
H2O + C — ^ CO + H2
»*
'*
=> D a p a n

X

X

X

2H2O + C

—

CO2 + 2H2

^

2y
y

2y
Ta C O : nx = 0,7 mol => 2x + 3y = 0,7

giai

^ % ^ 0,4.22,4 = 8,96 lit zz> Dap an D.

HuTdng dSn

Sd do piJ:
Fe
+
HNO3 ^ F e ( N 0 3 ) 2 + N O + NO2
a25m
07
a25m
^^5
,
, .
56
56
Ap dung DLBT nguyen to'N ta c6 : 0,7 = 2.^'^^"^ +0,25 m = 50,4 (g)

*

hhX

{CO,

H2} + CuO

CO + CuO
H2 + CuO

CO2
H2O

3Cu + 8HNO3 ^

0,6

^

+Cu
+Cu

Cu

3Cu(N03)2

+ HNO3 ^

(1)
0,4

mol

'
NO.
.^F .

+ 2N0 +

4H2O

^

'

"

0,4

Trong phan uTng khur oxi kim loai bdi CO, H2
Taco: n^^Q

= noarongcuo = no, =0,6 mol => 2x + 2y = 0,6 (2)

Tir(l),(2) =>x = 0,2; y = 0,l
Vay : % V c o = — . 1 0 0 % = 28,57% => Dap an B.
.
0,7
Cfiu 4: Hap thu hoan toan 2,24 lit CO2 (dktc) vao 100ml dung dich gom K2CO3
0,2M va KOH x mol/lit, sau khi cac phan ilng xay ra hoan toan thu diTdc dung
djch Y. Cho toan bp Y tac dung vdi dung dich BaCh (diT), thu diTdc 11,82 gam
ket tua. Gia tn cua x la:
A. 1,0
B. 1,4
C. 1,2
D. 1,6

(Trich de thi tuyen sinh Dai hoc khoi B ndm 2011)


PhUOng phap vl ky thujt giii nhanh BTTN H6a dgi cuang - vO CO - D 8 XuSn Hi/ng

HuTdng d§n giai

Cfiu 6: Hoa tan hoan toan m gam hon hdp X gom Na va K vao dung dich HCl dtf
thu du'dc dung djch Y. Co can dung djch Y thu du'dc (m + 31,95) gam hon hdp

11 82
Taco: 11^0^ = 0,1 mol; n^^^^^ = _ ^ = 0,06 mol, n^^co, - 0 , 0 2 mol

chat ran khan. Hoa tan hoan toan 2m gam hon hdp X vao niTdc thu dU'dc dung

Khi sue CO2 vao dung dich hon hdp gom K2CO3 va KOH, gia siif chi xay ra

djch Z. Cho tir tir het dung dich Z vao 0,5 lit dung djch CrCl31M den phan iJng

phan drng:

h6an toan thu dU'dc ket tua c6 kho'i liTdng la

CO2 + 2KOH

A. 54,0 gam.

K ^ O j + H2O

^

0,1

B. 20,6 gam.

= > " K J C O , (irong dung dich) = 0,1

+ 0,02

= 0,12

Hifdng dSi^giai

mol

Taco: m „ , u 6 - i = m K L + m
^

n;

cho = 0,06 mol

(Irong K2CO3) ^ "c

(trong B a C O j ) + "c

(trong K H C O 3 )

0,02 = 0,06 + a (a la so mol KHCO3)
KHCO3

0,06

0,06

0,06

CO2 + 2 K 0 H ^
0,04

1,5

"cr(OH)3

K2CO3 + H2O

Dap an B.

Hon hdp X gom Fe(N03)2, Cu(N03)2 va AgN03. Thanh phan % khoi

li/dng cua nitd trong X la 11,864%. C6 the dieu che diTdc toi da bao nhieu
gam hon hdp ba kirn loai tiT 14,16 gam X?

0,5

B. 7,68 gam

C. 3,36 gam

Hifdng d§n giai

Ta c6: % N = 11,864% ^m^
= 0,12 mol => n

= 1 4 , 1 6 . ^ - ^ ^ = 1,68 (g)
I vJL)

=0,12 mol
NO3

=>DapanD.

OH

''^i •

' •

= 0'5 - 0,3 = 0,2 mol ^

' "

'

'
\

mc,(OH)3 = 0,2.103 = 20,6g

'

D. 6,72 gam

2 lit dung dich X. Lay 1 lit dung djch X tac dung vdi dung dich BaCl2 (diT) thu
diTdc 11,82 gam ket tua. Mat khac, cho 1 lit dung djch X vao dung djch CaCl2
(dir) roi dun nong, sau khi ket thuc cac phan iJng thu difdc 7,0 gam ket tua. Gia
trj cua a, m ti/dng vlng la
A. 0,04 va 4,8.

_ = 14,16-0,12.62 = 6,72 (g)

NO3

B. 0,07 va 3,2.

C. 0,08 va 4,8.

D. 0,14 va 2,4.

(Trich de thi tuyen sinh Dai hoc khoi A nam 2010)

Trich de thi tuyen sink Dai hoc khoi B nam 2011}

Ta c6: mki = mx - m

Cl

Cfiu 7: Cho m gam NaOH vao 2 lit dung djch NaHC03 nong do a mol/1, thu du^dc
= 1,4M ^

0,1

nN

= n _ = 1,8 mol = n

=> D a p a n B .

0,08

A. 10,56 gam

^

= — - — = 0,9 mol

Cr(0H)3 + OH- ^ Cr(OH); (tan)
0,3
0,3

iiKOH = 0,14 mol => [KOH] = —

Cfiu 5:

n(K.Na)

Cr^^+ 3 0 H - > C r ( 0 H ) 3
0,5

=>a = 0,06
CO2 + KOH

Trong m gam hon hdp Na, K c6 n(K, N-D = n
Trong 2m gam hon hdp Na, K c6

Ap dung djnh luat bao toan nguyen to C ta c6:

z:>0,l+

=31,95 gam

cr

31 95

Vay trong phan ifng CO2 vdi KOH ngoai muoi K2CO3 con c6 muoi KHCO3.

"c

=:> m

cr

0,12

Ta thay; n; = 0,12

"C (trongCOj)

_<=> (m + 31,95) = m + m

ci

BaCOji + 2KC1

0,12

D. 51,5 gam.

(Trich de thi tuyen sinh Cao dang nam 2011)

0,1

BaCl2 + KiCOs

C. 30,9 gam.

^=0: "BaC03 =

Hil"(Jng d§n giai
mol, n(,^,co3 = 0'07 mol

PhiTdng trinh phan u-ng; NaOH + NaHCOj

t> nw!

1
r •: ^

> Na2C03 + H2O (1)

'

Ap dung DLBT nguyen to C, so' mol C trong hai ket tua phai b^ng nhau. Ma
ta thay 0,06 mol ;^tO,07 mol => Sau phan ifng (1) NaHCOjCon dir. Vay trong
dung djch X CO HCO3" va C03^'.
* Khi cho dung djch BaCl2 (dir) vao X:
NaOH + NaHCOj
, 0,06 mol 0,06 mol

> Na2C03 + H2O
0,06 mol

-

.

V-'v-j:,t^/;.

u:0.y'

> vt'v'


Phuang phip

Ba^* +
^m
*

ky thugt giSi nhanh BTTN H6a dgi cuong - v6 co - Dg XuSn Hung

COj^"

BaCOji

0,06

0,06 mol

^/

(2)

x= —
2

= 0,06.2.40 = 4,8g
) Na2C03 + COst + H2O

0,2mol

> 'I

..

I,

(3)

0,1 mol

CaCl2 + Na2C03

.100% = 25%
•
...

Cachll
Chpn so' mol cua hon hdp la 1.
Gpi so' mol cua N2 la x, thi ciia H2 la 1 - x, so' mol N2 phan lirng la a

Khi cho dung djch CaCl2 (dif) vao X roi dun nong:
2NaHC03

= 0,25 mol => H = —
I

N2 + 3H2

>-CaC03 + 2NaCl

(4)

2NH3
,

|j

Ban dau:

a

1 - a

PhaniJng:

x

3x

2x

Ta c6: so mol Na2C03 (4) = so mol Na2C03 (1) + so mol NazCOs (3)

Sau phan ufng:

a-x

l-a-3x

2x

=> so mol NazCOj (3) = 0,7 - 0,6 = 0,1 mol

HonhdpX:

=> so mol NaHCOj trong 1 lit dd = so' mol NaHC03 (1) + so mol NaHC03 (3)

Hon hdp Y c6 so' mol la: a - x + 1 - a - 3x + 2x = I - 2x

0,7 mol

0,7 mol

: ;;

28a + 2(1 - a) = 1,8.4 =^ a = 0,2

my = ( 1 - 2 x ) 2 . 4 m a mx = m Y ( D L B T K L )

= 0,6 + 0,2 = 0,8 mol

=> a = 0,8.2/2 = 0,8 mol/1 => Dap an C .

( 1 - 2 x ) 2 . 4 = 1,8.4

Cfiu 8: Hon hdp khi X gom N2 va H2 c6 ti kho'i so vdi He hlng 1,8. Dun nong

Hieusua'tphanu'ng= ^ ^ x l O O = : 2 5 %
0,2

Y CO ti khoi so vdi He bkng 2. Hieu suat cua phan iJng tong hdp NH3 la

=:>DapanD.

B. 36%.

C. 40%.

D. 25%.

Htfdng d§n giai

0i

nong den khi phan iJng hoan toan, thu difdc 8,3 gam chat r^n. Khoi liTdng
A. 0,8 gam.

2

^

C. 2,0 gam.

'

D. 4,0 gam.

(Trich de thi tuyen sinh Dai hoc khoi A nam 2009)

Siir dung sd do dudng cheo ta c6:

H2:

'

B. 8,3 gam.

Taco: M x = 1,8.4 = 7,2; M Y = 2.4 = 8

HvCdng dSn giai
Ap dung dinh luat bao toan khoi liTdng:

^5,2
2

^7,2

; ^

CuO CO trong hon hdp ban dau la

Cach 1;

28 ^

. ,

Cfiu 9: Cho luong khi CO (dU") di qua 9,1 gam hon hdp gom CuO va AI2O3 nung

(Trich de thi tuyen sink Dai hoc khoi A nam 2010)

N2:

.

x = 0,05.

X mot thdi gian trong binh kin (c6 bpt Fe lam xiic tac), thu du'dc hon hdp khi
A. 50%.

,;fi;d.>

_

5,2
20,8

]_

4

0

n . i h : : .

8

mo= 9,1-8,3 = 0,8 (g) => n o = - V = 0.05 (mol)
16

20,8 ^

,

V

=> ncuo=no= 0,05 (mol) => mcuo= 0,05.80 = 4 (g)
=> Gia sur ban dau c6 1 mol N2 v^ 4 mol H2
=> mhh X =

+ m,^^ = 1.28 + 4.2 = 36 (g)

=^DapanD.
Cfiu 10: Cho 3,68 gam hon hdp gom A l va Zn tac dung vdi mot liTdng vifa du

Ap dung D L bao toan khoi lifdng ta c6:

dung djch H2SO4 10% thu diTdc 2,24 lit khi H2 (d dktc). Kho'i liTdng dung djch

mx = my => 36 = Hy. M Y => 36 = nY.8 => ny = 4,5 mol

thu difdc sau phan iJng la
A. 101,48 gam.
B. 101,68 gam.

Pu' : N2 + 3H2

2NH,

D. 88,20 gam.

(Trich de thi tuyen sinh Dai hoc khoi A nam 2009)

Di/a vao puf ta c6 :

Hifdng d§n giai

1 mol N2 phan drng thi sau phan iJng so' mol h6n hdp giam 4 - 2 = 2 mol
Vay X mol N2 phan uTng thi sau phan tifng so' mol hon hdp giam 5 - 4,5 = 0,5 mol

C. 97,80 gam.

Ta c6: n „ „r, = n „
H2SO4

H2

=^^=0,lmol
22,4

''~m

l>'.'.

r

•


PhLiong

phap va

ky thugt

g\i\h BTTN H6a dgi cuang - vO co - D5 Xuan HLfng

_ 9,8x100
mddH2S04 =

= 98 gam
10
Ap dung dinh luat bao toan khdi liTdng:

=^ nb = 0,01n-0,01 (2)

^

Matkhdc: Ma+(2M+]6n)b = 2,9 => Ma + 2Mb + 16nb = 2,9

nih6nh(?pKL+ Tl^jj j^^SO^ - mddsaiiphantfng + T l p , ^
=> m j d sau phan iJng=nih6n hdp K L + m^jj^^^Q^ -

i;

Va: a + 2b = 0,5.0,04 = 0,02 => na + 2nb = 0,02n

m H2SO4 = 0,1x98 = 9,8gam

The (2) vao (3) ta diTdc:

'

Ma + 2 M b + 16(0,01 n - 0 , 0 ! ) = 2,9 ^

m,j^

= 3,68 + 9 8 - 0 , 1 x 2 = 101,48 gam

'

(3)

=> M(a + 2b) + 0,16n = 3,06

,

Ma + 2Mb + 016n = 3,06

0,02M + 0,16n = 3,06 ^

M + 8 n = 153

=> Dap an B.
Cfiu 11: Nung nong m gam PbS ngoai khong khi sau mot thdi gian, thu dU'cJc h5n
hdp rdn (c6 chtfa mot oxit) nSng 0,95 m gam. Phan trSm khdi lUdng PbS da bj
do't chay la

1

2

3

M

145

137

129

=:> M la B a n (Ba) ^

A. 74,69%

B. 95,00%

C. 25,31 %

D. 64,68 %

(Trich de thi tuyen sinh Dai hoc khoi A nam 2009)
HuT^ng dSn giai
m(g)

n

Cfiu 13: Khur hoan toan mot oxit s^t X cT nhiet do cao can vijfa du V lit khi CO (d
dktc), sau phan fj^ng thu du"dc 0,84 gam F e va 0,02 mol khi CO2. Cong thtfc

Ap dung D L B T K L ta c6:

A. F e O v a 0,224

B . F e j O j va 0,448

C. Fe304 va 0,448

D. Fe304 va 0,224

m o = m - 0,95m = 0,05m (g) => no = 3,125.10"^m (mol)
Ta c6:

npbsphiinifng= npbo = n o

=> %PbS (da bi ddt chay) =

(Trich de thi tuyen sinh Cao ddn^ khdi A,B nam 2009)

= 3,125.10"^m (mol)
3,125.10"'m.239.100%

Hifdng dSn giai
Ta c6:

= 74,69%

m
Cfiu 12: Hoa tan hoan toan 2,9 gam hon hdp gom kim loai M va oxit cua no vao
nirdc, thu di/dc 500ml dung dich chiJa mot chat tan c6 nong dp 0,04M va
0,224 lit khi H2 (d dktc). Kim loai M la
B.Ba

C. K

= 0,02 mol => V^.^)^ = 0,448 lit

^

Goi cong thiJc tdng quat cua oxit la FcxOy:
=
0,02

= 1 =^ X la Fe,04 =>
4

Dapanc'

Cfiu 14: Nung nong 16,8 gam hon hdp Au, Ag, Cu, F e , Zn vdi mot liTdng du" khi
O2, den khi cac phan lirng xay ra hoan to^n, thu diTdc 23,2 gam cha't rdn X.

