Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học phần 1

TS. NGUYEN VAN HAI (Chu bien)
NGUYEN NAM TRUNG - TRAN THE NGA - NGUYEN THj THU HA

Cam naligFluyen thi dai hoc
18 CHUYEN OE

^

HOA HOC
^ He thons cac phUdng phap siai nhanh bai tap hoa hoc.
^ Duns cho on tap va thi tot nshiep THPT.,.
^ Luyen thi vao Dai hoc va Cao dans.

rW

VJEWTIWHSINHTHUAN

D m MHA yi lAT RAN RAI HOC QUOC GIA HA NOl


Cty TNHH MTV DWH Khang Vi^t

r h u y e n de 1

,

C A C PHlrtl^NC P H A P G I A I N H A N H

MUC LUC
huyen di 1. C a c phucmg phap giai nhanh

?

1. P H l / O N G P H A P B A O T O A N K H O I L l / Q N G
3

huyen de 2. C a c axit v6 ca d i l n hinh

71

huydn de 3. T i n h chat c u a cac hidroxit

99

.

^. ,

a. N g i d u n g
« ^.
Tong khoi luthanh sau phan ung.
, m^i^^mm,
b. H# qua
Khoi lugng muoi = Tong khoi luong cac ion tao muoi.

-"^-''i''!'^

-

huySn di 4. T i n h chat cua cac mu6'i v6 c a

111

huyen d^ 5. T i n h chat c u a cac oxit

131

huyen di 6. C a c nguydn to phi kim di^n hinh

148

vi DU M A U

huydn d6 7. D a i cufotng v6 k i m loai

173

V i d u 1: H o a tan hoan toan 1,794 gam kim loai kiem M vao 400ml dung dich

huyen d^ 8. An mon k i m loai - D i ^ u chS' k i m loai

190

huyen de 9. K i m loai kiem - K i e m tho - Nhom

204

huyen de 10. Sat - C r 6 m - D o n g

231

huyen d l 11. C a c l i thuy^t c o ban ciia hoa hoc

253

huyen de 12. S u dien l i - A x i t , bazof, muoi

268

huyen de 13. A n c o l - Phenol

285

huyen

301

14. Andehit - A x i t cacboxylic

huyen de 15. Este - L i p i t

325

huyen d6 16. Cacbohidrat

341

huyen d6 17. A m i n - A m i n o axit - Peptit

352

huyen di 18. Polime v a vat lieu polime

370

e thi thir dai hoc

378

Khoi luong dung dich sau phan ung = Khoi \ugng dung dich truoc phan
ling - (khoi luong chat ket tua + chat bay hoi)

H C l 0,1M. C o can dung dich sau phan ving thu du(?c 3,86 chat ran khan Y .
K i m logi kiem M la
A. N a .

y,,?j,, ,
B. K.

C . Rb.

D. Li.

y v .

Laigidi:

, ,

Cac phan ung hoa hoc:

M + HCl

> M C I + -Hi
2

M +H2O

> M O H + ^Hi
2

-

.

Nhan xet: Bao toan khoi lu^ng -> m y = mM+ m J . + m^^^.
^

m ^ ^ . =3,86-1,794-0,04.35,5 = 0,646 gam ^

•

n^^^. = 0,038 mol.

1 794
- n M . n H a - n „ „ . = 0,078 ^ M . ^ . 2 3

(Na).^

^a:,-.,

—> D a p an A.
Vi

2: Cho 100ml dung dich H3PO4 a molA vao 100ml dung dich K O H 2 M
thu dugc dung dich Y c6 chua 15,44 gam hon hop muoi. G i a tri cua a la
A. 0,75.

B.1,00.

C.0,50.
Lai

D . 0,80.

gidi:

Nhan xet: V i Y chua hSn hop muoi - > chac chan c6 chiia muoi axit
- > K O H phan ung het.
S a d o p h a n u n g : H3PO4 + K O H

> Muoi + H2O.

Bao toan khoi luong: 0,la.98 + 0,2.56 = 15,44 + 0,2.18
• *

- > Dap a n D .

,

i

a = 0,80.^•' ' ^^
|

.

.


Cty TNHH MTV DWH Khanq Vigt
dm nangflnluy$n thi d
Vi

H6a hpc - Nguygn Van H5i

3: Dot chay hoan toan 17,4 gam h6n hop Mg, Z n va A l trong khi O2 (du)
thu du(?c 30,2 gam hon hgp oxit. The tich k h i O2 (dktc) da tham gia phan
ling la
A. 4,48 lit.

D. 11,20 lit.
Lmgidi:
'''''''
6 bai nay, cac em khong the giai bang vi?c viet phuong trinh phan utig va
datsomol. : ,
,'
,
,
"

O day, cac em can ap dung bao toan khoi lugng:
m k i m M

->

+

B. 8,96 lit.

moxi =moxit

->

= ^ ^ = 0'4 mol

-17,4

= 12,8

gam.

= 8,96 lit ^ Dap an B.

^
'

^

m o x i = m o x i t —> m o x i = 3,33

12
= TIT = 0,0375 mol

— 2,13

= 1,2

gam.

;

+ 2H^

>

Truoc

het, c a n x e m axit

c6

p h a n l i n g het h a y

-

^^^^

= 0,5.1 + 2.0,5.0,28 = 0,78

Khoi

lugng muoi

= m^|3+ +

se

gap kho k h a n

b a n g e a c h s o s a n h so
> H2.

mol.

, , ,

n^+ het -> X c h i c h i i a c a c m u o i .

^^i-

^^^^2+ +

, '

+ ^^^2-

'

= 7,74 + 0,5.35,5 + 0,14.96 = 38,93 g a m . , ; j v ^ ^ ^ y . , _
-¥ Dap a n A.

'
> V i dv 7: Nung h o n h g p bgt g o m 15,2 g a m Cr203 v a m g a m A l a nhi?t d g cao.
Sau k h i p h a n u n g h o a n toan, t h u d u g c 23,3 g a m h o n h g p r a n X. Cho X p h a n
l i n g v o i axit H C l ( d u ) thoat r a V lit k h i H2 (6 d k t c ) . Gia tri c i i a V l a

A. 7,84.

B.4,48.

C. 3,36.

D. 10,08.

Lim

t r u o c he't cac e m c a n t i n h d u g c so m o l

Al

ban dau.

y rang, trong p h a n u n g n h i f t n h o m , t h u o n g a p d u n g d i n h luat bao

t o a n k h o i l u g n g , cia t h e :
mc^03 + i"Al=mx

m A i = 23,3 -15,2 = 8,1 g a m ->

n ^ r 0,3 m o l .

.0
VHCI

= 0,075 lit = 75ml.

-> Dap an C.
V i dv 5 (A-12): Hoa tan hoan toan 2,43 gam hon hgp gom M g va Z n vao mpt
, lugng vira d i i dung djch

khong

m o l H " b a n d a u v a s o m o l k h i H2 b a y r a theo ti 1$: 2H*

H2O.

Suy ra: n^+ = 0,15 mol -> nnci = 0,15 mol ->

H2SO4

loang, sau phan ling thu dugc 1,12 lit

(dktc) va dung dich X chiia m gam muoi. Gia tri ciia m la
A. 4,83 gam.

n e u cac e m g i a i theo p h u o n g t r i n h p h a n l i n g

v i can viet d e n 4 p h u o n g trinh.

Nhan xet: Bai n a y

no = 0,075 mol

V - ( o x i t ) = 0-075 mol.

