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tap:
ban

D i e u nay giup ta giai bai toan h6a hoc mot each ddn gian, nhanh chong

PHUNG QUOC BAO

:

uO

Bien

tham gia phdn iing bang tong khoi lU0ig cdc chat tao thdnh "

chii y den cac chat ket tua, bay hdi, dac biet la khoi li/dng dung djch).

ie

bien

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

ban:

2. Cac d a n g bai toan thifoTng gap

KHANG V I E T



1, NOi d u n g

ro

hdnh:

/g

phdt

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

//e ^Ma i ; B a i toan: K i m loai + axit -> muoi + khi
'^aniontaomud'i

bo
ok

^

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

fa
ce

- 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



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 ,
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ILAJ

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-> khoi lUdng m u o i

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In xong va n o p

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

- 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)

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

qua
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)

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)

a25m
07
a25m
^^5
,
, .
56
56
Ap dung DLBT nguyen to'N ta c6 : 0,7 = 2.^'^^"^ +0,25 m = 50,4 (g)
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
»*
'*

oc

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

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


^ "02 ^^"^'^

.c
om

Ap dung DLBTKL ta c6: m^^ = moxu -

giai

01

=> D a p a n

4. C a c btfdfc giai

Hi^dng dSn

/

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.

*

hhX


'
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:


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

+ 0,02

= 0,12

Ta thay; n; = 0,12

n;

iH

cho = 0,06 mol

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

uO

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

Trong 2m gam hon hdp Na, K c6

(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)

fa
ce

Cfiu 5:

.c
om

0,06

up

KHCO3

ro

CO2 + KOH

s/

=>a = 0,06

li/dng cua nitd trong X la 11,864%. C6 the dieu che diTdc toi da bao nhieu
A. 10,56 gam

B. 7,68 gam

ww
w.


OH

''^i •

' •

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

' "

'

'
\

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

'

=> D a p a n B .
Cfiu 7: Cho m gam NaOH vao 2 lit dung djch NaHC03 nong do a mol/1, thu du^dc
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.

B. 0,07 va 3,2.

C. 0,08 va 4,8.
C r ( 0 H ) 3

Ta

z:>0,l+

"c

iL

"C (trongCOj)


Da

^

(Trich de thi tuyen sinh Cao dang nam 2011)

Hifdng dSi^giai

mol

Taco: m „ , u 6 - i = m K L + m
BaCl2 + KiCOs

D. 51,5 gam.

01

0,1

C. 30,9 gam.

oc

0,1

B. 20,6 gam.

/


BaCOji

01

^m

COj^"

oc

Ba^* +

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

iH

Phuang phip

0

n . i h : : .

8

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

,

V

C. 97,80 gam.

Ta c6: n „ „r, = n „
H2SO4

H2

=^^=0,lmol
22,4

''~m

l>'.'.

r

•


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:

^

Goi cong thiJc tdng quat cua oxit la FcxOy:

A. 600ml.

2M + 2nH20 ^

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

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

M 2 0 „ : b mol

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



bo
ok

Cfiu 12: Hoa tan hoan toan 2,9 gam hon hdp gom kim loai M va oxit cua no vao

B.Ba

^j,;
B . F e j O j va 0,448

s/
.c
om

m
=> Dap an A .

A.Ca

129

D a p an B.

up
= 74,69%

ro

= 3,125.10"^m (mol)

=> 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

01

PhLiong

(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.

,

Q^U


Ta CO sddo: 2FeS2

• ^

0,12
CujS a

,

0,06

+HNO3

2CUSO4

2a

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.

nT

hi

Sd do phan iJng:

11,36 + (3x + 0,06).63 = 242x + 0,06.30 + (1,5x + 0,03). 18
= 0,16 mol

'"feCNO^jj

Hi^dng d i n giai
Ta c6: n Cr203

15,2

= 0,1 mol

152
1«' k;-

81

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

HAI =

0,1

,

. ,,

mol

mbl

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

0.1

0,15

;|

Cr + 2HC1 - > CrCl2 + H2t

= 0'28.0,5 = 0,14 mol

0,2
Ap dung dinh luat bao toan khoi Iifdng:
mj^^5o^ = mmuoi +

0,1

.AI2O3: 0,1


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)

nHci

D. 3,36 lit.

Ap dung djnh luat bao toan kho'i liTdng:

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

bo
ok

=> X

C. 10,08 lit

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

s/

=ln

up

Diravkosc(d6tathay: n

a

dung djch X (chi chilfa

/

diTcfc 1,344

diTdc

01

loang du", thu

du, thu

oc

HNO3

H N O 3 viTa

iH

dung djch

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

Da



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
2Na^ + 2 C r + H2
39,4
2X + 2nHCl
> 2X"^ + 2nCr + nH2
.