HuTdng d§n giai
M : amol

The tich dung dich HCl 2M vuTa du de phan iJng vdi cha't rdn X la

M 2 0 „ : b mol

A. 600ml.

2M + 2nH20 ^

B. 200ml.

2M(0H)„ + nHz
na

C. 800ml.

D. 400ml.

(Trich de thi tuyen sinh Cao ddn^ khdi A.B nam 2009)

.0 n

Hi^dng dSn giai

Y

Theo D L bao toan khdi liTdng :

M 2 0 n + n H 2 0 ^ 2M(0H)n

T a c o : n„^ =

n^Q^

V a : nFe= 0,84:56 = 0,015(mol)

THe: ^ = ^
y

D. Na

(Trich de thi tuyen sinh Dai hoc khoi B nam 2009)

b

=

ncophanung

Mat khac: no(,rongoxi.)= ncophantfng = n^^o^ = 0,02 mol

=> Dap an A .

A.Ca

^j,;

cua X va gia tri V Ian lu'dt la

0,95m (g) hh (PbO va PbS di/) + SO2

)

D a p an B.

niQ^ = 23,2-16,8 = 6,4 (g)

2b

^''
no = 6,4/16 = 0,4 mol

Phan urng cua HCl vdi chat riln X CO the diTdc bieu dien vdi sd do:
= 0,01

na = 0,02

(1)

O'0,4

+ 2H^
0,8

H2O
1,'


PhUdng phap va ky thujt giii nhanh BTTN H6a dgi cuong - v6 co - D5 Xuan Hung

(lit) = 4 0 0 (ml) => Dap an D.

VHCI = 0,8:2 = 0,4

,

Q^U

C S u 15: Cho 11,36 gam hSn hdp g6m Fe, FeO, Fe^O,, va Fe304 phin iJng he't vdi

dung djch

HNO3

loang du", thu

diTcfc 1,344

lit khi NO (san pham khi^ duy nha't)

d (dktc) va dung dich X. Co can dung djch X thu diTcfc m gam muo'i khan. Gia
trj cua m la:
A. 38,72

B. 35,50
•

•

i

Hrftfng d i n giai:
1,344

H N O 3 viTa

du, thu

NO- Gii trj cua
A. 0,04

dung djch X (chi chilfa

2 muo'i

mol CujS vao

axit

sunfat) va khi duy nha't

la:
B. 0,075

C. 0,12

^ D . 0,06.

Hrfdilg d§n giai
> Fe2(S04)3

Ta CO sddo: 2FeS2

• ^

,

0,06

0,12
CujS -

+HNO3

a

2CUSO4

2a

Sd do phan iJng:

Ap dung djnh luat bao toan nguyen to S:

Fe, FeO, Fe203, Fe304 + HNO3 ^ Fe(N03)3 + NO + H2O

0,12.2 + a = 0,06.3 + 2a -> a = 0,06 mol =^ Dap an D.
C S u 18: Nung hon hdp bot gom 15,2g Cr203 va m(g) A l d nhiet do cao. Sau khi

Gpi X la so mol Fe(N03)3

phan iJng hoan toan, thu du^dc 23,3g hon hdp ran X. Cho toan bp h6n hdp X

Ap dung djnh luat bao toan nguyen to N, ta c6:
"NCtrongHNOj) = " N (trong F e ( N 0 3 ) 3 ) + "N(trongNO) =

+ ^,06)

phan uTng vdi axit HCl diT thoat ra V(/) khi H2 (dktc). Gia trj cua V la:

mol

A. 4,48 lit
Diravkosc(d6tathay: n

a

(Trich de thi tuyen sinh Dai hoc khoi A nam 2007)

= 0,06 mol

22,4

a

diTdc

D. 34,36.

(Trich de thi tuyen sink Dai hoc khoi A nam 2008)

-''^•:v*^''''^
Ta c6: n NO

C. 49,09

17: Hoa tan hoan toan hon hcJp gom 0 , 1 2 mol FeSa va

=ln

B. 7,84 lit

C. 10,08 lit

D. 3,36 lit.

(Trich de thi tuyen sinh Dai hoc khoi A nam 2007)

= ^(3x + 0,06) = (l,5x + 0,03) mol

Hi^dng d i n giai
Ap dung dinh luat bao toan khoi iiTdng, ta c6:
Ta c6: n Cr203

mhh + m ^ ^ o ^ = mpg(^o^j^ + " 1 ^ 0 + m ^ ^ o

=> 11,36 + (3x + 0,06).63 = 242x + 0,06.30 + (1,5x + 0,03). 18
=> X

= 0,16 mol

'"feCNO^jj

15,2

= 0,1 mol

152
1«' k;-

Ap dung djnh luat bao toan kho'i liTdng:

= 0.16.242 = 38,72 (g) =^ Dap an A.

81

mAi= nihh- mcr203 = 2 3 3 - 15,2 = 8,1 (g)

HAI =

C S u 16: Hoa tan het 7,74 g hon hdp bpt Mg, AI blng 500ml dung djch hon hdp
HCl

I M va

H2SO4

0,28M thu diTdc dung djch X va 8,736 lit

H2

d (dktc). Co

Phanu-ng: 2A1 + Cr203
0,2

can dung dich X thu du^dc lu'dng muoi khan la:
A. 38,93 gam

B. 103,85 gam

C. 25,95 gam

0,1

.AI2O3: 0,1

,

. ,,

0,15

0,2
mj^^5o^ = mmuoi +

;|

Cr + 2HC1 - > CrCl2 + H2t

= 0'28.0,5 = 0,14 mol

+ mHci +

mbl

2A1 + 6HC1 - > 2AICI3 + 3H2t

0.1

=0,5.1 =0,5 mol

Ap dung dinh luat bao toan khoi Iifdng:
mhh

0,1

mol

Hon hdp X + dung djch HCl:

Taco: n^. - ^ - ^ =0,39 mol
"2
22.4

"H2SO4

0,2

03

Vay hon hdp X gom: . Cr: 0,2 mol

Hifitng d i n giai

nHci

+ AI2O3

=

fAl dir: 0,3-0,2 = 0,1 mol

D. 77,86 gam.

(Trich de thi tuyen sinh Cao dang khoi A,B nam 2008)

2Cr

27

m^^^

=> m , w i = 7,74 + 0,5.36,5 + 0,14.98 - 0,39.2 = 38,93 gam=> Dap an A.

=> n „

0,2
2

= 0,15+ 0,2 = 0,35 mol ^

V

H2

=7,84 lit

=> Ddp an B.
13


PhuOng

ph&p

va ky

thujt

g i i i ntianh

BTTIM

H6a dgi cuong -

vO c o -

D5 X u a n

Hung

Cau 4: Sue het mot liTdng khi clo vao dung dich hon hdp NaBr va Nal, dun nong thu
Cau 1: Hoa tan hoan to^n 2,8Ig hon hdp gom FcjOj, MgO, ZnO trong 500ni| diTdc 2,34g NaCl. So mol hon hdp NaBr va Nal da phan iirng la:
B. 0,15 mol
C. 0,02 mol
D. 0,04 mol.
dung djch H 2 S O 4 0,1M viTa du. Co can dung dich sau phan iJng thl thu diTOc A.O.lmol
HUdngdSngiai
bao nhieu gam muoi khan:
Ap dung djnh luat bao toan nguyen to Na ta c6:
A. 6,81g
B.4,81g
C. 3,81g
D. 5,81g.
2,34

(Trich de thi tuyen sink Dai hoc khoi A nam 2007j UNaBr + " N a l - i N a C l = 0,04 mol => Dap an D.
58,5
Hif(}ng dan giai
cau 5: Cho 16,3g hon hdp 2 kim loai Na va X tac dung het vdi HCl loang, dtf
Ta c6: n H , 0 ~ " H 2 S O 4 = 0,1.0,5 = 0,05 mol
thu di/dc 34,05g hon hdp muoi khan A. The tich H2 thu du'dc la bao nhieu lit?
Ap dung djnh luat bao toan khoi lU'dng: mhh + ITIH2SO4 ~ "^H^o
A. 3,36
B.5,6
c! 8,4
D. 11,2.
=^ m = 2,81 +0,05.98-(0,05.18) = 6,81 (g) ^ Dap an A.
HifcJng din giai
c1*
Cau 2: Cho 24,4g hon hdp NajCOj va K2CO3 tac dung vCfa du vdi dung dicli Ta c6: m„,u6'i = m k i m i o a i + m^^_
•1
BaCl2. Sau phan iJng thu du'dc 39,4g ke't tua. Loc tach ket tua, c6 can dun^
=:>m = 3 4 , 0 5 - 1 6 , 3 = 17,75 =^ n _ = 0,5 mol
djch thu di/dc m gam muoi clorua. Gia trj cua m la:
cr
ci
A. 2,66
B. 22,6
C. 26,6
D. 6,26.
Phu'dng trinh phan iJng:
'"' '
Hifdng din giai
2Na + 2HC1 > 2Na^ + 2 C r + H2
39,4

2X + 2nHCl
> 2X"^ + 2nCr + nH2
.
Ta c6: n'BaCl2
n.^, ~
= "n^B a C O j ~
= 0,2 mol

C. BAI TAP AP DgNG

Theo phu'dng trinh phan iJng ta c6: n^,^ =
= "O'S = 0,25 mol
Ap dung dinh luat bao toan khoi lU'dng: mhh + 'TiBaCi2 ~ " ^ k e t t i i a + m
= 5,6 (lit) Dap an B.
=> m = 24,4 + 0,2.208 - 39,4 = 26,6 gam => Dap an C.
Cau 3: Cho 0,52 gam hon hdp 2 kim loai Mg va Fe tan hoan toan trong dung
Cau 6: Hoa tan 10,14g hdp kim Cu, Mg, Al bing mot liTdng vHa du dung dich
djch H 2 S O 4 loang, dU' thay c6 0,336 lit khi thoat ra (dktc). Khoi lifdng hon hdp
HCl thu diTdc 7,84 lit khi A (dktc) va l,54g chat ran B va dung djch C. Co can
muoi sunfat khan thu dufdc la:
dung djch C thu dufdc m gam muoK Gia tri cua m la:
A. 2 gam
B. 2,4 gam
C. 3,92 gam
D. 1,96 gam.
A. 33,45
B. 33,25
C. 32,99
D. 35,58.

Hifdng dan giai

Ta CO muo'i thu du'dc gom MgS04 va FeS04.
Theo dinh luat bao toan kho'i liTdng: m„u,^i = m k i , „ i o a i + m , (1)
Tac6 :n„ =

so|

=0,015 mol

22,4
PhiTdngtrinhphanurng: Mg +

H2SO4

> Mg^* + 804^" + H z t

Fe +

H2SO4

> Fe^* +

"2

804^"

Theo phUdng trinh phan tfng ta c6: " ^ ^ 2 - ~ " H 2 ~
mol
Tfif (1) = > m n , u ^ i = 0,52 + 0,015.96 = 1,96 gam => Dap an D

+ H2t

Hifdng d§n giai

7,84 = 0,35 mol ^ n =2n,. =2.0,35 = 0,7 mol
Taco n„ = —
"2
22,4
cr
"2
Theo djnh luat bao toan kho'i liTdng:
m = m(Ai + M g ) + m^^_ = (10,14 - 1,54) + 0,7.35,5 = 33,45 gam

=> Dap an A.

cau 7: Hoa tan 28,4g hon hdp 2 muo'i cacbonat cua 2 kim loai thuoc nhom lA
bang axit HCl thu diTdc 6,72 lit khi (dktc) va dung djch A. Tong kho'i li^dng 2
muoi clorua trong dung dich thu diTdc la:
A. 3,17
B. 31,7
C. 1,37
D. 7,13.
15


Phuong phap va ky thujt g'A't nhanh B T T N H6a d^i cuong - vfl cO - D 5 XuSn Hmg

Hrfdng dan giai

Hifdng din giai

Goi cong thiJc chung cua 2 muoi cacbonat kim loai nhom lA la R2CO3

.f

2RCI + CO2 + H2O

R2CO3 +2HC1
Ta c6: n H 2 0 ~ " c O i

'2

6,72
22,4

= 0,3

mol;

DHCI = 2 n(.Q

= 0,6

T a c 6 : ng^ci^-Hg^so^)j

mol

ri2>.

= ^8,2 + 0,3.208 - 69,9 = 30,7 (g)
, f | iJ's./Jtv.;;!

CSu 11: Cho 6,2g hon hdp gom mot s6' kim loai kiem vao dung djch HCl du" thu

=>m = 28,4 + 0,6.36,5 - (0,3.44 + 0,3.18) = 31,7 (g) => Dap an B.

diTdc 2,24 lit H2 (dktc). Co can dung djch thu diTdc sau phan ilng se thu diTdc

CSu 8: Tron 5,4 gam nhom vdi 6,0 gam FcaOs roi nung nong de tht/c hien phan

bao nhieu gam chat ran:
A. 1,33
B.3,13

iJng nhiet nhom. Sau phan ufng ta thu difdc m gam hon hdp chat ran. Gia tri
cua m la:
B. 9,40 gam

Ap dung djnh luat bao to^n khoiJurdng: mhSnhdp + mBaCi2 ~

=>DapanB,

nihh + niHci = m + m , , ^ + m^^ Q

A. 2,24 gam

1

= 0,3 mol

=> m = mnan H^P + ^n.a, ' n^-'.

Ap dung djnh luat bao loan khoi liTcJng ta c6:
J2

69,9

C. 10,20 gam

C. 13,3

D.3,31.

Hvldng d§n giai:

D. 11,40 gam.

)1

Htf^ng d i n giai

2,24
Taco nH^=22^^ = 0,1 mol

Theo djnh luat bao toan khoi lufdng
Ta c6: mnn sa., = mnn tnrdc = 5,4 + 6,0 = 11,4

Ma: n

gam

CSu 9: Thoi mot luong khi CO dU' qua o'ng siJ di/ng m gam hon hdp gom CuO,

=2nH = 2.0,1 = 0,2 mol

=> m„.5-i = m k i „ „ o , i + ni^^_ = 6,2 + 0,2.35,5 = 13,3 (g) =^ Dap an C .