Nhan xet: Bai n a y

D. 77,86 g a m .

Lai giai:

Khi cho oxit bazo tac dung vai axit tao ra nuoc:
02-

B. 103,85 g a m .
C. 25,95 g a m .
Lai giai:

Mat k h a c : n u , = 0,39 m o l ->

Lai gtat:
Nhan xet: bai nay cac em nhat thiet phai ap dung bao toan khoi lugng de
tinh kho'i lugng oxi tham gia phan ling:
+

A. 38,93 g a m .

nj^+ = H H C I + 2nH2S04

V i dv 4 (A-08): Cho 2,13 gam hon hgp X gom Mg, Cu va A l 6 dang bot tac
dyng hoan toan voi O2 thu dugc hon hop Y gom cac oxit c6 khoi luong 3,33

gam. The tich dung dich HCl 2M vira dii de phan ling het voi Y la
A. 60ml.
B. 45ml.
C. 75ml.
D. 80ml.

mkimioai

H2 ( d k t c ) . Co c a n d u n g d j c h X t h u d u g c l u g n g m u o i k h a n l a

C. 17,92 lit.

m o x i = 30,2
V02

V i dv 6 (CD-08): Hoa t a n het 7,74 g a m h o n h g p bgt Mg, A l b a n g 500 m l d u n g
d i c h h o n h g p HCl I M v a H2SO4 0,28M t h u d u g c d u n g d i c h X v a 8,736 lit k h i

B. 5,83 gam.

C. 7,33 gam.

^
D. 7,23 gam.

Lai giai:
O bai nay, cac em c6 the giai chi tiet dua theo phan ling hoa hgc.
Nhan xet: nH2S04 = r'H2 = 0,05 mol.
So do phan ling: K i m l o a i + H2SO4
> Muoisunfat + H2

Bao toan kho'i lugng cho so do tren:
^ 6I>''•
->• m = 2,43 + 0,05.98 - 0,05.2 = 7,23 gam '
ii£0tOB>;
->DapanD.
, , ^'^ ^^-^^^ ' ^

H2

Phuong trinh h o a h g c : Cr203 + 2A1
Mol: ai
0,2
X g o m : A l d u = 0,1 m o l ; Cr = 0,2 m o l .
3
Khi X tac d \ i n g v o i axit: A l ^
->

n H 2 = 0,35

mol

VH2=

V i d\ 8 (B-09): Cho 100ml
0,5M,

lit - »

dung djch

2Cr + Al2C)3.
0,2
0,1

(j^p.^v,-.^

'

'

.

^^-^

v a C r - > H2

- H 2
7,84

>

Dap

a n A.

K O H 1,5M

vao

200ml

d u n g d j c h H3PO4

t h u d u g c d u n g d j c h X. Co c^n X t h u d u g c k h o i l u g n g c h a t r a n k h a n l a

A. 15,5

gam.

B. 18,2

gam.

C. 12,8

gam.

D. 16,4

gam.

Lai giai:
" K O H = 0,15 m o l ; nH3P04 = 0,10 .

Nhan xet: Cac em c6

the xet ti le m o l

t r i n h p h a n u n g va d a t so m o l de g i a i .

K O H va

H3PO4, s a u

do v i e t 2

phuong


V i d u 11 (B-12): Hon hgp X gom 0,15 mol vinylaxetilen va 0,6 mol H 2 . Nung

Tuy nhien, ne'u ap dung bao toan kho'i lugng cho s o do:
H3PO4

+ KOH

:

nong hon hgp X (xiic tac Ni) mot thoi gian, thu dugc hon hgp Y c6 t i khoi so

> Muoi + H2O

Taco: mH3P04 + rnKOH=

voi H 2 bang 10. D i n hon hgp Y qua dung dich brom du, sau khi phan ung

niHjO

xay ra hoan toan, khoi lugng brom tham gia phan ung la

J^y^^yr^^

,

"H2O=0,15.

A. 0 gam.

B. 24 gam.

—> Dap an A.
V i du 9: D u n nong hon hgp khi X gom 0,06 m o l C2H2 va 0,04 mol H2 v o l xiic
tac N i , sau mot thoi gian thu dugc hon hgp khi Y. Dan toan bg Y Igi t u t u

D. 16 gam.

Bao toan khoi lugng: mY = mx = 0,15.52+ 0,6.2 = 9 gam.

5

Mat khac: M Y = 10.2 = 2 0

" '^

"2=

= 0/45 mol

,
•

dktc) CO t i kho'i so voi O2 la 0,5. Khoi lugng binh dung dich brom tang la

3.0,15 = 0,45 mol.

B. 1,32 g a m .

•

C. 1,64 gam.
Laigiai:

• ,,^.J^:n.

Gia su so mol H2 tham gia phan umg = a mol

D. 1,20 gam.
;;(;';

ling = a mol.
- > a = 0,75 - 0,45 = 0,3 mol.

,

- > So mol lien ke't 7i d u = 0,45 - 0,3 = 0,15 mol = So'mol Br2 phan ung.
Khoi lugng brom phan ung = 0,15.160 = 24 gam.

mz = 0,02.16 = 0,32 gam.
Luu y: Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thy.
Bao toan khoi lugng, ta c6:

Kho'i lugng binh brom tang = mv - mz = 1,64 - 0,32 = 1,32 gam. ,.j .j-^,^, •^i^^.j^.^
Dap an B.
V i d u 10 (A-10): Dun nong hon hgp khi X gom 0,02 mol C2H2 va 0,03 mol H2
trong mot binh kin (xuc tac Ni), thu dugc hon hgp k h i Y. Cho Y Igi t u t u
binh tang m gam va c6 280ml hon hgp khi Z (dktc) thoat ra. Ti kho'i aia Z so

ia08. Gia tri ciia m la

A.a585.

B. 0,620.

•
C. 0,205.

,iaw;.

: ,
D. 0,328.

Lbi gidi:
Nhan xet: Bao toan khoi lugng: mv = mx = 0,02.26 + 0,03.2 = 0,58 gam.
Mat khac: n z = ^
= 0,0125 mol va M z = 10,08.2 = 20,16
22,4
?
mz = 0,0125.20,16 = 0,252 gam.
Nhanthay:
'r^^.^
i

i,*!.'.a " ;
.m»s;c,ci ;,/•> •
Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thu.
Bao toan khoi lugng: m = mv - mz = 0,58 - 0,252 = 0,328 gam.
—>^ Dap a n D .

,

—> Dap an B.
V i d u 12: Hon hgp X gom 0,1 mol etilen, 0,2 mol axetilen va 0,5 mol H 2 . Nung
nong hon hgp X (xiic tac Ni) mgt thoi gian, thu dugc hon hgp Y c6 t i khoi
so v o i H 2 bang 10. Dan hon hgp Y qua dung dich brom du, sau khi phan
ling xay ra hoan toan, tha'y c6 m gam brom tham gia phan ung. Gia trj cua
A. 16.

B.32.

vao binh dung djch brom (du), sau khi ke't thiic cac phan ung, khoi lugng
voi H2 la

So mol lien ke't n phan

Mat khac, so mol khi giam = so mol H2 phan ung = n x - n Y nkfusi'^'^

22,4

,

,

Nhan xet: vinylaxetilen c6 chua 3 lien ke't 71 —> So' mol lien ke't K ban dau =

Nhan xet: Bao toan khoi lugng: mY = mx = 0,06.26 + 0,04.2 = 1,64 gam.
0 448
or,, Mat khac: n z =
= 0,02 mol va M z = 0,5.32 = 16

%

C. 8 gam.

;

qua binh dung dung dich brom (du) thi con lai 0,448 lit h6n hgp khi Z (6
V - A . 1,04 gam.

,

•.)