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
'2

6,72
22,4

= 0,3

mol;

DHCI = 2 n(.Q

= 0,6

)j


iL

Theo djnh luat bao toan khoi lufdng

Ma: n

gam

s/

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

Ta

Ta c6: mnn sa., = mnn tnrdc = 5,4 + 6,0 = 11,4

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

ie

Htf^ng d i n giai

D.3,31.

Hvldng d§n giai:

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

A. 2,24 gam

= ^8,2 + 0,3.208 - 69,9 = 30,7 (g)

oc

J2

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

/

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

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

no(trongoxit)

=nco =

^C02~

"CaCO,

moxit=

Hu

"77^" ^''^
lUU

ma: moxit= mki,„ioai + m o x i

;

.>

•

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


=> Dap an B . •

A. 3,07.; f ,

C. 1,12

,2-

Ta c6 sd do: H2SO4

ww
w.

Theo djnh luat bao toan nguyen to ta c6:

B. 1,008

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

fa
ce

MxOy

^,


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


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?

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

^

= 0,35

mol

"H, = n

D . 39,65.

^

= 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.

A . 20,6

H2 (dktc).

Co can dung djch thu diTdc m gam muoi khan.

iL
s/
D a p a n B .

B. 12,6g

mol

,

oc

»'v«> ''ifv

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^^^

A . 16,2g

= 0,02

56,5

, .



- vO

iH

CLiong

Da

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

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

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

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,

Dap an D.

CSu 20: Hoa tan hoan toan 20 gam hon hcJp gom Mg va Fe vao dung dich axit


m^

= 0,1. 27 = 2,7

•"Fe o = 9.66 - 2,7 = 6,96 gam => Dap an A.
= 2 . — = 0 , 1 mol
AI2O3

'


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 - ' -

^ 0 (trong F e , 0 y

= —

0,03

AI2O3 -H 3H20


> A1(0H)3 + NaHCOj

Do do:

? , . « ~

> NaA102 + - H 2

NaAlOz + CO2 + 2H2O

khi sau phan iJug qua dung djch Ca(0H)2 diT, thay tao ra 30 gam ket tua. Kho'i
^

(1)
r'

0,02

CSu 22: Thdi 8,96 lit CO (dktc) qua 16 gam FcxOy nung nong. Dan toan bo liTdng
lUUng s^t thu duUc la:

' '

0,02

2Ai(OH)3

•



thay thoat ra 14,56 lit Ha (dktc). Kho'i liTdng hon hdp muoi clorua khan thu

:

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

^. ,

uO

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

A. 48,75 gam

gmb r;

b. CongthiJccuaoxitsatla:

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

nT

.(:m^

A. 6,96g va 2,7g

•^•^'^ ^ ^ "^"^

Ap dung djnh luat hio toan khoi lUOng: m i d +
CO2
Hco =

-1) ir .

( H ozon hoa) = 0,039375.32 = 1,26 (g) =^ Dap an B.

CO2 + Ca(0H)2 ->• CaCOj + H2O


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 m = 1 (g)
M R = 64 - > R la Cu => D a p a n D .

33

%mKCi03(A)

#r

•


Hi/(}ng d§n g i a i

Ca(C103)2

r7.

uO

CaCl2 + 2O2

(3)

ie

(2)

=»DapanB.

KC1(^,

KCl

s/
up

CaCl2
(A)

KCl


D . 0,448 l i t va 16,48 gam.

CO + O

>

H2 + O

>



UQ =

^— = 0,02


= 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

C a C l j + 3O2

CaCl2

H6nhdpBJ0,18

Ta

KC103

B. 2,48 gam.

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

3
KCl + - 0 ,
2 2

t"

Da

iH

D . 58,55%.

hi

B. 56,72%.

nT

A . 47,83%.

/



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

/

01

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

hi

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

mo = 1,6 gam.


[y = 0,03 mol

—

0,01x72x101

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

, : i f d i it.