Fe203, FeO, AI2O3 nung nong thu diTdc 2,5g chat ran. Toan bp khi thoat ra sue
vao nu'dc voi trong diT tha'y c6 15 gam ket tua trang. Khoi liTdng cua hon hdp
oxit kim loai ban dau la:
A. 7,4 gam

B. 4,9 gam

C. 9,8 gam

dung djch H2SO4 loang thu diTdc V lit khi d (dktc) va 7,48g muoi sunfat khan.
Gia trj cua V la:

D. 23 gam.

B. 1,008

+ yCO — - — > x M + yC02

Ca(0H)2 + C O 2 - > CaCOj + H2O

Hu

^C02~

"CaCO,

"77^" ^''^
lUU

ma: moxit= mki,„ioai + m o x i
moxit=

;

.>

•

2,5+ 0,15.16 = 4,9 gam

C&u 10: Mot dung dich chiJa 38,2g hon hdp 2 muoi sunfat ciia kim loai kiem A
va kim loai kiem tho B tac dung viTa du vdi dung dich BaCl2 thu duTdc 69,9g
ket tua. Lpc b6 ke't tua va c6 can dung djch sau phan tfng thu du'dc bao nhieu
gam muoi khan:

: ^ ©: 30X- '••''''^ ; C. 7,03

'

=n

H2 + SO^

5,76
,2J4

so

96

= 0,06 mol
'

V

= 1,344 (lit)
H2

=> Dap an A.
Cfiu 13: Cho 2,81 gam h§n hdiJ A g6m 3 oxit Fe203, MgO, ZnO tan viTa du trong
300ml dung djch H2SO4 0,1M. Co can dung djch sau phan ijTng, kho'i liTdng hon

=> Dap an B . •

A. 3,07.; f ,

D. 3.36.

,2-

Ta c6 sd do: H2SO4

Theo djnh luat bao toan nguyen to ta c6:
=nco =

C. 1,12

Htfdng dSn giai:
m ,_ = 7 , 4 8 - 1,72 = 5,76 (g) ;
Ta c6: mn,u6'i= mki,„ioai + m
so
so,2-

Cac phiTdng tnnh h6a hoc:

no(trongoxit)

^,

A. 1,344

Hxidng d§n giai

MxOy

Cfiu 12: Hoa tan het 1,72 gam hon hdp kim loai gom Mg, A l , Zn va Fe b^ng

D. 70,3.

hdp cac muoi sunfat khan tao ra la:
A. 3,81 gam

B. 4,81 gam

C. 5,21 gam

D. 4,8 gam.

Hi^dng dSn giai
dung djnh luat bao tohn kho'i lifdng: moxii+ ni^^so^ ="^H20 +
mn„,fi-i
= moxii + mH 2 S O 4
'muoi =

"^HjO

THl/ VIEN TINH B!NH THUAN


Phiiang ph^p va ky thugt giSi nhanh BTTN H6a dgi

CLiong

- vO

CO -

B5 Xuan Hung

Cfiu 17: Sue k h i clo vao dung dich NaBr va N a l den phan tfng hoan toan ta thu^

Trong do: nj.,^o = n^^^Q^ = 0,3.0,1 = 0,03 m o l

Vay:

mn,u6-i=

(Ji/dc l , 1 7 g N a C l . X a c djnh so m o l hon hdp NaBr va N a l c6 trong dung djch

2,81 + 0 , 0 3 . 9 8 - 0 , 0 3 . 1 8 = 5,21 gam =:> D a p a n C .

..ban dau?

Cfiu 14: Hoa tan het lOg hon hcJp muoi cacbonat M g C O j , CaCOj, NaaCOs, K 2 C O .

siA.O.lmol

b^ng dung djch H C l duf thu dugc 2,24 lit khi (dktc) va dung dich Y . Co can dung
B. 11,1

C. 11,8

2 24

n N a B r + " N a l = HNaci =

»'v«> ''ifv

T a c 6 : n H c i = 2 n c o 2 = 2 . ^ = 0,2mol;

n^^Q = n^o^ = 0 , 1 m o l

A p dung dinh luat bao toan khoi lu'dng: mhh + mHci = m + m^^Q^ + m^^^

m = 1 0 + 0 , 2 . 3 6 , 5 - ( 0 , 1 . 4 4 + 0,1.18)= 11,1 (g) =:> D a p a n B .

C. 13,2g

'

1. G i a t r i c u a V i a :

^ '»•'(!
B . 0,112 l i t

C. 5,6 l i t

D . 0,224 l i t .

B . 15,8 gam

C. 2,54 gam

D . 25,4 gam.

2. Gia trj cua m l a :
A . 1,58 gam

Htfdng dan giai
1. Ta nhan thay, khi k i m loai tac dung vdi oxi va H2SO4, so mol
n„ = n

Trong do: mo =

Sau phan ^ng k h o i lu'dng trong ong suT giam 5,6g. Co can dung djch A thu
difdc m(g) muo'i. Gia trj cua m l a :

"H, = n

D . 39,65.

^

Hvtdng d§n giai
Kh6'i liTdng dng si? giam chinh la k h o i liTcJng ciia nguyen to o x i

~

" H O O ~ " o (trong oxit) =

^

= 0,35

mol

= SO]' hay

=nr,

T

dung dich A . L i / d n g k h i H2 tao thanh dan vao o'ng stj" difng CuO dU' nung n6ng.

" H ,

H2 (dktc).

Co can dung djch thu diTdc m gam muoi khan.
A . 2,24 l i t

Cfiu 16: H o a tan 14,8g hon hdp A l , Fe, Z n bang dung dich H C l viTa du thu diTdc

Ap dung djnh luat bao to^n nguyen to' ta c6:

D.

Phan 2: T a n hoan toan trong dung djch H2SO4 loang thu diTcJc V l i t

i::>m = 1 1 , 5 + 0 , 2 . 3 6 , 5 - ( 0 , 1 . 4 4 + 0,1.18) = 12,6 (g) => D a p a n B .

C. 27,575

an

-

"H2O ~ " c O j ~

B.28,8

Dap

Phan 1: B j o x i hoa hoan toan thu diTdc 0,78gam hon h d p oxit. " ~

D . 12,3g.

A p dung djnh luat bao toan khoi lufdng: mnn + mHci = m + m^^^^ + m^^^Q

A . 20,6

=>

-J

-

Hifdng d i n giai

Ta c6: nHci = ^^co2 ~ ^'^

mol

,

b i n g nhau:

thu diTdc 2,24 lit CO2 (dktc). Khoi li/cJng muoi clorua tao thanh la:
B. 12,6g

= 0,02

56,5

, .

Cfiu 18: Chia 1,24 gam hon hcJp hai k i m loai c6 hoa trj khong do'i thanh 2 phan

Cfiu 15: Cho 1 l , 5 g hon hcfp gom ACO3, B2CO3, R2CO3 tan het trong dung djch HCl
A . 16,2g

D . 0,02 m o l .
' ' i . '

A p dung djnh luat bao toan nguyen to' natri, ta c6:

D . 14,2.

Hifdng dSn giai
ur;

C. 0,015 m o l

JHrfdng dSn giai

djch Y thu diTdc x gam muoi khan. Gia trj cua x la:
A . 12

B . 0,15 mol

moxit - m i c i n , i o a i

2- = " o = ^

SO4

1 24
= 0,78 - ^
= 0,16gam.

g.^l.l -

= 0,01 m o l => V = 0,01.22,4 = 0,224 l i t .

16

=> D a p a n D .

0

^-

nimuoi

= mi(in, loai + m

,
so|

1 24
= - — + 0,01.96 = 1,58 gam => D a p a n A ,
2

Cfiu 19: H o a tan hoan toan .10 gam hon hdp M g v a Fe trong dung djch H C l dir

thay tao ra 2,24 l i t k h i hidro (dkTc). Co can dung djch sau phan ilng thu diTdc

=> n

= 2 n„

= 2.0,35 = 0,7 mol

muoi khan. K h o i liTcJng m u o i khan thu difdc l a :
A. 1.71 gam

B . 17,1 gam

D . 34,2 gam.

Hifdng dSn giai

=> m„„=> D a p &n D .

C. 3,42 gam

Tac6:n

= 2 n „ =2.( —
cr

"2

) = 0,2mol

'22,4

,


Phuong ph^p va ky thugt giSi nhanh BTTN H6a dji cuong - vO co - D 5 XuSn Hung

Ma:

16 + 28.0,3 = mpe + 0,3.44 ^ mpe = 1 1 , 2 gam

+ m^^_ = 10 + 0,2.35,5 =17,1 gam=> Dap an B,

m,„ua-i = mki,„ioai

Dap an D.

CSu 20: Hoa tan hoan toan 20 gam hon hcJp gom Mg va Fe vao dung dich axit
HCl dirthafy c6 11,2 lit khi (dktc) thoat ra va dung djch X. Co can dung djch X
B. 45,5 gam

C. 55,5 gam

'r«,KB - :u
^ ^""2

mn,u6i = n i k i o a i + m n c i - m h i d r o

.(:m^

'..

mnci

a. Khoi liWng cua Fe^Oy va A l trong X la:^,^ cMa Wlii'm gnon
gmb r;

A. 6,96g va 2,7g

•^•^'^ ^ ^ "^"^

Ap dung djnh luat hio toan khoi lUOng: m i d +

,5,.; ^

D loc ket tiiava nung den kho'i lUWng khong doi diTcfc 5,1 g chat r^n.

• ^' " ^

""2 ^22 4 " ^'^ ^

nhiet nhom vdi 9,66 gam hon hdp X gom Fe^Oy va

dung djch D 0,672 lit khi (dktc) va chat khong tan Z. Sue COaden diT vao dung dich

D. 65,5 gam.

Hxidng dSn giai

112

ufhg

nhom, thu dUdc hon hdp ran Y. Cho Y tac dung vdi dung dich NaOH diT, thu diTdc

thi khoi li/dng muoi khan thu diTdc la:
A. 35,5 gam

CSu 23: Thirc hien phan

=

C. 2,52gva7,14g

m„,u6-i+ mhidro

D. 4,26g va 5,4g

,ii .^.i . :

b. CongthiJccuaoxitsatla:

=20+1.36,5 - 2.0,5 = 55,5 gam.

.:r,fiv',

A.FeO

£>ap an C,,^,|j |^^.;

:

B. 5,04g v^ 4,62g

^. ,

•

B. Fe203

C. Fe304

D. Khong xac djnh du'dc

Cfiu 21: Hoa tan het 38,60 gam gom Fe va kim loai M trong dung djch HCl du'

Hi/dng d i n giai

' '

thay thoat ra 14,56 lit Ha (dktc). Kho'i liTdng hon hdp muoi clorua khan thu
a. 2yAl + 3 Fe^Oy

di/dc la:
A. 48,75 gam

B. 84,75 gam

C. 74,85 gam

D. 78,45 gam.

yAljO, + 3xFe

(1)

3

Al + NaOH + H2O

r'

> NaA102 + - H 2

n i

.uf.^on

>

Hrfdng dan giai
Ta

c6:

m,„ucri = m k i , „ i o a i +

m

_

ci
Trong do: n^^_ = nHa = 2nj^^ =

'^^

0,02

0,02

NaAlOz + CO2 + 2H2O

> A1(0H)3 + NaHCOj

AI2O3 -H 3H20

2Ai(OH)3

Do do:

UAJ(band^u)= 2 n
AI2O3

khi sau phan iJug qua dung djch Ca(0H)2 diT, thay tao ra 30 gam ket tua. Kho'i
lUUng s^t thu duUc la:

^

? , . « ~

C. 9,6 gam

rt

M

r,nco=

nAi(bandiiu)=2n

DK.*.

xFe + yCOz (1)

=0,4 mol " -''^

~'

^ 0 (trong F e , 0 y

= —

= 0,3 mol ^

= 0,3 mol

mol =>

m^

= 0,1. 27 = 2,7

'

gam

.

=0,1. 27 = 2,7 gam

mAi

102
n A i ( i ) = n A i ( b a n d 5 u ) - n A i ( 2 ) = 0,1 -0,02

= 0,08

mol

Theo djnh luat bao toan khoi liTdng nguyen to' oxi, ta c6:

^ ' V - ' -

CO2 + Ca(OH)2 ^ CaCOa + H2O
CaC03

= 2. —
= 0,1
102

= 2 . — = 0 , 1 mol

"Ai(band5u)= n A i ( i ) + nAi(2) =>

22,4

n

"^W,;"^^

'

•"Fe o = 9.66 - 2,7 = 6,96 gam => Dap an A.

AI2O3

^

(3) ,m

D. 11,2 gam.

Hufdng d i n giai
yCO + FexOy —

9

Nhan xet: Ta't ca liTdng Al ban dau chuyen het ve AI2O3 ( 4 ) .

CSu 22: Thdi 8,96 lit CO (dktc) qua 16 gam FcxOy nung nong. Dan toan bo liTdng

B. 6,4 gam

v

= 1,3 mol

=^m = 38,6 + 1,3.35,5 = 84,75 gam => Ddp an B.

A. 9,2 gam

0,03

i

^nl A

„
6,96-0,12.16
,
npe =
=0,09 mol

Theo dinh luai bao toan Idio'i Ii/dng c6: mp^ Q + mco = m p e + ^002

1,5.0,08 = 0,12 mol
s-

'

^^

n c-(,./«/V.. ^
t •

56

100

=> nco > n^Q^ -> CO diT va Fe^Oy het.

trong A l 2 0 3 ) =

"Fe: no = 0,09 : 0,12 = 3 : 4 => CTPT la Fe304 => Dap an C. ' '
Cfiu 24: Khur hoan toan 32g hon hdp CuO va Fe203 bang khi H2dirthay tao ra 9g
H"20. Kho'i liTdng hon hdp kim loai thu diTdc la:
A. 12 gam

B. 16 gam

C. 24 gam

D. 26 gam.


Phuang ph^p

ky thujt giai nhanh BTTN H6a dgi cuang - vO cO - D 8 Xufln Hung

Hrftfng d i n giai

27: Hai binh c6 the tich b^ng nhau, nap oxi vao binh thiJ nha't, nap oxi da

Vi H2 lay oxi cua oxit kim loai tao thanh H2O

(ji/dc ozon hoa vao binh thuf hai, thay khoi liTdng 2 binh khac nhau 0,42 (g)

9
Nen

ta

c6:

no(,rongoxio =

"HIO =

(nhiet do va dp suat d 2 binh nhiT nhau). Khoi lu'dng oxi da difdc ozon hoa la:

77"

= ^'^

^

l'l6g

B. l,26g

18

mo = 0,5. 16 = 8 gam =>

mk:m,ioai=

C. l,36g

D.2,26g.

HUdngdSngiai

32 - 8 = 24 gam=> Dap an C.

' | •)

KhS'i lu'dng khac nhau d 2 binh'la do kho'i liTdng oxi trong ozon:

Cflu 25: Thoi mot luong khi CO du" di qua ong diTng hon hdp 2 oxit Fe304 va
CuO nung nong den khi phan i?ng xay ra hoan toan thu du'dc 2,32g hon hdp

=>

"ocrongoj)

^^^^^^^^ ^.^.MismimrL'^'--^^

,6

kim loai. Khi thodt ra du'dc du'a vao binh diTng dung djch Ca(OH)2 diT thay c6
5 gam ket tiia tr^ng. Khoi lifdng hon hdp 2 oxit kim loai ban dau la:
A. 3,12 gam

B. 3,21 gam

C. 4 gam

D. 4,2 gam.

Hrftfng d§n giai
Cdc phan ifng: Fe304 + 4 C O
3Fe + 4CO2

dung dich X c6 pH = 13. Co can dung djch X du'dc m gam chat r^n. Gia tri
cua m la:

CO2 + Ca(0H)2 ->• CaCOj + H2O

A. 4,02 gam

CO lay oxi trong oxit -> CO2
no (trong oxit)

=

Hco =

m o x i , = miciniioai

^C02~

( H ozon hoa) = 0,039375.32 = 1,26 (g) =^ Dap an B.
Cfiu 28: Cho 2,22 gam hon hdp kim loai gom K, Na va Ba vao nU'dc du'dc 500ml

Cu + CO2

CuO + CO

B. 3,45 gam

C. 3,07 gam

D. 3,05 gam

Hxidng d i n giai

"CaCO^

+ Hioxi trong oxit

= 0>05 mol

-1) ir .