Lai gidi:

mx = 0,1.98+ 0,15.56-0,15.18 = 15,5 gam

I'

1TV DVVH Khang Vi$t

Cty TKi,

Ca'm nang 6n luyfn thi d^i hpc 18 ctiuySn dg H6a hpc - Nguygn Van HSi

C.8.
Laigiai:

D.24.
" ' , ' ; ,,s*^->r::

':^:s,rf'^~

Bao toan khoi lugng: mY = mx = 0,1.28 + 0,2.26 + 0,5.2 = 9 gam.
Mat khac: M Y =10.2 = 20

n z = =

0,45 mol

"

'

Nhan xet: etilen c6 chiia 1 lien ke't 71, axetilen c6 chiia 2 lien ke't 71 - > So
mol lien ke't 7t ban dau = 1.0,1 + 2.0,2 = 0,5 mol.
Gia su so mol H2 tham gia phan ung = a mol - > So' mol lien ke't n phan
ling = a mol.
Mat khac, so mol khi giam = so mol H2 phan ung = nx - nY

^ ^

- > a = = 0,8 - 0,45 = 0,35 mol.

- > So mol lien ke't 7t d u = 0,5 - 0,35 = 0,15 mol = So mol Br2 phan ung.
Khoi iugng brom phan ung = 0,15.160 = 24 gam.
—> Dap a n D .

'
7


Cty TNHH MTV DWH Khang Vijt

dm nang On luygn thi dgi hpc 18 chuySn dg H6a hpc - Nguygn van HJi

V i dv 13: Cho 2,1 gam h6n hgp X gom hai amin (no, don chiic, dong dang ke
tiep) phan ling het voi dung dich HCl (du), thu du(?c 3,925 gam hon hgp
muoi. Cong thuc cua hai amin trong X la
A. CH3NH2 va C2H5NH2.

B. C2H5NH2 va C3H7NH2.

C. C3H7NH2 va C4H9NH2. i '> •

D. C H 3 N H 2 va (CH3)3N.

Y^ic

Tuy nhien, bai nay dugc giai nhanh chong khi ap dung bao toan khoi
lugng:

maxlt + mbazo= mmuol + mnuoc

_^ mH20= 3,6+ 0,06.56+ 0,06.40-8,28 = l,08gam^> nH20= 0,06 mol.
Mat khac: n H , o =
^

S 0 >.

Lot gidi:

36
= 60
0,06

=

Max.t

Axit la C H 3 C O O H

• . .V

-> Dap an B.

Nhan xet: Loai D vi hai amin khong dong dang ke tiep. Cac phuong an con

V i du 16: Dot chay hoan toan hon hgp X gom hai este dong phan can diing

lai deu la amin don chuc, bac I.

7,84 lit khi O2, thu dugc 6,72 lit khi CO2 va 5,4 gam H2O. Neu cho m gam

Khi cho X tac dung voi axit HCl:

X tac dung vua dii voi dung dich NaOH, c6 can dung dich sau phan ung

R-NH2 + HCl

> R-NH3CI

Bao toan kho'i lugng:

mamin

' i<-

+ mnci = mmuoi

•

>. '

ling ete hoa hoan toan Z, thu dugc m gam ete. Cac the tich khi deu do 6

;

dieu ki?n tieu chuan. Gia tri cua m la

mnci = 3,925 - 2,1 = 1,825 gam -» nnci = 0,05 mol = namm
^

A. 2,65.

Mamin = 7 ^ = 42 ^ R - N H 2 = 42.
0,05
*

Dap an A.
V i dij 14: Dot chay hoan toan m gam hon hgp X gom 3 ancol (don chuc, thugc
Cling day dong dang), thu dugc 26,4 gam khi CO2 (dktc) va 19,8 gam H2O.
Neu thuc hien phan ling ete hoa m gam X (hieu suat 100%) thi tong khoi
lugng ete thu dugc la
B. 14,80 gam.

C. 10,92 gam.

D. 12,90 gam.

Lai gidi:
Nhan xet: n^jo = 1/1 > ^C02 = 0,6 ^ X gom cac ancol no, mach ho va:
"X =nH20 -rtC02 =1,1-0,6 = 0,5mol -> n(0)x =nx =0,5
Bao toan kho'i lugng:

+

nHjO

i^H20 = 0,25 mol.

D. 3,25.

este la C n H 2 n 0 2 .

= 0,3 mol.

^^^^ este deu no, dan chuc -> Cong thiic cua 2
*

'

•

" ' ••.:

Bao toan nguyen to'oxi: 2n ^^te 2n
_> n = ^^^22. =
Heste

=3

= 2n CO2

^ H2O

~ ^ " este

0,1 mol.

2 este trong X c6 cong thuc phan tu la C3H6O2.

0,1

-> Cong thuc cau tao 2 este la: HCOOC2H5 va CH3COOCH3.
Phan ling hoa hgc:
HCOOC2H5 + NaOH — ^

HCOONa + C2H5OH

C H 3 C O O C H 3 + NaOH —!—> CHsCOONa + C H 3 O H
nNaOH = neste

.'

= nz = 0,2 mol.
meste+mNaOH

^

= mv + mz

-> 0,1.74 + 0,1.40 =7,85 + mz ^ mz =3,55 gam.

Ht:k{OCC;i

Nhan xet: Trong phan ung ete hoa ta c6:
mete

= 17,4 - 0,25.18 = 12,90 gam -> Dap an D.

V i d u 15 (B-08): Cho 3,6 gam axit cacboxylic X (no, don chiic) tac dung hoan
toan voi 500 ml dung dich gom K O H 0,12M va NaOH 0,12M. Co can dung
dich thu dugc 8,28 gam chat ran khan. Cong thiic cua X la

A.C2H5COOH.

Nhan xet: UQQ^ = nH20

Bao toan kho'i lugng:

Bao toan kho'i lugng:
"^x = r^ete

C. 5,15. ' " "

n 02 = 0,35 mol; n CO2 = 0,3 mol; n

De tha'y:

mx = mc + niH + mo = 0,6.12 + 2,2.1 + 0,5.16 = 17,4 gam.
Trong phan ung tao ete thi: nx = Zn^jO

B.3,70.

^

Lai gidi:

-> R = 26 -» Hai goc hidrocabon la C H 3 - va C2H5-.

A. 8,40 gam.

thi thu dugc 7,85 gam chat ran khan Y va hon hgp ancol Z. Thyc hi|n phan

B.CH3COOH.

C.HCOOH.

D.C3H7COOH.

Lai gidi:
Nhan xet: Theo bai, axit phan ung hoan toar\i baza c6 the con du, do
vay neu giai dua vao phuong trinh phan ung se rat kho khan.

1
„
,
H2O = T n z n H20 = 0,05 mol.

2
Bao toan khoi lugng: mz =

mete

n ,mw>'^''iOfU nAoi • i

; y 7.- .

+ m H20

-> m = mete = 3,55 - 0,05.18 = 2,65 gam -> Dap an A.

,s :sv

V i d^ 17: Xa phong hoa hoan toan 0,1 mol este X (dan chiic, mach ho) bang
100 g^m dung dich M O H 11,2% (M la kirn loai kiem). Co can dung dich sau
phan ling, thu dugc 15,4 gam chat ran khan, dong thai ngung tu phan hoi
bay ra tha'y tao thanh 92 gam chat long. Cong thiic cua X la
9


Ca'm nang On luygn thi dgi hpc 18 chuyen dg H6a hpc - Nguygn VSn Hii

Cty TIMHH MTV DVVH Khang Vigt

A. CH3COOCH3.

B. CH3COOC3H7.4

C. CH3COOC2H5.

D . C2H5COOC2H5.

Lmgidi:
Ggi cong t h u c cua chat beo la (RCOO)3C3H5.