%Fe203 = 86,96%

Ipc bo ket tua diTdc dung dich X. Tiep tuc cho 50 gam dung djch H2SO4 9,8%
vao dung djch X thay ra 0,448 lit khi (dktc). Biet cac phan iJ-ng xay ra hoan
toan. Nong do % cua dung dich Na2C03 va khoi li/dng dung djch thu du'dc sau
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

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

Dap an C

Na 2C03

0,07x106
100

'

+ 2NaCl


B. 35,2 g a m .

,1! jU

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

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

FeO + CO

iH

Da

nc02

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 .

^

nT

> H2O.

.



Hi^ngdSngiai

2 24

Ta c6: n hh(CO+H2)

C. 13,04%.

xlOO% =7,42%

•m.umo.


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
' ' fcfi

D. 16,04 gam.

M+
•••fott'i •fJ^'G.ti"-

^^I'l'O '

• Ap dung djnh luat bao toan nguyen to'Fe:
1 =:>3n = 0,3

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

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

'

0;


=> Dap

* r^n B gom 4 chat. Hoa tan chat r^n B b^ng dung djch HCl diT thay thoat ra

=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,

i A. 0,006.

(1)

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

"pe (Fe304) ~ " F C (FeS04) "

•f B so mol oxit sat tir bang 1/3 tong so mol sit (II) oxit va s^t (III) oxit.

M20„

—>

iH

iMfljO')*''.

' n mol

^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

oc

tang giam kho'i liTdng va bao toan khoi lUdng la 2 anh em sinh doi. Tuy nhieti,
PhUdng phap tang giam khoi lUdng thUdng dUdc suf dung trong cac bai toan
hSn hdp nhieu chat.

Xem xet sU tang hoac giam cua AM va Am theo PhU'dng trinh phan tfng va

up

.

ro

Lap phUdng trinh toan hoc de giai.

/g

HQA

.c
om

CSu 1: Cho 9,125 gam muoi hidrocacbonat phan iJug het vdi dung dich H2SO4
muoi hidrocacbonat la
B.Mg(HC03)2

C. Ba(HC03)2

bo


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

D. Ca(HC03)2

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

ww
w.

Htfdngd§ngiai

0,03

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

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

/

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

s/

theo dffkien bai toan

;


Vay: xmol Cu(N03)2
Theoptpi?:

Xac dinh dung moi quan he ty le giffa chat can tim va chat da biet (nhd van

-

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

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

^mMihmmn.

3. C a c bride giai
-

Ta c6: Imol Cu(N03)2

Da

-

iH

tuy tifng bai tap ma phUdng phap nay hay PhU'dng phap kia se la \iu viet hdn.

hi

-


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)
,
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)


hi

nT

Fe(N03)2 + Cu

'"^^

Ta

iL

ie

uO

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

s/

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

8 61
Ta c6: nNaci = nAgci = —— = 0,06 mol => niNaci = 0,06.58,5 = 3,5 Ig
143,5

= V2

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

0,03

ro

85

/g

if'frvf

/

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

01

Phi/Dng phAp

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
'a khong dang ke).


Htfdng din giai

a

r-'

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

.

/

a
4

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

'

Wdng muo'i tang: 71 - 60 = l i g

oc

• •'"
,
A a

Fe203 + 2CO2 (2)

0,5M.

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

ie

uO

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

iL

C. 85,30%

D. 12,67%.

Hrfdng dSn giai

a

b


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

.c
om

. w; .tf,;,

Zn + C U S O 4

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

C.BAITAPAPDVNG
Cfiu 1: Cho 2,81 gam hon hdp gom 3 oxit Fe203, MgO, ZnO tan viTa dii trong
300m! dung djch H2SO4 0,1M thi khoi iiTdng hon hdp cac muo'i sunfat tao ra la:
A. 3,81 gam

B. 4,81 gam

C. 5,21 gam

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
^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 =^


Hiidng din giai

Ta

up

ket thtic cac phan i?ng, loc bo dung dich thu diTdc m (g) chat r^n. Thanh phan

s/

CSu 8: Cho m (g) hon hdp bpt Zn va Fe vao li/dng dU dung djch C U S O 4 . Sau khi

A. 90,27% .

B. 1,28 gam

"

2: Dem nung mot kho'i liTdng Cu(N03)2, sau mot thcJi gian thi tha'y kho'i
•i'