Ta c6: U Q (bi ozon hoa) = - n ^ = - .0,02625 = 0,039375 mol

^

= 2,32 + 0,05.16 = 3,12 gam => Dap an A.

C&u 26: Cho 7,8g hon hdp 2 kim loai Mg va A l tac dung vdi dung djch H2SO4
loang du". Khi phan i?ng ke't thiic, thay khoi lu^dng dung djch tang 7g. Khoi

'f

T a c 6 : p H = 1 3 - > p O H = 1 4 - 1 3 = 1 =^ [ O H - ] = 0 , 1 M

^

=> n

= 0,1.0,5 = 0,05 mol
on
Ap dung djnh luat bao toan khoi lu'dng:
Ta c6:

mbaz<, =

nikin.ioai

liTdng moi kim loai trong hon hdp ban dau la:

+ m

° '''
•'

= 2,22 + 0,05.17 = 3,07 (g) => Dap an C .
OH

A. 2,4g Mg va 5,4g Al

B. 2,4g Mg v^ 4,5g Al

C. 4'2g Mg va 5,4g Al

Cfiu 29: Cho m (g) hon hdp 3 kim loai Fe, Al, Cu vao mot binh kin chiJa 0,9 mol

D. 4,3g Mg va 5,6g A l

oxi. Nung nong binh mot thdi gian cho den khi so mol oxi trong binh chi con

H\i6ng d i n giai
Ap dung djnh luat bao toan kho'i lu'dng:
mhh+ r"ddH2S04

-'"ddsauptr

0,865 mol va chaft ran trong binh c6 khoi liTdng 2,12g. Gid tri cua m la:
A. Ig

B. l . l g

C.2g

, ,

D.2,lg.

Hiring d i n giai

+

,

So mol oxi ke't hdp vdi kim loai de tao thanh oxit:

,

^ O j = 0,9 - 0,865 = 0,035 mol
7

= 7,8 - m

Ap dung djnh luat bao tohn khoi liTdng: m + 0,035.32 = 2,12 => m = 1 (g)

=> m,^^ = 0,8 (g) => n^^^ = 0,4 mol

Ta

CO

Bap an A.

24x + 27y = 7,8

x = 0,l

•"Mg=2,4(g)

x + l,5y = 0,4

y = o.2

m^, =5,4(g)

he: <

Dap an A.

'
.

, , . ,

Cfiu 30: Nhiet phan hoan toan 9,8g hidroxit kim loai hoa trj II khong ddi thu
^^(ic hdi ni/dc va 8 (g) chat r^n. Hidroxit do la:
AFe(OH)2

B.Zn(0H)2

C. Mg(0H)2

D. Cu(OH)2.

Hrftfng d§n giai
Ap dung djnh luat bao toan khoi liTdng:

m h i d r o x i t = iTicha^,ri„ +

m^^o
01


Phuang ph^p va ky thujt giii nhanh BTTN H6a dgi cuong - vO co - D5 XuSn Hung
=>

"IH^O

= 9,8 - 8 = 1,8 (g) => n^^o = 0,1 mol

R(OH)2

-> R O + H2O

Theophanu-ngd):

0,1
9,8
MR(OH)2 = ^

^

-m^c,

„^ = m ^ p ,

^rr,

= 3 8 , 7 4 - 8 , 9 4 = 29,8 gam.

29 8
m K c , 0 3 = : ^ x ' 2 2 , 5 = 49gam.

_f9><100^33^^^^^^^

= 9 8 - > M R = 64 - > R la Cu => D a p a n D .

33

%mKCi03(A)

#r

•

M<

I M

53

C § u 3 1 : Hon hdp A gom KCIO3, Ca(C102)2, Ca(C103)2, CaCb va K C l nang 83,68

C&u 32: Hoa tan 3,28 gam hon hdp m u o i M g C h va Cu(N03)2 vao niTdc diTdc

gam. Nhiet phan hoan toan A ta thu di/dc chat ran B gom CaCb, K C l va 17,472

dung dich A . N h i i n g vao dung djch A mot thanh sat. Sau mot khoang thdi gian

lit k h i (d dktc). Cho chat ran B tac dung vdi 360ml dung djch K2CO3 0,5M (viTa

la'y thanh sat ra can lai thafy tang them 0,8 gam. Co can dung djch sau phan

du) thu di/dc ket tua C va dung dich D . LiTdng K C l trong dung djch D nhieu gap

i?ng thu di/dc m gam m u o i khan. Gid trj m la

22/3 Ian lifdng K C l c6 trong A . % khoi liTdng KCIO3 c6 trong A la

A . 4,24 gam.

A . 47,83%.

B. 56,72%.

C. 54,67%.

KC103
Ca(C103)2
83,68 gam A
1

I

Ca(C102)2

r7.

'°

(1)

> C a C l j + 3O2

(2)

CaCl2 + 2O2

(3)

khoi lUdng cua thanh Fe bang dp giam khoi liTdng cua dung djch m u o i . Do
do: m = 3,28 - 0,8 = 2,48 gam.
rt«

CSu 33: T h d i tiT tif V l i t hon hdp k h i (dktc) gom CO va H2 di qua mot ong diTng
— ^

16,8 gam hon hdp 3 oxit: CuO, Fe304, AI2O3 nung nong, phan uTng hoan toan.

CaCl2

CaCl2

Sau phan uTng thu di/dc m gam chaft r^n, mot hon hdp k h i va hdi nang hdn

KC1(^,

KCl

kho'i liTdng cua h6n hdp V la 0,32 gam. Tinh V va m .

(A)

A. 0,224 l i t va 14,48 gam.

B. 0,448 l i t va 18,46 gam.

C. 0,112 l i t va 12,28 gam.

D . 0,448 l i t va 16,48 gam.
HuTdng d§n g i a i

=> mn = 83,68 - 32x0,78 = 58,72 gam.

CaCl2 + K2CO3
H6nhdpBJ0,18

m KCl
=> m

(B)

(D)

< - 0,18

— ^

1^-'

= m g - mc,ci2

(4)

0,36 mol
KCl

(B)

CO + O

>

H2 + O

>

<-

H2O.

;() )

(D)

22

•„ •

0 32

mo = 0,32 gam. =>

UQ =

^— = 0,02

mol

,

r^tiiH

«

'

=> ( " c o +

"H

mol.

=> 16,8 = m + 0,32

moxit

.

Dap

an

D.

,/

= nich^, ran + 0,32

=> m = 16,48 gam.

^ - \ h r r m i i > = 0 ' 0 2 x 2 2 , 4 = 0,448 l i t
• iici ihuimih...

idff"

*

'^P dyng djnh luat bao toan k h o i l u p n g ta c6:

X 65,56 = 8,94 gam

ov

CO2

hon hdp D

(B)

= 58,72 - 0 , 1 8 X 1 1 1 = 38,74 gam

3

u

•wpng ciia nguyen tir oxi trong cac oxit tham gia phan ung. D o vay:

= "1 KCl (B) + "1 KCl (p,4) = 38,74 + 0,36 X 74,5 = 65,56 gam
_

"

Kh6i lupng hdn hop k h i tao th^nh ndng hon hon hop k h i ban dau chinh la khoi

C a C O , ; + 2KC1
^

•!<" ^ J ' -

Thyc c h i t phan ung khOr cac oxit tren la

Son.

Cho chat ran B tac dung vdi 0,18 mol K2CO3

KCl

'

=»DapanB.

A p dung djnh luat bao toan khoi liTdng ta c6: m A = me + m^^

m

D . 1,49 gam.

A p dung djnh luat bao toan khoi liTdng: Sau mot khoang thdi gian dp tang

3
KCl + - 0 ,
2 2

nQ^ = 0 , 7 8 m o l .

KCl

C. 4,13 gam.

Hxidng d a n g i a i

Hi/(}ng d§n g i a i
t"

B. 2,48 gam.

D . 58,55%.

^ . ^ ^ ^ ^^^^

^.^^^.^ ^^^y.

>
_^

_

^ . ,^

-


Phuang ph^p

ky thugt giSi nhanh BTTN H6a

cuang - v6 ca - D5 Xuan Hi;ng

Cfiu 34: Thoi ra't cham 2,24 lit (dktc) mot hon hdp khi gom CO va H2qua mot

36- Cho mot luong CO di qua ong si? difng 0,04 mol hon hdp A gom FeO
ong siJ dirng hon hdp AI2O3, CuO, Fe304, FezOj c6 khoi lU'dng la 24 gam dir C&»
^.-^ ^^^g s^u khi ket thiic thi nghiem thu diTdc B gom 4 chat nSng
dang diTdc dun nong. Sau khi ket thuc phan u-ng khoi liTdng chat rSn con lai
^o/gam. Khi di ra khoi ong stf cho hap thu vao dung djch Ba(OH)2 dir thi
4,784 gam.

trong ong su" la:
B. 11,2 gam.

A. 22,4 gam.

C. 20,8 gam.

thu difdc 9,062 gam ket tua. Phan tram khoi liTdng Fe203 trong hon hdp A l a

D. 16,8 gam.

A"86,96%'

B . 16,04%.

Hifdng d i n giai

0,04 mol hSn hdp A (FeO va FejO,) + CO ^

^'^^• = 0,1 mol
22,4

Vay:

> CO2

H2 + O

> H2O.

diTng m g a m h o n hdp X nung n o n g . Sau khi ke't thuc thi nghiem thu diTdc 64

20,4. Tinh g i a trj m.
B. 35,2 g a m .

C. 70,4 gam.

D. 140,8 gam.

Hrfdng dfin giai

fx =0,01 mol

72x + 160y = 5,52

[y = 0,03 mol

—

0,01x72x101

,^^.0,'

= 13,04%

, : i f d i it.

%Fe203 = 86,96%

-> 2Fe304 + C02

-> 3FeO + C02
Fe + C02

Ipc bo ket tua diTdc dung dich X. Tiep tuc cho 50 gam dung djch H2SO4 9,8%
toan. Nong do % cua dung dich Na2C03 va khoi li/dng dung djch thu du'dc sau

(1)
(2)

cung la:
A. 8,15% va 198,27 gam.

B. 7,42% va 189,27 gam.

C. 6,65% va 212,5 gam.

D. 7,42% va 286,72 gam.
Hi^dng d§n giai

(3)

khong quan trong va viec can bang cac phu'dng trinh tren cung khong can thiet,
quan trong la so mol CO phan iJug bao gicf cung bhng so mol C O 2 tao thanh.

"Bacij = 0,05 m o l ; n^^^o^ = 0,05 mol
BaCl2 + Na2C03 - > BaCOj i

0-05
•=^0,5 mol.

Goi X l a so m o l c i i a C O 2 ta c6 phuTdng t r i n h v e kho'i liTdng c i i a B:

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

0,05

0,05 .

0,1

Dung dich B + H2SO4 - > khi => dung dich B c6 Na2C03 diT
NazCOs
+ H2SO4 -> Na2S04 + C02t + H2O
0,02
0,02
i I ^

Theo D L B T K L ta c6: m x + m c o = m A + rn CO,

"Na^coj bandiu = 0,05 + 0,02 = 0,07 mol

X

44 - 0,4 x 28 = 70,4 gam => Dap an C

Na 2C03

0,07x106
100

'

+ 2NaCl

n h a n dUdc x = 0,4 m o l va do c u n g c h i n h la so m o l CO t h a m g i a p h a n lifng.

• m = 64 + 0,4

Dap an A

vao dung djch X thay ra 0,448 lit khi (dktc). Biet cac phan iJ-ng xay ra hoan

' Nhir vay chat dn A c6 the gom 3 chat Fe, FeO, Fe304 hoac it hdn, dilu do

22,5

,1! jU

Cfiu 37: Cho 50 gam dung dich BaCh 20,8 % vao 100 gam dung djch Na2C03,

Cdc phan ufng khuf sat oxit CO the c6:

11,2

x + y = 0,04

•%mFeo =

g a m chat r a n A trong o n g suT va 11,2 lit khi B (dktc) c6 t i khoi so vdi H2 la

-

^

Bkt npeo = X mol, np^^^^ = y mol trong hon hdp B ta c6:

Cfiu 35: Hon hdp X g o m Fe, FeO va Fe203. Cho mot luong CO di qua o n g su"

FeO + CO

4,784 gam hSn hdp B + CO2.

= > m A = 4,784 + 0,046x44 - 0,046x28 = 5,52 gam.

mo = 1,6 gam.

=> Dap an A.

Fe304 + CO

.

nc02

Khoi lu'dng chat ran c o n lai trong ong su la: 24 - 1,6 = 22,4 gam.

BFezOj + CO

.

.

Ap dung dinh luat bao toan khoi liTdng ta c6: m A + mco = me + m^^Q^

= n^^^ + nj^^ = 0,1 mol.

A. 105,6 g a m .

*

CO2 + Ba(OH)2d.
> BaCOji + H2O
= 0,046 mol
= nBaC03 =0.046 mol va n^o
Thirc chat phan ung khOr cac oxit la:
CO + O

: £;D.6,01%.

Hi^ngdSngiai

2 24

Ta c6: n hh(CO+H2)

C. 13,04%.

xlOO% =7,42%

•m.umo.


Phuong

phap

ky thujt giSi nhanh B T T N H6a

DLBTKL:

itidd sa., cOng

•t

a?!

cuong - vO ca - B 5 Xuan HtJng

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

Dap an B
Cfiu 38: Khijf het m gam Fe304 bkng CO thu diTcJc hon h(?p A gom FeO va Fe. A

^gu 40'
^ ^""^
b^ng oxi du" thu diTdc 44,6 gam hon hdp
fiCl thu di/dc dung dich D. Co can dung
^ 9 9 , 6 gam.

B. 49,8 gam.

*^*-''Htfdng d i n giai

tan vuTa du trong 0,3 lit dung dich H 2 S O 4 I M cho ra 4,48 lit khi (dktc). Tinh m?
A. 23,2 gam.

B. 46,4 gam.

C. 11,2 gam.

Goi M la kim loai dai dien cho ba kim loai tren vdi hoa tri la n. >' ' fcfi

D. 16,04 gam.

M+
Fe304

->

(FeO,Fe)

^

3Fe^*

^ •
iMfljO')*''.

' n mol

•••fott'i •fJ^'G.ti"-

^^I'l'O '

• Ap dung djnh luat bao toan nguyen to'Fe:

—>

M20„

MjOn + 2nHCl
> 2MC1„ + nHzO
T h e o p h i « n g t r i n h ( l ) , (2)
n^^^^ =4.no

^.o'lftgO'"',j,o'

"Pe (trongFeS04) = "sO^" " ^ ' ^

1 =:>3n = 0,3

1^

'

^^"^
iioan toan 28,6 gam A
oxit B. Hoa tan het B trong dung dich
djch D diTcJc hon hdp muo'i khan la

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

(1)

^^^^^^

(2)

fa'*
xai

.

Ap dung djnh luat bao toan khoi lu'cJng -> mQ^ =44,6-28,6 = 16 gam

"pe (Fe304) ~ " F C (FeS04) "
=> no
"2

n = 0,l => mpg^o^ =23,2 gam=>DapanA

=0,5mol->

= > mnuirfi = m h h k i

Cfiu 39: Cho mot luong khi CO di qua ong diTng 0,01 mol FeO va 0,03 mol FejO,
(hon hdp A) dot nong. Sau khi ket thuc thi nghiem thu diTcfc 4,784 gam chat

=> Dap

UHCI

= 4x0,5 = 2 mol => n
= 2 mol
ci

+ m^,- = 28,6 + 2x35,5 = 99,6 gam.