Lot gidi:

(RCOO)3C3Hs + 3 N a O H
" " '

''

•

- > kho'i l u a n g d u n g m o i (nuac) = 100 - 11,2 = 88,8 g a m .
Phan u n g hoa hoc:

gfi;

RCOOR'

rfsB<.m
S^'-

M o l : 0,1

a

Mol:
r i f / v i •«

iv;y£;f{;-^,';f0<;j;

+ MOH

^

0,1

0,1

^MROH

^

gam.

= 32^>R' = 15(R'lanh6mCH3-)".''*^'

^

Meste = 7 4 R C O O C H s

d u , sau p h a n u n g , k h o i l u g n g muo'i t h u d u g c la 764,6 g a m . Kho'i l u g n g
K O H d a t h a m g i a p h a n u n g la
D . 350 g a m .

Laigiai:
^"^^^ ^ 8==

^"^g^

rn^hatbeoCg)

m ^ o H ^ 5600 m g = 5,6 g a m
Phan l i n g hoa hoc:
'

RCOOH

Mol:
*

0,1

Mol:

3x

->

+ C3H5(OH)3

<-

0,01

V i d u 21: A m i n o a x i t X chua m g t n h o m -NH2. C h o 10,3 g a m X tac d u n g v o i axit
H C l ( d u ) , t h u d u g c 13,95 g a m muo'i k h a n . Cong thuc ca'u tao t h u g g n cua X
A . CH3CH2CH(NH2)COOH.

B. H2NCH2CH2COOH.

C. C H 3 C H ( N H 2 ) C O O H .

D . H2NCH2COOH.

+ HCl

Bao toan kho'i l u g n g :

^

Mx = —

- j||||vi.^^^

CIH3N-R-COOH

mammoaxit

+

^

.CJ^jinV

mHci = mmuoi
UHCI

=

0/1

=

= 103 ^ - H 2 N - R - C O O H = 103.

-

0,1

mKOH = (0,1 + 0,8.3).56 = 140 g a m .

^g,; >

f'^^,

-> mHci= 13,95 - 10,3 = 3,65 g a m - >

3x

•

R = 42 —>• Goc hidrocacbon la -C3H6-

'^-'^

••••vv,;::fv;„

D a p an A .

V i d u 22: D u n n o n g m g a m h o n h g p g o m a m o l tetrapeptit mach h o X va 2a

19 (B-08): Xa p h o n g hoa hoan toan 17,24 g a m chat beo can vvra d u
p h o n g . Gi a t r i ciia m la
A . 16,68.

+ C3H5(OH)3

-> D a p an D .

H2N-R-COOH

0,06 m o l N a O H . C o can d u n g d i c h sau p h a n u n g t h u d u g c m g a m xa
^

! aijfi

> 3RCOOK

K h i cho X tac d u n g v o i axit H C l :

+ H2O

D a p an A .
Vi

•>• m. :,3f'

0,03

:k'^'

Bao toan k h o i l u g n g : m + 0,03.56 = 9,58 + 0,92 - > m = 8,82 g a m .

- > 700 + (5,6 + 56.3x) = 764,6 + 92.x + 0,1.18
- > x = 0,8 m o l

D . 8,82.

VinV

ggi cong t h u c cua X la H2N-R-CC)OH.

> 3RCOOK

' "

Laigiai:

nmn = 0,1 m o l .

> RCOOK

* "• '

Nhan xet: Cac d a p an d e u cho aminoaxit chua m g t n h o m - C O O H

Bao toan kho'i l u g n g : mchat beo + m K O H = mmuoi + mgUxeroi + m
'

C. 10,50.

700

0,1

(RCOO)3C3H.s + 3 K O H

B. 9,38.
Laigiai:

'

+ KOH

D a p an C. '

0,92 g a m g l i x e r o l va 9,58 g a m h o n h g p muo'i cua axit linoleic va axit oleic.

(RCOO)3C3H5 + 3 K O H

V i dv 18: C h o 700 g a m cha't beo c6 chi so axit la 8 tac d u n g v o i d u n g d i c h K O H

C h i so axit cua chat beo =

= 17,24 + 0,06.40 - 0,02.92 = 17,8 g a m

Mol:

C. 280 g a m .

C3Hs(OH)3
0,02

ngiixe((ji = 0,01 m o l

= 74 ^ R la C H 3 -

B. 175 gam.

+

mch.nt beo + m N . n O H = m x a phong (muoi) + mgiixeroi

.„

^ Este la C H 3 C O O C H 3 ^ D a p an A .

A . 140 g a m .

m x . i phong

Gia t r i cua m la
A . 9,94.

Bao toan k h o i l u g n g : meste = 15,4 + 3,2 - 11,2 = 7,4 g a m
^

i,

1,.

V i dv 20: Xa p h o n g hoa hoan toan m g a m m o t t r i g l i x e r i t bang K O H t h u d u g c

N/ian f/ifli/: Cha't l o n g sau k h i n g u n g t u g o m nuoc va ancol nen:
mancoi = 93,4 - mnuoc = 92 - 88,8 = 3,2

> 3RCOONa

0,06

A p d u n g bao toan kho'i l u g n g :
->

)• R C O O M + R ' O H fti ..fO !d>i Oi

0,1

< ' i - b tfW

K h i cho chat beo tac d u n g v o i N a O H :

Nhan xet:
K h o i l u g n g M O H bang 11,2 g a m

,

B. 18,24.

-'"^^^ *
C. 17,80.

D . 18,38.

m o l t r i p e p t i t m a c h h o Y v o i 600ml d u n g d j c h N a O H I M ( v u a d u ) . Sau k h i
cac p h a n u n g ke't thiic, c6 can d u n g d i c h t h u d u g c 72,48 g a m m u o i khan
cua c^c a m i n o axit d e u c6 m g t n h o m - C O O H v a m g t n h o m - N H 2 t r o n g
p h a n t u . Gia t r i a i a m la
A. 51,72.
B. 54,30.

•

• •

C. 66,00.
•
•

,

D . 44,48.

11


nang On luy?n thi dgi hpc 18 chuySn 6i H6a hpc - Nguygn Van HJi

dm

Cty TNHH MTV D W H Khang Vi^t

Lai gidi:

I,I

2. P H L f O N G P H A P B A G T O A N N G U Y E N T O

nhom - C O O H (cua amino axit dau C ) tham gia phan ung.

a. N Q i d u n g

Trong cac phan ung hoa hpc, cac nguyen to luon dupe bao toan.

Theo bai, a mol tetrapeptit mach h a X c6 a.3 = 3a mol lien ket peptit.

b, H ? qua

2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit

Tong so mol nguyen tu ciia mpt nguyen to c6 trong cac chat truoc phan ung

Tong so'mol nhom c a c b o x y l - C O O H bang a + 2a =3a.
Cac phan ling nit ggn cua lien ke't peptit va nhom - C O O H la:
-CO-NH-

'yinit

> -COONa + H2N-

va sau phan ung luon bang nhau.

'

truoc va sau phan ling.

+ NaOH

3a

> -COONa + H2O

3a

,

3a

Theo bai: nwaOH = 7a + 3a = 10a = 0,6

^

M'-.A
' i(..>.:-i(jin

-> m = 72,48 + 0,06.3.18 - 0,6.40 = 51,72 gam
D a p an A.

tripeptit mach ho Y voi 400ml dung djch H C l I M (vira du). Sau khi cac
phan ung ket thuc, c6 can dung dich thu dugc 48,1 gam muoi khan cua cac
amino axit deu c6 mpt nhom - C O O H va mpt nhom - N H 2

trong phan ttr.

