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HA

Pi6l

Npi

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Chiu
Gidm

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

khoi lUdng m u o i


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 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 .

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

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


+ 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


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
nw!

1
r •: ^

> Na2C03 + H2O (1)


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



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

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

= 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

'

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
%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

^



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


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


(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



,

. ,,

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


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


(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
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.

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

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>.

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

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



•

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^

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

''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^^^

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.


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

.(: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


:

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

_

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:

^

? , . « ~

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-

'
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


^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.
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 Dap an C .
OH
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)


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,



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.



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

— ^



0 32

mo = 0,32 gam. =>

UQ =

^— = 0,02

mol

,

r^tiiH

«

'

=> ( " c o +

"H

mol.

=> 16,8 = m + 0,32

moxit

.

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.


-> 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.

Dap an C

Na 2C03

0,07x106
100


-

^

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

.


•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.


•••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
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


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

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?^ ^

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


-

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

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


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
,^
=—

> 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 =
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

'"^^
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

(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:


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


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.