.

Mfc # 3

an A

trtdb df^ll)

,
u'l:?:'

sji'o?

* r^n B gom 4 chat. Hoa tan chat r^n B b^ng dung djch HCl diT thay thoat ra
I 0,6272 lit H2 (d dktc). Tinh so mol oxit sat tiT trong hon hdp B. Biet rang trong
•f B so mol oxit sat tir bang 1/3 tong so mol sit (II) oxit va s^t (III) oxit.
bsuA

i A. 0,006.

fi&oti m

Y.B.?,

B. 0,008.

snt) nHv

C. 0,01.

PHLfdNG PHAP TANG GIAM KHOI IU0MG

D. 0,012.

Htf^ng d§n giai

rjftt

x rbjb j^nub

0;

•
"""^ + CO -> 4,784 gam B (Fe, FczOj, FeO, Fe304)
[^^303 :0,03mol
"'^Hi.

' ttWngtfngvdis^molla:a,b,c,d(mol).

FeCl2 + H 2

^ => a = 0,028 mol.
'

DUNG P H l / d N G

PHAP

., s
'

• , ,

•'^

•

,,,,
'

T h e o d l u b a i : np^^o^=i(np^o + np^^03) "> d = l ( b + c)

H>

.

Moi sir bien doi hoa hoc (diTcJc mo ta bang phiTdng tnnh phan iJng) deu c6
lien quan den siT tang hoac giam khol liTdng cua cac cha't.

.mi^STSJ?^ ^

Hoa tan B bKng dung djch HCl diT thu diTcJc n^^ = 0,028 mol.
Fe + 2HC1 ^

NOI

1. Npidung

" Hon hdp A r

•

A.

,

+ DiTa v^o sir tang hoac giam kho'i lirdng khi chuyen 1 mol chaft X thanh 1
hoac nhieu mol chaft Y (c6 the qua cac giai doan trung gian) ta de dang
tinh dirdc so' mol cua cac chaft va ngu'dc lai, ttr so mol hoac quan he ve so

(1)
(2)

: Tong mB la: (56.a + 160.b + 72.c + 232.d) = 4,78 gam.
'
' (3)
So mol nguyen tuT Fe trong hon hdp A bkng so mol nguyen tijf Fe trong hoU

mol cua 1 trong cac cha't ma ta se biet dirdc sir tang hay giam kho'i liTdng
cua cac chat X, Y.
+ Ma'u cho't cua phiTcfng phap la:

Xac dinh dung mo'i lien he ti 16 giffa cdc chat da biet (chat X) vdi chat can
^ac djnh (cha't Y) (c6 the khong can thiet phai viet phirdng trinh phan iJng, ma
^hi cin lap sd do chuyen h6a giiJa 2 cha't nay, nhiTng phai duTa vao DLBT

•

Ta c6: npe (A) = 0,01 + 0,03 x 2 = 0,07 mol
cO,0 ^v.
npe(B) = a + 2b + c + 3d
•
, • •••• • •
=^a + 2b + c + 3d = 0,07
(4)
: T i r ( l , 2 , 3,4) ^ b = 0,006mol; c = 0,012 mol; d = 0,006 mol => Dap an A

^ nguydn to' de xdc dinh ti le giffa chung).
x6t khi chuyen tiT cha't X thanh Y (hoac ngiTdc lai) thi khoi lirdng tang
^ '^n-hay gi^m di theo ti le phan lirng va theo de cho.
^au cdng, dira vao quy t^c tam suaft, lap phU'dng tnnh toan hoc d^ giai.


PhL/ong p h ^ p va k y t h u j t g i i i nhanh BTTN H6a dgi cUdng - vO co - D5 XuSn Hung

2 . Danh gia phifrfng phap tang giam khoi Ivtifng
-

Cfiu 21 Nung 6,58 gam Cu(N03)2 trong binh kin khong chuTa khong khi, sau mot

PhU'dng phap tSng giam khoi luTdng cho phep giai nhanh diTdc nhieu bai todn khj

thdi gian thu dUdc 4,96 gam cha't rSn va hon hdp khi X. Hap thu ho^n to^n X

biet quan he ve khoi lUdng

vao nUdc de dUdc 300ml dung djch Y. Dung djch Y c6 pH bhng

ti le moi cua cac chat trUdc va sau phan iJng.

Dac biet, khi chUa biet ro phan tfng xay ra la hoan toan hay khong hoan toan

A.
A. 2.

B. 3.

thi viec suf dung phUdng ph^p nay ckng giup ddn gian hoa bai toan hdn.
-

(Trich de thi tuyen sinh Dai hoc khoi A nam 2009)

Cac bai toan giai b^ng phUdng phap tang giam khoi lUdng deu c6 the gij,
di/dc theo phUdng phap bao toan khoi lifdng, vi vay c6 the n6i phUdng phap
tang giam kho'i liTdng va bao toan khoi lUdng la 2 anh em sinh doi. Tuy nhieti,
tuy tifng bai tap ma phUdng phap nay hay PhU'dng phap kia se la \iu viet hdn.

-

PhUdng phap tang giam khoi lUdng thUdng dUdc suf dung trong cac bai toan
hSn hdp nhieu chat.

-

Xac dinh dung moi quan he ty le giffa chat can tim va chat da biet (nhd van
dung DLBTNL).
Lap sd do chuyen hoa cua 2 cha't nay.

-

Xem xet sU tang hoac giam cua AM va Am theo PhU'dng trinh phan tfng va
theo dffkien bai toan
Lap phUdng trinh toan hoc de giai.

B. B A I T A P M I N H

.

Ta c6: Imol Cu(N03)2

1 moi CuO thi A M giam = 188 - 80 = 108 (g)

Vay: xmol Cu(N03)2

x moi CuO thi Am giam = 6,58 - 4,96 = 1,62 (g)

Theoptpi?:

(du), thu dUdc dung dich chiJa 7,5 gam muoi sunfat trung hoa. Cong thi?c cua
muoi hidrocacbonat la

C u ( N 0 3 ) 2 ^ C u O + 2NO2 + 1/2O2

;

/

0,03

^

4NO2 + O2 + 2H2O - ^ 4 H N 0 3
0,03

-

0,03

^

=> [ H i = [HNO3] = 0,03 : 0,3 = 0,1(M) ^

pH = 1 zi^Dap an D

CSu 3: Nhung mot thanh sat nang 100 gam vao 100ml dung dich hon hdp ggm

'

B.Mg(HC03)2

C. Ba(HC03)2

sach lam kho can dUdc 101,72 gam (gia thie't cac kim loai tao thanh deu bam
het vao thanh sat). Kho'i lUdng sat da phan tfug la :
A. 2,16 gam

B. 0,84 gam

C. 1,72 gam

Hridng d§n giai
c6: nc„(N03)2 = " AgNOj = 0.02 moi

Htfdngd§ngiai

D. 1,40 gam

(Trich de thi tuyen sinh Dai hoc khoi B nam 2009)

D. Ca(HC03)2

(Trich de thi tuyen sinh Cao dang nam 2010}
V

' ^

Cu(N03)2 0,2M va AgNOj 0,2M. Sau mot thdi gian lay thanh kim loai ra, riJa

HQA

CSu 1: Cho 9,125 gam muoi hidrocacbonat phan iJug het vdi dung dich H2SO4

A.NaHC03

PhUdng trinh phan ifng: Cu(N03)2 - » CuO + 2 N O 2 + - O;

0,015

-

-

Hri^ngdSngiai

=> x = 1,62: 108 = 0,015 moi

^mMihmmn.

3. C a c bride giai

D. 1.

C.4.

:

Goi cong thiJc muoi hidrocacbonat: M(HC03)n

Khoi lUdng thanh s^t tang: Am tang = 101,72 - 100 = l,72g

PtpiJ: 2M(HC03)n + nH2S04

Khi cho thanh sat vao dd gom AgN03 va Cu(N03)2, Fe phan rfng \d\d

>M2(S04)n + 2nC02 + 2nH20

Ta tha'y:

AgNOj trireme. Gia sOr AgNOj phan iJng het.

2mol M(HC03)n - > linol M2(S04)n thi khoi li/dng giam: 2.61n - 96n = 26n (g)

Fe + 2AgN03-^Fe(N03)2 + 2Ag

Vay xmol M(HC03)„ - > M 2 ( S 0 4 ) „ thi khoi lUdng giam: 9,125 - 7,5 = 1,625 (g)

0,01

1,625.2
=>x=
26n

0,02

=>Do tang kho'i lUdng thanh s^t: Ami tang = 0,02.108 - 0,01.56 = I,6g < l,72g

0,125
,
^,
9,125
,^
=—

m o l = > M + 61n= —
= 73n =:>M= 12n
n
0,125

=> c6 xay ra phan ung giSa Fe voi Cu(N03)2.
V I phan ung nay lam tang khoi lugng: Am2 tang = 1,72 - 1,6 = 0,12 g

n
cap nghiSm phil hdp 1^: n = 2 v^ M = 24 (Mg)
Cong thiJc cua muo'i hidrocacbonat la Mg(HC03)2 => Dap an B .

0,02

(1)

Pe + Cu(N03)2-^Fe(N03)2+Cu (2)
,


Phi/Dng phAp

ky thujt giai nhanh BTTN H6a dgi cuang - vO CO - 05 XuSn Hung

GQ'I nFepir(2) = X iTiol; ta

c6: A m 2 tang = 64x - 56x = 0,12 => x = 0,015
Theo ptpur (1),(2) ta c6: npepc = 0,01 + 0,015 = 0,025 mol
= ^ mpepcr = 0,025.56 = 1,4 (g)
Dap an D.

> X'usf a^:,^
Cfiu 4: Cho dung djch chifa 6,03 gam hon hcfp gom hai muoi NaX va NaY (X, Y
la hai nguyen to' c6 trong tii nhien, ct hai chu ki lien tiep thuoc nhom VIIA, so
hieu nguyen ttf Zx < Zy) vao dung djch AgNOs (dU'), thu dU'cJc 8,61 gam ket
tua. Phan tram khoi lu"dng cua NaX trong hon hcfp ban dau la
A. 58,2%.
B. 52,8%.
C.41,8%.
D. 47,2%.
(Trich de thi tuyen sink Dai hoc khoi B nam 2009)
¥L\idng d§n giai
Goi NaR la cong thuc chung ciia 2 muoi NaX va NaY.
. . ,-•1^ | ... ^v..
NaR + AgNOj - AgRi + NaNOj
1 mol NaR - AgR khoi lugng tang: AM tang = 108 - 23 = 85g
Vay: x mol NaR - AgR khoi lugng tang: Am tang = 8,61 - 6,03 = 2,58g
z:> X = ^
= 0,03 mol => M NaR = — = 201
if'frvf

85

0,03

=> M R = 201 - 23 = 178 khong c6 2 halogen nao thoa man.
Vay X, Y Ian \iigt la F va CI; ket tua la AgCl
8 61
Ta c6: nNaci = nAgci = —— = 0,06 mol => niNaci = 0,06.58,5 = 3,5 Ig
143,5
=> mNaF =

6,03 - 3,51 = 2,52g => %NaF = — . 1 0 0 % = 41,8%
i.',
6,03
=^DapanC.
CSu 5: Nhiing mot la kim loai M (chi c6 hoa tri hai trong hcfp chat) c6 khoi
iMng 50 gam vao 200ml dung djch AgNOs IM cho den khi phan ifng xay ra
hoan toan. Loc dung dich, dem c6 can thu dUdc 18,8 gam muoi khan. Kim
loai M la
A. Mg
B. Zn
C.Cu
D. Fe
(Trich de thi tuyen sinh Cao dang khoi A,B nam 2009)
Hifdng d§n giai
Ta c6 : n^g^Oj = O.il = 0,2 (mol) => m^^^^^ = 170.0,2 = 34 (g).
, PtpiJ: M + 2AgN03-»M(N03)2 + 2Ag
Ta c6: 2 mol AgNOj - 1 mol M(N03)2 thi AM giim = (2.108 - M) (g)
Vay: 0,2 mol AgNOa - 0,1 mol M(N03)2 thi Am giam = 34-18,8 =15,2 (g)

f'"':/
./"^''l
(0,^.(216 - M) = 15,2 X 2 => M = 64 (Cu)
=>DapanC
^^^^ '
Cfiu 6: Tien hanh hai thi nghiem sau:
Ir^
Thi nghiem 1: Cho m gam bpt Fe (dU) vao Vi lit dung dich Cu(N03)2 IM.
Thi nghiem 2: Cho m gam hot Fe (dU) vao Vj lit dung djch AgN03 0,1M.
Sau khi cac phan iJng xay ra hoan toan, khoi liTdng chaft r^n thu diTdc d hai thi

nghiem deu b^ng nhau. Gid trj cua Vi so vdti V2 \h:
A.V,

= V2

B.V,

= 10V2

C.V,

D. V, = 2 V 2 .

=5V2

(Trich de thi tuyen sinh D^ii hoc khoi B nam 2008)
Htfdng d i n giai -,
Thinghi$ml: ncu(N03)2 = " ^ o '
Fe + Cu(N03)2

Fe(N03)2 + Cu

Vimol Vimol
Vimol
Fe di/nen Cu(N03)2 he't =:> ncu= Upe = ncu(N03)2 ~
=> Do tang kho'i liTdng: (64 - 56).Vi = 8V| mol.
Thi nghiem 2: n^g^^^ = 0,1 .V^ mol
Fe

'"^^

'^

+

2AgN03 ^ Fe(N03)2 + 2Ag
0,05V2mol
0,l.V2mol
O.l.Vzmol
Fedu-nen AgNOj het => n A g = n.
= 0 , 1 . V 2 mol
V ^ " F e = ^nAgNO3 =

^0-I-^ =

0,05.V2mol

=> Do tjing khoi liTdng: lOS.O.lVj - 56.0,05V2 = 8V2 mol.
\
Theo de bai: sau phan iJng khoi li/dng chat r^n thu diTdc bkng nhau.
Dp tang khoi lu'dng d hai thi nghiem ciing bang nhau.
^'
Hay: 8 V , = 8V2 => V, = V2
, :,\„H ^ => Dap an A.
*
^Su 7: Nung mot hon hdp r^n gom a mol FeC03 va b mol FeS2 trong binh kin
chu-a khong khi (di/). Sau khi cac phan ij-ng xay ra hoan to^n, di/a binh ve
"hiet dp ban dau, thu diTdc cha't ran duy nhat la Fe203 va hon hdp khi. Biet ap
suat khi trong binh tru'dc va sau phan liTng bang nhau, mo'i lien he giila a va b
(biet sau cac phan iJng, lUu huynh d miJc oxi hoa +4, the tich cdc cha't r^n

'a khong dang ke).
^•a.= 0,5b.
B. a = b.
C. a = 4b.
D. a = 2b.
(Trich de thi tuyen sinh D(^i hoc khoi B nam 2008)
33


Phuang ph^p va ky thuat giSi nhanh BTTN H6a dgi ciiong - vO cd - D8 Xuan Hong

Cfiu 9:*H6a tan 14 gam hon hdp 2 muo'i MCO3 va N2(C03)3 bing dung dich HCl
diT, thu diTdc dung dich A va 0,672 lit khi (dktc). Co can dung djch A thi thu
(Ji/dc m gam muo'i khan. Gia trj ciia m 1^:
A. 16,33 gam
B. 14,33 gam
C. 9,265 gam
D. 12,65 gam.