G i a tri ciia m la
B. 29,0.

' ' ,

C . 30,8.

D . 28,1.

Lai gidi:

Nhan xet: K h i cho cac peptit tac dung voi H C l thi c6 cac lien ket peptit va
nhom H 2 N - (ciia amino axit dau N ) tham gia phan ung.
Theo bai, a mol dipeptit mach ho X c6 a.l = a mol lien ket peptit.
2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit
Cac phan ung riit gpn ciia lien ket peptit va nhom H 2 N - la:
-CO-NH-

Mol:

+ HCl + H2O

5a
H2N-

3a

(du) vao X, lupng ket tua thu dupe la 46,8 gam. G i a tri cua a la
A. 0,55.

B.0,60.

C.0,40.

5a
+ HCl

)--COOH

Lai gidi:
C a c h 1: 6 bai nay, cac em c6 the giai dua vao cac phuong trinh phan ung:
Al + K O H + H2O

+ •|H2

> KAIO2

AI4C3+ 4 K O H + 4H2O

> 4KA102
> Al(OH)3

KAIO2 + CO2 + 2H2O

C a c h 2: So do phan ung:^

.

Al, A I 4 C 3

^^02^^20

) KAIO2

+ 3CH4

r H - f f - + cllV^-:

+ KHCO3

, .

^

^

j

^

^ Al(OH)3

Bao toan nguyen to A l :

,

.

j w , . . ,0-!:

^,

" A l + 4 n A i 4 C 3 = I^Al(OH)3

+

Theo bai:

nAi

n ^ i +nAi4C3

,

= 0'3

^^A\4C3=

=0,2;

nAi4C3

= 0'l-

'

V a y : a = | n A i + 3 n A i 4 C 3 = ^,6

D a p an B.

V i d v 2: H o a tan hoan toan hon hpp gom 0,12 mol FeS2 v a a mol CU2S vao axit

+ CIH3N-

H N O 3 (vua dii), thu dupe dung djch X (chi chiia hai muoi sunfat) v a khi

5a
>

D . 0,45.

>
' ff

Tong so mol nhom amino H 2 N - bang a + 2a = 3a.

Mol:

V i d u 1 (A-08): H o a tan hoan toan 0,3 mol hon hpp gom A l v a AUCs vao dung

. ,,

V i d\ 23: D u n nong m gam hon hop gom a mol dipeptit mach ho X v a 2a mol

,

, ,

VIDVMAU
dich K O H (du), thu dupe a mol hon hpp khi v a dung dich X. Sue khi C O 2

—> a = 0,06 mol.

Bao toan khoi lugng: m + mNaOH = mmuai + m

A . 33,5.

o , ^ , . . r '

LKU y: C a n xac dinh diing va day dii cac chat c6 chiia nguyen to dang xet 6

,

7a

-COOH
Mol:

+ NaOH

7a

• nnMQh u

duy nhat N O . G i a tri cua a la
A. 0,075.

CIH3N-

B.0,12.

Bao toan khoi lupng: m + m H c i + m

C.0,06.

"

D.0,04. '

Lai gidi:

3a

Theo bai: nHci = 5a + 3a = 8a = 0,4 —> a = 0,05 mol.

--V'

Nhan xet: Cac em luu y la dung djch X chi chua hai muoi sunfat - > sau cac
phan ung, S nam het a dang goc sunfat.

= mmuoi

T a CO cac so do chuyen hoa:

m = 48,1 - 0,4.36,5 - 0,25.18 = 29 gam.
—> D a p an B.

^ FeS2
'-vi'.- ./\

10

./.5„ A

Nhan xet: K h i cho cac peptit tac dung voi N a O H thi c6 cac lien ket peptit v a

• Mol:

0,12

;

i
i'ouU-,,r!>! •

> iFe2(S04)3

0,06 'ih.^1bm^l^^.,

:

.

^; ].,>f:v., r v - ; v : - ' -


C^m nang 6n luygn Ihi dgi hpc 18 chuyfin ii

Cu2S
Mol-

Cty TNHH MTV D W H Khang Vi$t

Hoa hpc - Nguyin van H i !

> 2CuS04

a

2a

Bao toan nguyen to P theo so do:
^^^^

,

Bao toan nguyen to S, ta c6: 2 npeSj + "CuzS = "504

,

.

j,|,|,j3i);,

Ca(H2P04)2 <

ii^,

Khoi lupng mol:

- H 1. a

-> 0,12.2 + a = 0,06.3 + 2a -> a = 0,06 -> Dap an C.
V i d u 3 (A-12): Cho 18,4 gam hon hop X gom Cu2S, CuS, FeS2 va FeS tac dung
het voi HNO3 (dac nong, du) thu duoc V h't khi chi c6 NO2 (6 dktc, san
pham khu duy nha't) va dung dich Y. Cho toan bo Y vao mot lugng du
dung dich BaCl2, thu dugc 46,6 gam ket tiia; con khi cho toan bp Y tac
dung voi dung dich NH3 du thu dugc 10,7 gam ket tua. Gia tri cua V la

A. 38,08.

B. 24,64.

C. 16,8.

D. 11,2.

NMn xet: Dung dich Y chiia cac ion: F e ^ Cu^^ SO4 ,

.

rt:.

,

'

,.

tap chat khong chua kali) dupe san xua't tu quang xinvinit c6 dp dinh
duong 55%. Phan tram khoi lupng ciia KCl trong loai phan kali do la
A. 95,51%.

B. 65,75%.

C. 87,18%.

D. 88,52%.
< *= -

khoi lupng cua K2O.

Kho'i lupng mol:
% khoi lupng:

„,,

"

Cu(OH)2 + 4NH3 — ^ [Cu(NH3)4](OH)2
10 7

v;.. V . - .

74,5 gam
x%
<-

47 gam
55%

>
..,,!'

r .

= 87,18 -> Dap an C.
* ' ' '
47

^
Vi dy 6: Dot 5,6 gam Fe trong khong khi, thu dupe hon hpp chat ran X. Cho X
^

Luu y: Cu(OH)2 tan trong NH3 du tao thanh phuc chat:

Bao toan nguyen to Fe: npe (X)= nFe(OH)3

,

Vi dvi 5 (A-12): Mot loai phan kali c6 thanh phan chinh la K C l (con lai la cac

KCl <—> -!-K20

Khi cho Y + dung djch NHa du:
> Cu(OH)2i

,

Ap dung bao toan nguyen to'K, ta CO so do:
mol.

Cu2* + 2NH3 + 2H2O

% khoi lupng:
69,62%
x%
69,62.142
v r ^ ''
—>x=—

= 42,25 - > Dap an C.

v >

/'1

> Fe(OH)3i

1'

o. '"'!(,

6 day, cac em can nho la dp dinh duong cua phan kali dupe tinh theo %

va NO3.

Ba2* + SO^" — ^ BaS04i

Fe3* + 3NH3 + 3H2O

^ i, n
142 gam

Loigidi:

Khi cho dung dich Y + dung dich BaCh:
Bao toan nguyen to'S: ng (X) = nBaS04 ^

,^

> P2O5

234 gam

X=

tac dung voi dung dich FiNOs loang (du), thu dupe khi NO (san pham khu
'

=

duy nha't) va dung djch chua m gam muo'i. Gia trj cua m la
A. 18,0.

B.22,4.

C. 15,6.

D. 24,2.

;

Bao toan khoi lugng: mx = m^u + mpg + mg

npg=0,lmol.