Hif^ng din giai
4FeS2 + IIO2 — ^

2Fe203 + 8SO2

H

2b

ii^

(D

Htfdng din giai
2FeC03 + ^02 — ^
• •'"
,
A a

a
4

Fe203 + 2CO2 (2)

van dung phu'dng phap tang giam khoi liTcfng. Theo phU'dng tnnh ta c6:

a

r-'

•

CiJ 1 mol muo'i C03~-> 2 mol CI^ + 1 mol CO2,

.

'

Wdng muo'i tang: 71 - 60 = l i g
Theo de so mol CO2 thoat ra la 0,03 thi kho'i luTdng muo'i tang:

Phan tfng (1) lam giam: ^

- 2b = 0,75b mol khi

11.0,03 = 0,33g

,:.

Vay mn,uoi ciomn = 14 + 0,33 = 14,33 (g) => Dap an B.

Phan uTng (2) lam tang: a - - = 0,75a mol khi

Cfiu 10: Nhung 1 thanh nhom nang 45 gam vao 400ml dung djch

CUSO4

0,5M.

Theo de bai ap suat triTdc va sau phan ufng khong thay ddi

Sau mot thdi gian la'y thanh nhom ra can nang 46,38 gam. Kho'i liTdng Cu

=> so mol khi tang va giam bang nhau.

thoat ra la:

Hay: 0,75a = 0,75b => a = b =:> Dap an B.

A. 0,64 gam

B. 1,28 gam

C. 1,92 gam

D. 2,56 gam.

Hiidng din giai

CSu 8: Cho m (g) hon hdp bpt Zn va Fe vao li/dng dU dung djch C U S O 4 . Sau khi
ket thtic cac phan i?ng, loc bo dung dich thu diTdc m (g) chat r^n. Thanh phan

CiJ 2 mol Al ^ 3 mol Cu; khoi liTdng tang 3.64 - 2.27 = 138 gam

% theo khoi lifdng cua Zn trong hon hdp ban dau la:

Theo de: n mol Cu; khoi lu-dng tang 46,38-45 = 1,38 gam

A. 90,27% .

B. 82,20%

. w; .tf,;,

C. 85,30%

D. 12,67%.

(Trich de thi tuyen sink Dai hoc khoi A nam 2007)
Hrfdng dSn giai
> ZnS04+ Cu

a

(1)

b

> FeS04 + Cu

300m! dung djch H2SO4 0,1M thi khoi iiTdng hon hdp cac muo'i sunfat tao ra la:
A. 3,81 gam

a

Fe + CUSO4

C.BAITAPAPDVNG
Cfiu 1: Cho 2,81 gam hon hdp gom 3 oxit Fe203, MgO, ZnO tan viTa dii trong

Goi a,b Ian lifdt la so mol cua Zn va Fe.
Zn + C U S O 4

no, = 0,03 mol => mc, = 0,03.64 = 1,92 (g) => Dap an C .

(2)

b

Phan iJng (1) l^m giam kh6'i li/dng hon hdp kim loai, phan tfng (2) lam tang
khoi liTdng kim loai.

B. 4,81 gam

C. 5,21 gam

Hrfdng din giai
Ap dung phiTdng phap tang giam kho'i lUdng:
CiJ 1 mol H2SO4 phan tfng, de thay the O (trong oxit) b^ng SOj" trong cdc
loai, khoi IiTdng tang 96 - 16 = 80g

Theo de so mol H2SO4 phan iltig la 0,03 thi khoi lu'dng tang 80.0,03 = 2,4 g

Ma theo de bai ban dau c6 m (g). Sau phan ifng c6 m (g) chat tin.

^Sy kho'i liTdng muo'i khan thu diTdc la: 2,81 + 2,4 = 5,21 g

Vay: At = A4 => (65 - 64)a = (64 - 56)b =^

=>DapanC.

^

%Zn =

—
100% = 90,27%
65.8 +56.1

=> Dap an A.

D. 4,86 gam.

"

2: Dem nung mot kho'i liTdng Cu(N03)2, sau mot thcJi gian thi tha'y kho'i
•i'1^ baonhieu?
^-•^.Sg.

B.0,49g

C.9,4g

D.0,94g. ;


PhUOnp phap va ky thujt giai nhanh

BTTN

H6a dgi cuong - va CO - D 8 Xufln HJng

Htfdng d i n giai
Cu(N03)2 — ^

CuO +2NO2 + - O 2

Hi^dng dSn giai:

• -m:-<'ifmM:

Ta CO sd do phan uTng: 104,25(g) hon hdp NaCI, Nal
Cilrl88(g)Cu(N03)2 — ^

CuO,khoi liTdng chat ran giam 188-80= 108(g)

> 58,5 (g) NaCI

P,;a vao sd do ta thay:

,

1 moi Nal phan uTng thi khoi liTdng giam 150 - 58,5 = 91,5 (g)
Vay: ciir a (g) Cu(N03)2
a=

^i^iM

'° > CuO, khoi liTdng chat ran giam 0,54 (g)

= 0,94 (g) ^

Vay X moi Nal phan uTng thi khoi liTdng giam 104,25 - 58,5 = 45,75 (g)
45,75

Dap an D.

108

Cfiu 3: Nhiet phan hoan toan 9,4g mot muoi nitrat kim loai thu difcJc 4g oxit r^n.

Cong thiJc muoi da dung la:
A. Fe(N03)3.

C. Cu(N03)2.

D. AgN03.

=>

%NaI

NaBr den phan iJng hoan toan. Co can dung djch thu di/dc 25,55g muo'i khan.
Kho'i liTdng cua NaBr trong hon hdp dau la:

*° ) R20„ + 2nN02 + - O2
2

B. 5,15 gam

9,4.108n = 5,4.2(MR + 62n)

M R = 32n

tfng phSn hu^.
B. 40%.

C. 27,5%

D. 50%.

HuTdng d i n giai

NaF
hon hdp NaCl-

^° ) PbO + 2NO2 + - O 2
2

2-^25,55 (g) hon hdp

NaF
NaCI

Theo sd do ta thay:
1 moi NaBr phan u'ng thi khoi liTdng giam 103 - 58,5 = 44,5 (g)
Vay X moi NaBr phan u'ng thi khoi lu'dng giam 30 - 25,55 = 4,45 (g)
4,45
44,5

= 0,1 moi =>

mNaBr

= 0,1.103 = 10,3 (g) => Dap an A .

Cfiu 7: Nung hon hdp A gom CaC03 va CaS03 tdi phan lirng hoan toan diTdc chat

^° > PbO, khoi liTdng chat r^n giSm

331-223= 108(g)

Vay: a (g) Pb(N03)2

+C1

NaBr

=> X =

Cur 331 (g) Pb(N03)2

D. 12 gam.

Ta c6 sd do phan u'ng:
30(g)

Cfiu 4: Nung nong 66,2g Pb(N03)2 thu du-cfc 55,4g chat r^n. Tinh hieu suat phan
A. 25%.

C. 6gam

Hifdng dSn giai:

=> chon n = 2 => M R = 64 => Cong thuTc muoi la Cu(N03)2 => Dap an C .

r^n B CO kho'i liTdng bhng 50,4% khoi IiTdng cua hon hdp A. Thanh phan phan
'

^° > PbO, khoi liTcJng chat r^n giam

tram ve kho'i IiTdng cac chat trong hon hdp A:

A. 40% va 60%

B. 25% va 75%

66,2-55,4=10,8 (g)

C. 30% va 70%

D. 20% va 80%

Hi^dng d i n giai

10,8.331
, , ,
„
33,1.100%
, ^
=> a = —^
=33,l(g)=> H = —:
=50% ^ Dap an D.
108
^
66,2
CSu 5: Hoa tan 104,25g hon hcJp cac muoi NaCI, Nal v^o niTdc. Cho du khi clo

36

133 Ujfa ;;•„';,;•!'(;•

/fed? •'){•'.

9,4 (g) mu6'i — ^ — > R20„, khoi liTdng chat dn giam 9,4 - 4 = 5,4 (g)

Pb(N03)2

0,5.150 = 75 (g)

= 71,94% va %NaCl = 28,06% = > D a p a n C .

A. 10,3 gam

Cur 2 ( M R + 62n) (g) muoi — R j O n , khoi liTdng chat r^n giam 108n (g)
vay:

mNai=

CSu 6: Sue khi C I 2 diT vao dung djch chtfa 30g hon hdp 3 muo'i NaF, NaCl v^

B. A1(N03)3.

Hi^dng d§n giai
2R(N03)n

= 0,5 moi ^

91,5

CaC03 ->CaO + C02

,

CaS03 ->CaO + S02

(1)
(2)

Theo p t ( l ) , (2) ta c6:

di qua roi c6 can. Nung chat ran thu dU'dc cho den khi het mau tim bay ra. Ba

• OOg CaCOj ^ 56g CaO tiTdng u'ng 56% mcacbona.

r^n con lai sau khi nung nang 58,5g. % khoi liTdng moi muo'i trong hon hdp

' 20g CaSOj

thudiTdcla:

Goi X la thanh phan phan t r i m ve kho'i liTdng cua CaC03. Ta c6 :

A. 29,5% va 70,5%.

B. 65% va 35%.

C. 28,06% va 71,94%

D. 50% va 50%.
• .»

56g CaO tiTdng u'ng 46,67% msunm

56x + 46,67(1 - X ) = 50,4 <^ x = 0,4
Dap an A.

«

% m CaCO = 40%


PhUOng pMp

va ky thujt g\&\h BTTN H6a dgi cuong - vO CO - D8

Xufln Hung

C§u 8: Nhiing mot thanh kem va mot thanh s^t vao cCing mot dung djch CUSO4.
Sau mot thdi gian lay hai thanh kim loai ra thay trong dung djch con lai c6 nong
do mol ZnS04 bang 2,5 Ian nong do mol FeS04. Mat kh^c, khoi liTcfng dung
dich giam 2,2g. Khoi liTdng dong bam len thanh kem va thanh sat Ian liTdt la:
A. 12,8g; 32g
B. 64g; 25,6g C. 32g; 12,8g D. 25,6g; 64g
Uridng din giai
Do Cling nhung hai thanh kim loai vao mot dung djch nen dung djch c6n lai
j cung the tich
CM (ZnS04) = 2,5CM (FeS04) -> n^^^^^ - 2,5np^so^
Zn + C u S 0 4 ^ ZnS04 + Cui
(1)
2,5x ^ 2,5x ^ 2,5x -> 2,5x
Fe + CuS04-> FeS04 + Cui
(2)

X

<-

X

<— X

->

X

Dp giam khoi lifdng cua dung djch la: mcu (biSm) - mzn(tan) - mpe(tan)
<-> 2,2 = 64(2,5x + x) - 65. 2,5x - 56x -> x = 0,4 (mol)
- > mcu
leii
Zii = 64g ; mcu
len thanh Fe = 25,6g => Dap 8 1 1 B.
C&u 9: Hoa tan 12g muoi cacbonat kim loai bang dung djch HCl dU" thu diTPc
dung dich A va 1,008 lit khi bay ra (dktc). Khoi liTdng muoi khan thu diTPc khi
CO can dung dich A la:
A. i2,495g.
B. 12g.
C. ll,459g
D. I2,5g.
Hif(tng din giai
R2(C0.,)n + 2nHCl ^ 2RC1„ + nCOj + nHjO
Ti( R 2 ( C 0 3 ) „ -> 2RC1„ thi khoi lu-dng tang 11 n (g) tao ra n mol C O 2 .
Vay R2(CO.0n ^ 2RC1„ thi khoi liTcfng tang a (g) tao ra
= 0,045 mol C O 2 .

hnm

=> a =

Ihiinh

biiiii

0,045.1 In
n

= 0,495g

Ta c6 sd d6:

A2CO3
+HC1
23,8(g) hon hdp< BCO,
> hon hdp 2AC1 + CO,
BCL
2
TiT hh mu6\t hh muoi clorua khoi IiTdng giSm 11(g) tao ra 1 mol CO2.
=e> hh muoi cacbonat
hh muoi clorua khoi liTdng giam a(g) tao ra 0,2 mol C O 2
a = 0,2.11 =2,2 (g)

VSy: nimuo'i cioma = 23,8 + 2,2 = 26 (g)

Dap an A.
Cfiu 11: Nung nong lOOg hon hdp NaHCOs va Na2C03 den khoi lU'dng khong

d6\u diTdc 69g hon hdp ran. % khoi luTdng cua NaHCOs trong hon hdp la:
' A . 80%.
B. 70%.
C. 80,66%.
D. 84%.
HU'dng dan giai
Phan tfng: 2NaHC03 — - — > NuiCOi + C O 2 + H 2 O
Cur 2 mol NaHCOaphan u-ng thi khoi liTdng giam 2.84 - 106 = 62 (g)
Vay X mol NaHCOsphanu'ng thi kho'ili/dng giam 100-69 = 31 (g)
;
31.2
,
,
=> x=
= 1 mol
"lNaHC03 = 84 (g) =:> % NaHCO^ = 84%
62
=> Dap an D.
CSu 12: Khi la'y 16,65g muoi clorua cua mot kim loai IIA vao mot muoi nitrat
cua kim loai (c6 cilng so mol vdi 16,65g muoi clorua) thi thay khac nhau
7-,95g. Kim loai IIA la:
A.Mg.
B.Ba.
C.Ca.
D. Be.
Hrf(Jng din giai
;,
Cu" I mol R C I 2 R ( N 0 3 ) 2 khoi li/dng tang 53 (g)
Vaiy X mol R C I 2 R ( N 0 3 ) 2 khoi lufdng tang 7,95 (g)
=> X = 0,15 mol =:> M R C I , = 1 1 1

=> m„„,6iciorua = 12 + 0,495 = 12,495 (g)- Dap an A.
CSu 10: Hoa tan hoan toan 23,8g hon hdp mot muoi cacbonat cua kim loai hoa
tri I va mot muoi cacbonat cua kim loai hoa tri II b^ng dung dich HCl tha'y
thoat ra 4,48 lit khi C O 2 (dktc). Co can dung djch sau phan drng thi liTdng muoi
, khan thu difcJc la:
A. 26g.
B.28g
C.26,8g
D. 28,6g.
Hvldng din giai
Ta c6: npQ^ = 0,2mol
^

'

= > M R = 40 (Ca) => Dap an C.

^fiu 13: Cho dung djch AgNOj du" tac dung vdi dung dich hon hdp c6 hoa tan
6.25g hai muoi KCl va KBr thu diTdc 10,39g hon hdp ket tua. Xac djnh so' mol
cfia hon hdp dau?
* ^
,
A. 0,08
B. 0,06.
C. 0,055
D. 0,03
Hifdng din giai
Ta CO sd do:
6,25 (g) hon hdp fKCl , A g N 0 ,

,
[AgCl
hdp
KBr — " ^ > 10,39
. ghSn
e
.F
K g^.