-» mcu = 18,4 - 0,1.56 - 0,2.32 = 6,4 gam -> ncu (X)= — = 0'^ i"ol-

Nhan xet: Bai nay ne'u dua theo phuong trinh phan ung se rat dai dong va

Qui doi X ve hon hgp gom cac don chat:

ton nhieu thoi gian. O day, cac em can su dung so do phan ung:
A iKt/tij:

Fe = 0,1 mol; Cu = 0,1 mol va S = 0,2 mol.
-» ne (X) = 2ncu + Snpe + 6ns = 1,7 mol

n^o^ =

(x) = 1,7 mol

-> V = 1,7.22,4 = 38,08 lit ^ Dap an A.

mm

Fe

va ap dung dinh luat bao toan nguyen to'Fe:
Fe

Vi dv 4 (B-10): Mot loai phan supephotphat kep c6 chua 69,62% Ca(H2P04)2, con
lai gom cac chat khong chiia photpho. Dp dinh duong ciia loai phan Ian nay la
A. 48,52%.

B. 39,76%.

C. 42,25%.

D. 45,75%.

Loigidi:
I

6 day, cac em can nho la dp dinh duong cua phan supephotphat dupe tinh
theo % khoi lupng cua P2O5.

•;• NMn xet: 1 mol Ca(H2P04)2 hay 1 mol P2O5 deu chua 2 mol P . , » , u

MM.

> X (Fe, FeO, Fe203, Fe304)

Mol:

> Fe(N03)3

> Fe(N03)3
; j ,0
„ ,
.0 f:5

0,1
0,1
m =0,1.242 = 24,2 gam ^ Dap an D.
V i du 7: Cho 31,2 gam hon hpp gom Al, Cu va Ag tac dung vua dii voi 900ml
dung dich HNO3 1,5M, thu dupe dung djch chua m gam muoi va 4,48 lit
hon hpp khi X (dktc) gom NO va N2O. Ti khoi ciia X so voi H2 la 16,75. Gia
tri ciia m la

' ' '
A. 98,3.
B.97,2.
C. 96,3.
D. 91,0.


Cty TNHH MTV DWH Khang Vi§t
Ca'm nang fln luygn thi dji hgc 18 chuy6n dg H6a hpc - Nguyin Van Hai

Laigidi:
b a i nay, t r u o c het cac e m can t i m so' m o l m o i k h i t r o n g X de t h u d u g c ket
qua: nNo=

hlhan xet: K h i cho p h a n 1 tac d y n g v o i N a O H , tat ca A l va AI2O3 d e u tac

m o l ; n ^ j o " ^ O'OS i ^ o l -

Al,Cu,Ag

d y n g v a c h u y e n t h a n h NaA102.

) M u o i n i t r a t + N O + N2O

Chatoxihoa:

N^^ + 3e

> NO;

v a CO the xay ra ca qua t r i n h :

ZN*'

2N*5

+

+ 8e

„

ge

/, /

D o vay, bao toan n g u y e n to A l , ta c6: n A i = njvjaOH = 0/3 n i o l .

'

N2O

>

_^ Ban d a u : nAi = 2.0,3 = 0,6 m o l

> NH4NO3 (a m o l )

K h i cho k i m loai + H N O s :

Cr203

n ^ ^ Q - ( m u o i ) = n g trao J6i = 3 nfyio + 8 nivj20 +

Bao toan n g u y e n to N :

nHNOa =

8 nfvjH^NOs ~

+ " N O +

0,9.1,5 = 0,85 + 8a + 0,15 + 2.0,05 + 2a ^

2

mol. '

0,2

2A1

^

— ^

a4:r,^

AI2O3

+

0,2 m o l . .

2Cr

,

OA

,R>0,2;;..

,

: ^

,

Cac chat t r o n g p h a n 2: n ^ i = 0,1 m o l ; n c r = 0,2 m o l ; 0^1203 = 0,1 m o l .

nN20 + 2nNH4N03

a = 0,025

Bao toan k h o i l u g n g : m = m ^ i , cu, Ag + m ^ ^ ^ +

Mol:

0'^^

+

"€1203= ^^^-r^^^=

K!)

iTs->

mNH4N03

->

HHCI =3nAi

+2ncr+6nAi203 = l ' 3 m o l .

^DapanB.

W d v 10: H o a tan hoan toan 8,16 g a m h o n h g p g o m Fe304 va FeS2 t r o n g d u n g

'

d i c h axit HNO3 (dac, d u ) , t h u d u g c 4,032 l i t k h i NO2 (dktc) v a d u n g d j c h X.
C h o X tac d u n g v o i d u n g d i c h Ba(OH)2 d u , Igc ket tua va n u n g t r o n g k h o n g

= 31,2 + (0,85 + 8.0,025).62 + 0,025.80 = 98,3 gam.
-> D a p an A .

k h i d e n k h o i l u g n g k h o n g d o i t h u d u g c m g a m cha't r a n . Gia t r j cua m la

NMn xet: Bai nay da " g i a u d i " san p h a m NH4NO3.

A . 12,66.

V i d u 8: H o a tan het 7,8 g a m h o n hgp

g o m A l va AI2O3 b a n g d u n g d i c h H C l

( d u ) , t h u d u g c V l i t k h i H2 (dktc) va d u n g d j c h X. N h o d u n g d i c h N H s d u

= ^

i';

'

C. 5,60.

D . 4,48.

G
'

So d o p h a n u n g :
, AlAl^Og

'Atii

./v

^
)AlCl3

vM^b

Fe304 - l e

> 3Fe^3

FeS2-15e

> Fe^^ + 2S^

*~

- - ,0'' ^
- li^a.W/
. .

" N O z = nFe304 +15"FeSz

a + 15b

= ai8

- » a = a03;

3

nH2 = - H A I
2'

= 0,15 m o l ^

= 0,2.

Fe304

V = 3,36 l i t .

D a p an B.

Mol:
' '

0,03
FeS2

V i d v 9 (B-12): N u n g n o n g 46,6 g a m h o n h g p g o m A l va Cr203 (trong d i e u
Mol:

I h g p t h u d u g c sau p h a n u n g t h a n h hai phan bang n h a u . Phan m o t p h a n

-> m = 160.0,05 + 0,02.233 = 12,66

i u n g v u a d u v o i 300 m l d u n g d j c h N a O H I M (loang). De hoa tan het p h a n
hai can v i r a d u d u n g d i c h chiia a m o l H C l . Gia t r j cua a la
A . 0,9.

B. 1,3.

C.0,5.

D . 1,5.

Vi

0,01

0,005

'[

_

U m fiiJ5 ifi BfO > i

> ^Fe203
2
0,045
1
> - F e 2 0 3 + 2BaS04
2

J k i f n k h o n g c6 k h o n g k h i ) den k h i phan u n g xay ra hoan toan. Chia h o n

'',;t»r,'
b = aOl.

Bao toan n g u y e n to Fe v a S:

= 0,1; y = 0,05

.
,

Bao toan electron:

"NH3.H2O > A i ( O H ) 3 _ ' % AI2O3

Bao toan n g u y e n to A l : x + 2y = 2nAi203 -> x + 2y = 2 . ^ ^
X

n^B

Cac p h a n l i n g khvr:

'

Ldigidi:

.

0,18 m o l .

D . 20,66.
|>,s£' V - '

G g i so m o l : Fe304 (a m o l ) v a FeS2 (b m o l ) . Ta c6: 232a + 120b = 8,16.

10,2 g a m chat ran. Gia t r i cua V la
B. 3,36.

=

C. 11,86.
Laigiduii
^' '

22,4

vao X, IQC ket tua va d e m n u n g den kho'i l u g n g k h o n g d o i t h u d u g c
A . 2,24.

B. 11,06.

-8
808,£
( \.1.0 »

0,02
-> D a p an A.