PhUOng phAp v i ky thujt giSi nhanh BTTN H6a (J^i cuong - vO co - 05 Xuan Hung

Theo sd do ta thay:
Ca 1 mol hSn hdp dau tac dung v6i AgNOj khoi iuWng tang 108 - 39 = 69 (g)
=>a mol hon hdp dau tdc dung vdti AgN03 khoi lUdng tang
10,39 - 6,25 = 4,14 (g) = : > a = ^ = 0,06 mol ^ Dap an B.
Cfiu 14: Nhiing mot thanh kim loai hoa trj II vao dung dich CUSO4 dU'. Sau phan
iJng, khoi liTcJng thanh kim loai giam di 0,24g. Cung thanh kim loai do ne'u
nhung vao dung djch AgNOs thi khi phan ilng xong thay khoi liTdng thanh kim
loai tang len 0,52g. Kim loai hoa trj II la:
A.Pb.
B.Cd.
C. Sn.
D. Al.
Hxldng d§n giai
R

+ CUSO4

RSO4 +

Cu

a
a
a
a
=> Kho'i liTcJng thanh graphit giam: (MR - 64)a = 0,24 (g) (1)

R + 2AgN03 ^ R(N03)2 + 2Ag

Khoi lu-dng thanh graphit tang: (2.108 - MR)a = 0,52 (2)
L a y ( 2 ) : ( l ) => M R = 1 1 2 ( C d ) = > D a p a n B .
Cfiu 15: Ngam mot vat bang Cu c6 khoi liTdng 15g trong 340g dung dich AgNO^
6%. Sau mot thdi gian lay vat ra thay khoi lu"dng AgNOs trong dung dich giam
25%. Khoi lUcfng cua vat sau phan iJng la:
A. 3,24g.
B. 2,28g
C. 17,28g.
D.24,12g.
Hifdng dfin giai

Ta c6: m^^^^^ = ^

= 20,4 (g) =^ n^^^^^ = 0,12 mol

Khoi lUdng AgNOj giam 25% chinh la liTdng AgNOs phan uTng.
Vay: riAgNOaPhSn^jng =0,12.25% = 0,03 mol
Phan iJ-ng: Cu + 2AgN03 ^ Cu(N03)2 + 2Ag
0,015 0,03

0,03
=^ mva,= 15+ (0,03.108-0,015.64)= 17,28 (g)
Dap an C.
Cfiu 16: Khur hoan toan mot oxit sat nguyen chat b^ng CO dir d nhiet do cao. Ket
thiic phan iJng khoi lu'dng cha't rdn giam di 27,58%. Oxit da dung la:
A.Fe203
B. Fe304
C. FeO
D. Tat ca deu sai.
Hi/(?ng d§n giai
Dat cong thufc oxit FcxOy:
Khoi liTdng cha't rin giam 27,58% chinh la khoi liTdng ciia oxi trong oxit.

^
%o =
100% = 27,58% =>,. ll,5872y = 15,4448x '
Ta'^"56x + 16y
u^v
=
=>
oxit la Fe304. => Dap an B.
'
^
Hay y 4
Cfiu 17: Cho a gam hon hdp gom FeSj va FeCOj vdi so mol bing nhau vao m6t
binh kin chtfa li/dng oxi dir. Ap suat trong binh la Pi atm. Nung nong binh de
phan ifng xay ra hoan toan roi diTa binh ve nhiet dp ban dau, ap suat khi trong
binh luc nay la P2 atm. Biet r^ng the tich chat ran trong binh trUdc va sau
phan i?ng la khong ding ki. Ti le P,/P2 la:
A. 0,5

B.l
C.2
D.2,5.
Hrfdng d i n giai
4FeS2 + IIO2 — ^
2Fe203 + 8SO2
(D
Ilx
2x
2FeC03 +

^02

— ^ FezOj +

2CO2

(2)

-'

llx
Phan urng (1) lam giam
- 2x = 0,75x mol khi
Phan UTng (2) lam tang X - - = 0,75x mol khi
4
ir"
Ta thay liTdng mol khi tang va giam bKng nhau => so mol khi khong doi.
p
=> Pi = P2 h a y = 1 = > D a p a n B .

IT
Cfiu 18: Cho 3,78g bpt Al phan ilng viTa du vdi dung dich mu6'i XCI3 tao thanh
dung dich Y. Khoi lirdng chat tan trong dung djch Y giam 4,06g so vdi dung
djch XCI3. Xac djnh cong thtfc cua muoi XCI3.
A.InCl3
B.GaCb
C. FeCU
D. GeCU
Hifdng d i n giai:
Taco: n„= ^ = 0,14 mol
' '
'•
^' 27
Al

+ XCI3

> AICI3+ X

0,14 0,14
0,14 0,14
Un'<' \
Ta c6: (Mx + 35,5.3).0,14 - (133,5.0,14) = 4,06
=> Mx = 56
muoi do la FeCh => D a p an C.

;J!|:K:»


Phuang

phap

va ky thii$t glSi nhanh BTXU

H6a dgi cuong - vO CO - P g Xuan Himg

Cfiu 19: Nhiing thanh kem vao dung dich chtfa 8,32g CdS04. Sau khi khuT hoan
to^n ion Cd^"^, kh6'i lU'cJng thanh kem tang 2,35% so vdi ban dau. Hoi khoj
lu'dng thanh kem ban dau:
A.80g

B.72,5g.

C.70g

D. 83,4g. ; ;

HtfdngdSngiai

P2

= 95% P,. (Vdi P, = PdSu, P2 =

Trong cung d i l u kien, ta c6: ^

PhanuTng: Zn + CdS04 -> ZnS04 + Cd
0,04
^

HrfdngdSngiai
• Ap su^ giam 5% =>

Ta c6: n^^^^^ = 0,04 mol

=^ (II2-65).0,04 = — a
100

^^tfng tW ap
g'^"' ^"^^ ^° '^^^
^"^^
^ ^ ^ " ^ P^^"
cfla ozon trong hon hdp sau phan i?ng ]h:
^5%
B. 10%
C.15%
D.20%.

¥

Goi khoi Itfdng ban dau cua thanh kem la a (g)

0,04

22:*Sau khi ozon hda 100ml khi oxi, dUa nhiet d6 ve trang thdi trufdc phSn

=> Vhhgi=im

o 0,95 = - j ^ =^

" •

= 95 ml , ,.
a

= 100-95 = 5 m l

Phantfng:
3O2 - > 2O3
Theo phan ilng: Ci? 3 ml khi O2 phan iJng lam giam 1ml khi

a = 80 ( g ) o Dap an A.

Cfiu 20: Nhiing thanh kim loai R hoa tn I I vao dung djch

=^

Psau)

CUSO4.

,

Vay V ml khi O2 phan iJng lam giam 5ml khi

Sau mot thdi

?
'

: ,

gian la'y thanh kim loai ra thay khoi liTdng giam 0,05%. Mat khac nhiing thanh
kim loai tren vao dung djch Pb(N03)2, sau mot thdi gian thay khoi lu'dng tang

=>

phJntfng =

1.5

ml

VQ^

=>

uoU,Jnh =

7,1%. Biet ring so mol R tham gia d 2 triTdng hdp nhu" nhau. R la:
A.Cd

B.Zn

C.Fe

D. Sn.

-

10

ml

03:10ml

Vay hon hdp sau phan tfng gom:

mh'!

0 2 : 9 5 - 1 0 = 85ml

v5i :

a

Hi^dng dan giai
Goi a la kho'i liTdng ban dau cua thanh kim loai R.
R + CUSO4
.

%0

RSO4 + Cu

>

X

100% = 15% => Dap an C.

thay thoat ra V lit khi (dktc). Dung dich thu di/dc dem co can thu diTdc 5,1

=>(M^-64).x:.^a

(1)

gam muoi khan. Gia tri cua V la:
A. 1,12 lit

> R(N03)2 + Pb

=^ ( 2 0 7 - M „ ) . x = — a
100

(2)

Cfiu 21: Sau khi chuyen 1 the tich khi oxi thanh ozon thi thay the tich khi giam
5ml. The tich oxi da tham gia phan ufng la:
B. 15ml

B. 1,68 lit

C. 2,24 lit

M C O 3 + 2HCI ->

M C I 2 +H2O + CO2

M + 60

M + 71

4g

1 mol

5,lg

D. 17ml.
Cfiu 24: Cho 1,26 gam mot kim loai tdc dung vdi dung dich H2SO4 loang tao ra
3,42 gam muo'i sunfat. Kim loai do la:

CtJ 3 the tich O2 phan liTng lam gidm 1 the tich khi.

A.Mg

VQ^ phJni?ng = 5.3 =

15ml

M,a„g= 71 - 6 0 = 11 (g)

x m o l m , a n g = 5 , l - 4 = 1,1 (g)

Phan iJng: 3O2 - > 2O3
VSy V the tich O2 phan tfng lam gidm 5ml khi.

/

=>x=M=o,l(mol)=> V = 0,1.22,4 = 2,24 ( 0 => Dap an C.

C. 16ml

Hifdng d i n giai

^

B.Fe

C.Ca

^

D. A l .
i

Hi^ng d i n giai

*r
2R + nH2S04 ^

R2(S04)n +

nH2t

dung phiTdng phap tSng giam kho'i liTdng ta c6:
=> Dap an B .

D. 3,36 lit.

Hvfdng d i n giai

L a y ( 2 ) : ( l ) =i> M R = 65 (Zn) i ^ D a p a n B .

A. 14ml

1112
48.10 + 32.85

Cfiu 23: Hoa tan hoan toan 4 gam hon hdp MCO3 va M ' C O j vao dung djch HCl

X

R + Pb(N03)2

=
^

S

'J

:

-


Phuong ph^p Vci ky thujt giSi nhanh BTTN H6a djii cuong - vO ca - B 5 Xuan Hung

27' 1'«Jhung mot thanh graphit diTdc phu m6t Idp kim loai hoa trj (11) vao
Ctf 1 mol kim loai h6a trj n tac dung vdi H2SO4 loang tao thanh muoi SO^
khS'i li/dng tang len ^

^^dung '^''^^ C U S O 4 dir. Sau phan iJng kho'i liTdng cua thanh graphit gi&m di 0,24
m Cung thanh graphit nay ne'u difdc nhiing vao dung djch AgNOa th\i

= 48n (g)

phan i^ng xong tha'y kho'i liTdng thanh graphit tang len 0,52 gam. Kim loai hoa
Theo de khoi lu-cJng tang 3,42 - 1,26 = 2,16g
• V§y so' mol kim loai R la: ^
=
48n

tri (II) la •^'"^
^ pb.

(mol)
—

n

'

C. A l .

D. Sn.
'

0at kim loai hoa tn (II) la M vdi so' gam la x (gam).
= 28n => cap nghiem phu hcJp la n = 2 va M R = 56 (Fe)

+ CuS04du

Ci? 1 mol M
n
Dap an B .

Vay :

HCl ta thu durpc 12,71 gam muoi khan. Th6' tich khi H2 thu diTdc (dktc) la:
A. 0,224 lit

>

M S O 4 + Cu

:M

1 mol Cu, kho'i li/dng giam ( M - 64) gam

x mol M -> x mol Cu, kho'i lu'dng giam 0,24 gam
- 0'24.M

CSu 25: Hoa tan hoan toan 12 gam hSn hdp 2 kim loai X va Y bang dung dich
B. 2,24 lit

C. 4,48 lit

D. 0,448 lit.

Hufdng din giai

"^"^ '

> M(N03)2 + 2Ag

CiJ I mol M -> 2 mol Ag khoi liTdng tang (216 - M) gam
Vay :
=>x =

, • ,

216-M

.,

so mol CP phan iJng la 0,02 mol

nj^^ =i-n^^_ =0,01(mol)

Ta c6:

V=0,224(Z) i : ^ Dap an A.

;

x mol M ^ 2x mol Ag kho'i li/dng tang 0,52 gam
0,52.M

Ctf 1 mol C r sinh ra sau phan iJng khoi lu'dng muoi tang 35,5g
Theo de, tang 0,71 g

M-64

Matkhac:M + 2AgN03

Ap dung phiTdng phap tang giam khoi lu'dng:

M-64

= 0^2Jvl
216-M

^

,

M = 112 (kim loai Cd) => Dap an B .

Cfiu 26: Cho hon hdp bpt gom: 0,48 gam Mg va 1,68 gam Fe vao dung dich

Cfiu 28: Hoa tan 5,94 gam hon hdp 2 muoi clorua ciia 2 kim loai A,B (deu c6

CUCI2, roi khua'y deu den phan iJng hoan toan thu diTdc 3,12 gam phan khong

hoa tri II) vao niTdc diTdc dung dich X. De lam ket tua het ion CF c6 trong

tan X. So' mol CUCI2 tham gia phan iJng la:

dung dich X ngUdi ta cho dung dich X tac dung vdi dung dich AgNOs thu difdc

A. 0,03

B.0,05

C.0,06

D. 0,04.

Hrfdng dSn giai:
Ta c6:

nFe

= 0,03 mol va

gam hon hdp muoi khan. Gia tri cua m la:
A. 6,36 gam

= 0,02 mol.

nwg

17,22 gam ket tua. Lpc bo ket tua, thu diTdc dung djch Y. Co can Y diTdc m
B. 63,6 gam

C. 9,12 gam

D. 91,2 gam.

Hi^dng d§n giai

Cha't ran X c6 khoi liTdng tang so vdi kho'i lu'dng ban dau.
3,12-(0,48 + 1,68) = 0,96 (g)

Ap dung phiTdng phip tang giam kho'i liTdng:

Mg C O tinh khuT manh hdn, khuf tru'dc Fe. Neu chi c6 0,02 mol Mg phan itng,

Cur 1 mol M C I 2 ^ 1 mol M(N03)2 va 2 mol AgCl thi m tang 2.62 - 71 = 53 gam

khoi liTdng X tang: 0,02.(64 - 24) = 0,8 (g) * 0,96 (g)
.

-

Hyidng din giai

1 26
Vay Mj^ =

•
B. Cd.

=> Fe c6 phan iJng.

^mSimvM

Gpi X la so mol Fe phan lifng, do tSng cua X so vcli khoi lu'dng ban diu:
0,02.(64 - 24) + x.(64 - 56) = 0,96 => x = 0,02 mol.
Vi

Nen

: npephiintfng = 0,02

0,12 mol AgCl kho'i liTdng tang 3,18 gam

mol

<

npe ban d^u = 0,03

mol

Fe dir va C U S O 4 het.

Vay : so mol C U S O 4 da phan uTng la: 0,02 + x = 0,04 mol

=

mhh+ m,ang= 5,94 + 3,18 = 9,12 (g)

Y

=>DapanC.
CSu 29: Hoa tan 9,875g mot muoi hidrocacbonat vao niTdc, cho tac dung vdi

'lung djch H2SO4 vifa du roi dem c6 can thu diTdc 8,25g mot muo'i sunfat trung
hoa khan. Cong thtfc phan tuf cua muo'i la:
A.NH4HCO3

B.NaHCOj

C.Ca(HC03)2

D. K H C O 3

=> Dap an D .
45


PhUcmq phip

ky thu$t

giai

•

nhanh BTTN H6a djii ciiong - vO CO - D8 XuSn Hang

Hxidng d i n giai

.0

1

2 °2

Goi muoi hidrocacbonat: RCHCOj)^ vdi n la hoa tri cua kim loai trong muoi do

Mhh +

2R(HC03)„ +nH2S04 ^ R 2 ( S 0 4 ) „ +2nH20 + 2nC02 t

I^^,0 + 2 HCl ^ MhhCl2 + H2O

Theo phiTdng trinh: cu" 2 mol muoi hidrocacbonat chuyen thanh 1 mol muo'i
sunfat thi khoi liTcfng muoi giam: 61x 2n - 96n = 26n (g) va la khoi liTdng ciia
2n mol CO2.

Theo de: Amgij,,, = 9,875 - 8,25 = 1,625 (g)

,,

Z chifa muoi khan c6 khoi lU'dng Idn hdn khoi lu"dng oxit 55g. Do chinh 1^ d6
chSnh lech khoi liTdng ciia 2 anion CI'
MhhCl2 tang: 7 1 - 1 6 = 55(g)
Theo de: Am,a„g
Ta c6:

-(mol)
^''^^

=^"co2 = 1 - 6 2 5 x ^ = 0,125 (mol) =>n^,(HC03)„
T a c6 he thiJc tinh M R : M R =

9,875
•-61n = 18n
0,125

O^": 1 mol MhhO chuyen thanh 1 mol

moxu

=

55 (g)

Uoxit = nn,u^i

= 1 (mol)

i

= mM^^i^ + mQ

'

=> a = m^i^^ = m o x i t - m o = b - l x l 6 = b - 1 6

=> D a p a n A .

Cfiu 32: Cho hon hdp gom ba muoi MgCl2, NaBr, K I vdi so' mol tU'dng tfug la 0,2
mol; 0,4 mol va 0,2 mol. H6a tan hon hdp A tren v^o ni/dc tao ra dung djch X.
Dan V(l) CI2 sue vao dung djch X, c6 can dung dich sau phan uTng thu dU"dc

2

1

n

66,2g chat ran. Tinh V (dktc)
39

18

R

A. 2,24 lit

B. 8,96 lit

(loai)

(NH4)

C. 6,72 lit

D. 4,48 lit

KvCdng d i n giai

=> D a p a n A .

P T P U C O the

CSu 30: Mot binh cau dung tich 448ml diTdc nap day oxi roi can. Phong dien de
ozon hoa, sau do nap them cho day oxi roi can. Khoi lifdng trong hai trU'cJng

xay

ra

c i 2 + 2 r - • 2cr + i2(i)
Cl2+2Br" ->2Cr+Bi^(2)

hcJp chenh lech nhau 0,03gam. Biet cac the tich nap deu d (dktc). Thanh phan
Neu phan iJug (1) xay ra hoan toan, khoi lU'dng muoi giam:

% ve the tich ciia ozon trong hon hdp sau phan vtng la:
A. 9,375%

B. 10,375%

C. 8,375%

D. 11,375%.

0,2(127-35,5) = 18,3 (g)
Khi ca hai phan ufng (1) va (2) xay ra hoan toan khoi li/dng muoi giam:

Hvtdng d i n giai
The tich binh khong doi, do do khoi liTdng chenh lech la do siT ozon hoa.
Ci? 1 mol oxi difdc thay bkng 1 mol ozon khoi li^dng t3ng 16 g

0,2(127 - 35,5) + 0,4(80 - 35,5) = 36,1 (g)
Theo de bai, khoi liTdng muoi giam 93,4 - 66,2 = 27,2 (g)

1* ,

Ta thay: 18,3 < 27,2 < 36 => chiJug to phan iirng (1) xay ra hoan toan va c6
Vay khoi liTdng tang 0,03g thi so ml ozon (dktc) la: ^ . 2 2 4 0 0 = 4 2 ( m l )
16
% 0 , =—100%=9,375%
3
448

D a p a n A.

3 oxit Y (ZnO, PbO, NiO). Hoa tan b (g) Y tren trong dung djch HCl loang thu
diTdc dung djch Z. Co can Z diTdc hon hdp muoi khan c6 khoi liTdng (b + 55)
gam. Khoi lU"dng a (g) cua hon hdp X ban dau la:
B.a = b - 2 4

C.a = b - 3 2

Hvtdng dSn giai
C^c kim loai nhy c6 cilng hoa trj -> goi chung la Mhh

Goi n _

= x thi khd'i li/dng cua muoi giam: 18,3 + x(80 - 35,5) = 27,2

Br

c a u 31: Oxy hoa ho£kn toan a (g) hon hdp X (gom Zn, Pb, Ni) diTdc b (g) hon hdp

A.a = b-16

mot phin phan (Jng (2)

D.a = b - 8

<-> X = 0,2 (mol)

1

n ^ i , = ^(0,2 + 0,2) = 0,2 (mol)

=>Vc,^ = 4 , 4 8 ( 0
=>DapanD.
Cfiu 33: Cho h6a tan hohn toan a gam Fe304 trong dung djch H C l , thu diTdc dung
dich D, cho D tac dung vdi dung djch NaOH d\i, Ipc ket tua de ngoai khong
khi dd'n khoi lifdng khong do'i, thay khoi liTdng ket tua tang len 3,4 gam. Dem

47


Phuong fhAp

ky thugt gii\h BTTN H6a dgi ci/ong - v6 CO - D 5 XuSn Hung

nung ke't tua de'n kho'i liTcJng khong doi diTdc b gam cha't ran. Gia trj cua a, b
4Fe(OH)2+O2 —

lanliTdtla:
A. 46,4 va 48 gam

B. 48,4 va 46 gam

C. 64,4 va 76,2 gam

D. 76,2 va 64,4 gam.

^

2Fe203 + 4H20

Goi x,y la so' mol Mg va Fe phan ufng. SiT tang kho'i liTdng tif hon hdp A
pe) _> hon hdp B (gom Cu va Fe c6 the diT) la:
64x + 64y) - (24x + 56y) = 12,4 - 8 = 4,4

Hiring din giai
Fe304 + 8HC1 ^ 2FeCl3 + FeCl2 + 4H2O

Hay: 5x + y = 0,55(1)

F e C l 2 + 2NaOH ^ Fe(0H)2 + 2NaCl

'

Hay:x + 2y = 0,2 (2)

4Fe(OH)2 + O2 + 2H2O -> 4Fe(OH)3

-pjy (1) va (2) tinh dUdc x = 0,1 va y = 0,05

^

; '

Khoi li/dng cac oxit MgO va Fe203: m = 40x + 80y = 8

^

FeClj + 3NaOH -> Fe(0H)3 + 3NaCl

2Fe(OH)3 —

(gom Mg

Fe203 + 3H2O

niMg

'

= 24.0,1 = 2,4 g; m,.e = 8 - 2,4 = 5,6 g ^
0,15

Nhan xet: Ta thay Fe304 c6 the viet dang FeO.Fe203. Khi cho D tac dung vdi

Dap an C .

=0,75M => Dap an B.

dung dich NaOH, ket tua thu diTdc gom Fe(0H)2 va Fe(OH)3
c. Hon hdp B gom Cu va Fe di/, nc, = 0,15 mol;

De ngoai khong khi Fe(0H)2 -> Fe(0H)3
1 mol Fe(OH)2

"FeO =

"FejOj =

"Fe(0H)2

+3

.

Chatkhijrla Fe va Cu: Fe -

= 0,2(mol)

3e -> Fe

0,05

0,15

Cu -

2e -> Cu

0,15

0,3

+2

0,2 mol Fe304 - > 0,3 mol Fe203
a = 232.0,2 = 46,4 (g); b =160.0,3 = 48 (g)

Dap an A .

Cflu 34: Cho 8 gam hon hdp A gom Mg va Fe tac dung het vdi 200ml dung djch

+5

+2

CUSO4 den khi phan iJng ket thuc, thu dUdc 12,4 gam cha't r^n B va dung dich

- Chat oxi hoa la HNO3: N + 3e

D. Cho dung djch D tac dung vdi dung dich NaOH diT, loc va nung ket tua ngoai

3a

khong khi den khoi lifdng khong doi thu diTdc 8 gam hon hdp gom 2 oxit.
A. 4,8 va 3,2 gam B. 3,6 va 4,4 gam C. 2,4 va 5,6 gam D. 1,2 va 6,8 gam.
b. Nong do mol cua dung djch CUSO4 la:
B.0,75M

C. 0,5M

B. 3,36 lit

. C. 4,48 lit

a

\o phiTdng phap bao toan electron, ta c6: 3a = 0,15 + 0,3 => a = 0,15 mol
Cflu 35: Cho 4,48 lit CO (dktc) tac dung vdi FeO d nhiet dp cao mot thdi gian,
sau phan tJng thu du-pc chat ran X c6 khoi liTdng be hdn 1,6 gam so vdi kho'i

D.0,125M.

c. The tich NO thoat ra khi hoa tan B trong dung djch HNO3 diT la:
A. 1,12 lit

> N (NO)

= > VNO = 0,15.22,4 = 3,36 lit => Dap an B.

a. Khoi lu'dng Mg va Fe trong A Ian liTdt la:

, A.0,25M

= 0,1 - 0,05 = 0,05 mol

Khi tac dung vdi dung dich HNO3:

1 mol Fe(OH)3them 1 mol OH khoi liTdng tang 17g

0,2 mol Fe(0H)2-> 0,2 mol Fe(0H)3them 1 mol OH khoi liTdng tang 3,4g

n.-e

D. 6,72 lit.

liTdng FeO ban dau. Khoi lu'dng Fe thu difdc v^ % the tich CO2 trong hon hdp
khi sau phan iJng Ian liTdt la:
A. 5,6 gam; 40%

B. 2,8 gam; 25%

Hrfdng dSn glai
a. Cac phan uTng: Mg + CUSO4 -> MgS04 + Cu
Fe + CUSO4 -> FeS04 + Cu
Dung dich D gom MgS04 va FeS04, chat ran B bao gom Cu va Fe diT.

C. 5,6 gam; 50%

C. 11,2 gam; 60%

Hifdng dSn giai
FeO + C O — ^ F e + C02
•^gian,

=

mo(oxitdaphilni?ng)= ^

=

0.UniOl)

'•( •

MgS04 + 2NaOH -> Mg(OH)2 + Na2S04
FeS04 + 2NaOH -> Fe(0H)2 + Na2S04
Mg(OH)2

^°

) MgO + H2O

=> np^ = n^o^ = 0.1 ("lol)

™ F e = 0,1.56 = 5,6 gam (*)

Th'eo bao toan nguyen to: nhSnhctpkhisauphinang = nCO(ba„d^u) = 0,2 (mol)


Phuong ph^p va kg thujt giai nhanh BTTN H6a dji cuang - vO CO - D5 Xuan Hung

=^% thi tich khi C O 2 = ^\\00%

= 50%(**)

,,

TO (*) v M * * ) ^ D a p an C

' '^'^

C. 80%.

Trong mot dung djch tong dien tich diTdng ciia cac cation bllng tong dien tich
Do do ta c6: 2a + 2b = c + d => Dap an A .

D. 65%.

Vf du 2 : Mot dung djch c6 chiJa 4 ion vdi thanh phan : 0,01 mol Na"^, 0,02 mol

HrfdngdSngiai
2KMn04

F

Hrfdng dan giai
cua cac anion.

cha't r^n. % khoi liTdng kali pemanganat da bi nhiet phan la
B. 70%.

B. a + b = 2c + 2d
D. 2a + b = c + 2d.

'

Cfiu 36: Nung 47,40 gam kaH pemanganat mot thdi gian tha'y con lai 44,04 gan^

A. 50%.

/V. 2a + 2b = c + d
^ a + 2b = c + d

Mg^^ 0,015 mol SO^^ , x mol C P . Gia tri cua x la

'° > K2Mn04 + Mn02 + O2 T

Bo giam khoi liTcfng cua chat ran = mQ^ = 47,4 - 44,04 = 3,36 gam

3

n^^ = 3,36: 32 = 0,105 mol =0 m K ^ „ o ^ tham gia = 0,105.2 = 0,21 mol

A. 0,015.

B. 0,035.

C. 0,02.

D. 0,01. * ' ^ ^
a< c .

HiMngdSngiai
Ap dung djnh luat bao toan dien tich ta c6:

:\

0,01.1 + 0,02.2 = 0,015.2+ x . l ^ x = 0,02

^

% n i ^ „ o ^ phdn urng =

'^^'^^

. 100%= 70%

Dap an B

=> Dap an C
Dang 2: Ket hop vd'i djnh luat bao toan khoi lii^ng

Chuygn etc 5 .

PHLfONG PHAP AP DgNG DjNH LUAT BAO TOAN DIEN TfCH
A. NOI DUNG PHl/dNG PHAP

Vi du 3 : Dung djch A cMa hai cation la Fe^"": 0,1 mol

anion la C P : x mol va SO^" : y mol. Dem c6 can dung dich A thu diTdc 46,9
gam hon hdp muoi khan. Gia tri cua x va y Ian lu'dt 1^:
A. 0,6 va 0,1

I. N O I D U N G :

^

B.0,3va0,2

- He qua; Tong so mol cation = tong so mol anion
!S'^cation
II. PHAM V I A P

S''anion

DVNG:

- Dinh luat bao toan dien tich diTdc dp dung trong cac triTdng hdp nguyen lii,
phan tilf, dung dich trung hoa dien.
- Trong phan uTng trao doi ion ciia dung djch chat dien li tren cd sd cua djnh
luat bao toan dien tich ta thay c6 bao nhieu dien tich du'dng hoac am cua cac
ion chuyen vao trong ket tiia hoac khi tdch ra khoi dung djch thi phai tra lai
cho dung dich bay nhieu dien tich du'dng hoac am.

C. 0,5 va 0,15

D. 0,2 va 0,3

Hufdng dan giai

- Trong phan tfng trao doi ion, va trong mot dung djch:
Tong dien tich am = tong dien tich du'dng

AP"": 0,2 mol va hai

Ap dung djnh luat bao toan dien tich ta c6:
0,1.2 + 0,2.3 = x.l +y.2

x + 2y = 0,8 (*)

Khi C O can dung dich khoi li/dng muoi = I kho'i lu'dng cac ion tao muoi
0,1.56 + 0,2.27 + x.35,5 + y.96 = 46,9 => 35,5x + 96y = 35,9 (**)
Tilf (*) va (**)=^ x = 0,2; y = 0,3

•

=> Dap an D.
^' du 4 : Chia hon hdp X gom hai kim loai c6 hoa trj khong doi thanh 2 phan
i . ; tM).^

b^ngnhau.

Ph^n 1: Hoa tan hoan toan bhng dung dich HCl diTthu di/dc 1,792 lit H2 (dktc).
: Nung trong khong khi diT thu duTdc 2,84 gam hon hdp r^n chi gom cac
oxit. Khoi liTdng hon hdp X la

B. B A I T ^ P MINH HQA
Pang 1: Ap dung drrn thuan djnh luat bao toan di?n tich
Vf dy 1 : Trong mot dung dich c6 chiJc a mol Ca'*, b mol M'j'*, c mol Ci , d m<
NOi . Bieu thuTc iicn he giu.i a , b, c, d la:

A-1,56 gam.

B. 1,8 gam.

C. 2,4 gam.

D. 3,12 gam.

Hifdng dan giai
^ S n xet: Tong so mol x dien tich ion du'dng (cua hai kim loai) trong hai phan la
b^ng nhau

Tong so' mol x dien tich ion am trong hai phan cung bhng nhau
51