1 1 (A-08): H o n h g p X g o m p r o p a n , p r o p e n v a p r o p i n . T i k h o i cua X so
v o i H2 Ja 21,2. K h i d o t chay hoan toan 0,1 m o l h o n h g p X, t o n g k h o i l u g n g
c u a C C h va H2O t h u d u g c la
A . 20,40 g a m .

B. 18,60

.

SB^JHUZ^Ji^^^j^^^ff^fpf^^


Cty TNHH MTV D W H Khang ViQt
Ca'm nang On luyjn thi dgi hgc 18 chuyfin dg H6a hgc - Nguygn Van H5i

Vi

Laigiai:
NMn

xet: Cac e m can nh|in ra cac chat t r o n g h o n h o p X d e u c6 chiia 3

n g u y e n t u cacbon.
'

' Bao toan kho'i l u g n g , ta c6: mx = mc + mn. '>rn 6,(J " f' > . -

no, d a chiic, m a c h h o , c6 c u n g so n h o m - O H ) can v i r a d u V l i t k h i O 2 , t h u
d u p e 11,2 l i t k h i C O 2 v a 12,6 g a m H 2 O (cac the t i c h k h i d o a d k t c ) . G i a t r j
ciia V la

N h u vay, k h i do't chay 0,1 m o l X t h u dugc 0,3 m o l C O 2 - > nc = 0,3 m o l .
M a t khac: M x = 21,2.2 = 42,2 - > mx = 42,4.0,1 = 4,24 g a m .

14 (B-10): Do't chay hoan toan m p t l u p n g h o n hgip X g o m h a i ancol ( d e u

i

A . 14,56.

" ' '"*V

^ 1 /•-(.'" M^' «f^ '

- > m H = 4,24 - 0,3.12 = 0,64 g a m ^ nn = 0,64 m o l - > nn^o = 0/32 m o l .
T o n g k h o i l u g n g C O 2 v a H 2 O bang 0,3.44 + 0,32.18 = 18,96 g a m .

B. 15,68.

C . 11,20.

D . 4,48.

Laigiai:
11'2
,
12,6
,
"CO2=^=0'5"^ol;nH2O=^=0,7mol.

^

Theo b a i ra, X c h i i a 2 ancol n o - > nx = nH20 " ^COi ^ ^,2 m o l .

- > D a p an C.
V i d\ 12: Do't chay hoan toan m p t the tich k h i thien n h i e n g o m metan, etan,

-> So'nguyen tvr C t r u n g b i n h =

= 2,5 - > X chua m p t ancol d a chuc c6

p r o p a n bang o x i k h o n g k h i (trong k h o n g k h i , o x i chiem 20% the tich), t h u
d u g c 7,84 l i t k h i C O 2 (a dktc) va 9,9 g a m H 2 O . The t i c h k h o n g k h i (a dktc)

D o X chua h a i ancol c u n g so' n h o m chuc (hai chiic) - > n o = n o H = 2nx

.X n h o nhat can d i i n g de d o t chay hoan toan l u o n g k h i t h i e n n h i e n tren la
:gfi A . 70,0 l i t .

ir sv B B. 78,4 l i t .

C. 84,0 l i t .
Lmgidi:

D . 56,0 l i t .

> 100

•JIA

so n g u y e n t u C n h o h o n 2,5 - > ancol d o la C2H4(OH)2.

^

Ij

no = 0,4 m o l .

Bao toan n g u y e n to' O, ta c6: no (OH) + 2 no^ = 2 TXQQ.^ +

«;:.«,) jfW

2.0,5 + 0 , 7 - 0 , 4
,
no2=—
^
^ = 0,65 m o l

nc02 = - ^ = 0 3 5 m o l ; n H 2 O = ^ = 0 ' 5 5 m o l .
Nhan xet: Ban d a u , n g u y e n to' o x i 0 dang O 2 t u do, c o n sau p h a n l i n g chay
t h i chuyen he't v a o C O 2 v a H 2 O .

= ^'^^-^^^'^^ = 0,625 m o l
Vkhongkhi

+

n^jjo •

thupc c u n g d a y d o n g dang, t h u dugc

s

at

w

timA •

3,808 l i t k h i C O 2 (dktc) v a 5,4 g a m

B. C H 2 = C H C O O H v a C H 2 = C ( C H 3 ) C O O H .

C. C H s C O O H v a

' '

C2H5COOH.

D. C H 3 C O O H va C H 2 = C H C O O H .

B.5,72.

C. 4,72.,

//,4

10

Laigiai:

D . 7,42.

Laigiai:
' ^
ri one

KA
nc02 = ^ = 0^17 m o l ; n H 2 0 = ^ = 0,30 m o l .

=

.„

4

••

2.0,1 + 2 . 0 , 2 4 - 0 , 2
,
Y
^ = 0'24mol.

Nhan xet: rxyi^Q < rxQOj

H o n h p p X chua i t nhat m p t axit k h o n g n o - >

Logii A v a C.
-»

no = 0,13 m o l .

5' j Bao toan k h o i l u g n g t r o n g X, ta c6:

....lii^^

So n g u y e n t u C t r u n g b i n h = "^^^ = 2,4 - > X chua m p t axit c6 so

n g u y e n t u C n h o h o n 2,4 - > L o ^ i B ( H a i axit d e u chua so'cacbon > 2,4).

m x = m c + m H + m o = 0,17.12 + 0.60.1 + 0,13.16 = 4,72 g a m
t'-vs'jji'ii

•

^C02=—

0,30 - 0,17 = 0,13 m o l .

D o X chua cac ancol d a n chiic ^ no = noH = nx

H!"\,0-,.

Bao toan n g u y e n t o O, ta c6: 2nx + 2 n o , = 2 nco2 "H2O
• • ;loM

. .it

Cac e m can thay rang, k h i d o t chay X: n^^^Q > UQQ.^ - > X chiia 3 ancol no

- > ' D a p an C . M / » t j ; ; ,

•

A. H C O O H va C 2 H 5 C O O H .

;.::;;;»•*

H 2 O . Gia t r i ciia m la

n^Q^

V02 = 0,65.22,4 = 14,56 l i t ^ D a p an A .

axit la

V i d u 13 (A-10): D o t chay hoan toan m g a m h o n h g p X g o m ba ancol d o n chiic,

nx = nj^^Q -

'

0,1 m o l X can 0,24 m o l O2, t h u d u p e C O 2 v a 0,2 m o l H 2 O . C o n g t h i i c h a i

V Q J = 0,625.22,4 = 14,0 l i t .

= 5 V o 2 = 70,0 l i t D a p an A .

A . 5,42.

^

^^i^rW

V i d y 15: H o n h p p X g o m h a i axit cacboxylic d o n chuc. Do't chay h o a n toan

Bao toan n g u y e n t o O, ta c6: 2 n o j = 2 TYQQ^

Vay:

nyi^^Q

i,-.>.,..t-

—> D a p a n D .
19


Cty TNHH MTV DWH Khang Vi$t

Ca'm nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - IMguySn VSn H5i

V i dv 16 (A-12): Hon hg-p M gom mpt anken va hai amin no, don chuc, mach
" h6 X va Y la dong dang ke tiep (Mx < MY). Do't chay hoan toan mot lugng
r M can dung 4,536 lit O 2 (dktc) thu dugc H 2 O , N 2 va 2,24 lit C O 2 (dktc). Chat
Yla
A. Etylamin. ' '
C. Butylamin.

[ '2 :
.

B. Propylamin.
D. Etylmetylamin.

•'XlS:',U,'0
,

- ,(•) .A

Laigidi:

. , Laigidi:
n .i .
., . ii..t»
4536
224
n ;fom t ; , i . - - n o , = ^ ^ = 0,2025 m o l ; n r o , = — = 0,1 m o l .
.
• •
'! •

^

,

.'' .

22,4

22,4

^ 2

H H C I = 0,03 mol; n o 2 = ai425 mol.

M|it khac: n N = nNH2 = " H C I

^ H J O = 0/205 mol.

Nh^n thay, khi dot chay 1 mol anken thu du(?c XXQQ^ = n H 2 0 '
amin no, don chijfc thi n H j O " ' ^ C 0 2
CxH2x.3N

1'^

vol hai

ph"^°''^8 ^""^^'^'^"'^S-

^ X C O 2 + ( x + l , 5 ) H 2 0 + 0,5N2

+ O2 —

Dovay: n H 2 0 " ' ^ C 0 2 = l / 5 n a m i n - > namin=—^

'— =0,07mol.

-> So'nguyen t u C trung binh trong M = — ^ ^ < — ^ 2 2 _ =
"M

I^aniin

n^

Bao toan khoi lugng: mx = m ^ + mj^ + mo +T^-H
12a+ b = 1,81.

0,07

a. N p i dung

Dot chay hoan toan mpt lugng X can vua du V lit khi O 2 , thu duQC 15,68 lit
'' khi C O 2 va 18 gam H 2 O . Cac the tich do 6 dktc. Gia t r i cua V la

NMn

C. 16,24.
mt?iV6J

• *^' •

Bao t o a n n g u y e n to O: n o (ancol) + 2 n o 2 = 2 n c o 2 + n H 2 0
->

HQ^ =

2.0,7 + 1-0,3

,
= 1 , 0 5 mol

2
V 0 2 = 1,05.22,4 = 23,52 lit ^ Dap an D.

ffl>|

{

dau dugc thay the (ho^c cgng hgp) bang nguyen t u (nhom nguyen tit) mai
de tao thanh san pham.

^

Do do, khoi lugng cua chat tao thanh c6 the tang len hay giam di do chenh
Dua vao sy tang hay giam nay c6 the xac djnh so mol cac chat trong

1

phuang trinh hoa hgc, t u do c6 the giai nhanh nhieu bai toan.

"ancol = " H 2 O ~ ^C02 •

Do a n c o l d o n c h u c n e n U Q (ancon ^ "ancol =0,3 m o l .

,

l|ch khoi lugng mol cua cac nguyen t u (nhom nguyen tu).

D. 23,52.

xet: K h i d o t c h a y a n k e n t h u duQC UQQ^ = XIH^Q , c o n k h i d o t c h a y a n c o l

Do v ^ y : nancol = " H Z O - ^coz = 1 - 0,7 = 0,3 m o l .

,. .

Khi tham gia phan ling hoa hgc, nguyen t u (nhom nguyen tu) cua chat ban

Dap an A.

V i d\ 17: H o n hgp X gom mpt anken va hai ancol (no, don chiic, mach ho).

n c o 2 = 0,7mol; n H 2 0 = l m o l .

,.

-> Dap an A.

ii C H 3 N H 2 (X). O day cac em can l u u y: anken chua tir 2 nguyen t u C tra len!

B. 17,92.

'fl iiAA US'

Bao toan nguyen to C: n c ( X ) = n c o 2 = '^CaC03 = 0 , 1 3 ^ m c a C O a =13 gam.
3. PHl/ONG PHAP T A N G - G I A M K H O I L l / p N G

A. 22,40.

21/14 ~ 3 •

^ 0,1 + 2.0,1425 = 2a + 0,5b -> 4a + b = 0,77 ^ a = 0,13.
Y

'

^ Ggiso'mol: n ^ (x) = a i^ioJ va n H ( x ) = b m o l .

f, -> M chua mQt chat c6 so' nguyen t u C nho hon 1,43 -> do phai la amin
-> A m i n ke tiep la C 2 H 5 N H 2 (Y)

1

n p _ 80/16 10

Bao toan nguyen to O: n Q (x) + 2no2 = 2 n c o 2 + ^HjO

sot os8

~ 1^43.

C>D < - -

(x)= 0,03 mol -> no(X)= 0,1 mol.

-» m c + n i H = 3,83 - 0,1.16 - 0,03.14 = 1,81
^ '

0,205-0,1

•

^. r n o ^ S O
man xet: Ixx ti 1? khoi lugng:
21 ^

......£Dtjd3X,P'!J6do'?riT

Bao toan nguyen to'O, ta c6:

6: 2 no2 = 2 n c o j + " H J O

V i dv 18: Hon hgp X gom hai amino axit no (chi c6 nhom chuc - C O O H va - N H 2
trong phan tu), trong do ti 1§ mo : mN = 80:21. De tac d\ing vvra du voi 3,83
gam hon hgp X can 30ml dung dich HCl I M . Mat khac, do't chay hoan toan
3,83 gam hon hgp X can 3,192 lit O 2 (dktc). Dan toan bp san pham chay ( C O 2 ,
H 2 O va N 2 ) vao nude voi trong d u thi kho'i lugng ket tiia thu dugc la
A. 13 gam.
"
B. 20 gam.
C. 15 gam.
D. 10 gam.

C03
, „,.

Bien doi k h o i lugmg (tinh cho 1 mol)

S\f thay the (cpng hgfp)
0-2 (oxit)
CO

—

-OH
_COOH

-NH2

Tang 71-60 = 11 gam

> CI2

Tang 96-16 = 80 gam

""''^"^ > S O ^
^ CO2; H2

H2O

Tang 16 gam
Tang 23-1 = 22 gam

> -ONa
> -COONa
> -NHsCl

Tang 23-1=22 gam
Tang 36,5 gam
21


dm

Cty TNHH MTV DVVH Khang Vi§t

nang fln luygn thi dgi hgc 18 chuySn de H6a hpc - Nguyjn Van Hi\

VI DU MAU
V i d\ 1 (A-08): Cho V l i t hon hop khi (dktc) gom CO va H2 phan ung vol mot
lugng d u hon hop ran gom CuO va Fe304 nung nong. Sau khi cac phan ung
xay ra hoan toan, khoi lugng hon hgp ran giam 0,32 gam. Gia tri ciia V la
A. 0,448.

B. 0,112.

C. 0,224.

D. 0,560.

1 mol H 2
->

—*°

> CO2

— * ° > H2O

1 mol (CO va H 2 )

Theobai:

Kho'i lugng chat ran giam 16 gam.

^^^m

—> Kho'i lugng chat ran giam 16 gam.
^ ° > ( C O 2 va H 2 O )

0,02 mol

<-

> 1 mol SO l~

Khoi lugng tang 96 - 16 = 80 gam.
tang 52 - 22 = 30 gam.

_^ a = — = 0,375mol -> no (X) = 0,375 mol.
80
Khi cho 22 gam X tac dung voi CO thi: no

Lot giai:
1 mol CO

1 mol 0 - 2
a mol

(X)

'" '

ff
f. •

''

•' '

= n^oj = riCaCOa = 0375 mol.

- > m =0,375.100 = 37,5 gam.
—> Dap an B

'

'''t*''-t> i?>'; f vcr-f vi

KT *

•'•H'>I..

V i dv 4 (A-10): Dot chay hoan toan mgt lugng hidrocacbon X. Hap thu het san

giam 16 gam.

pham chay vao dung dich Ba(OH)2 du, tao ra 29,55 gam ket tiia, dung dich

giam 0,32 gam.

sau phan ung c6 khoi lugng giam 19,35 gam so voi ban dau. Cong thuc

- > V = 0,02.22,4 = 0,448 lit.

phan t u cua X la

—> Dap an A.
V i d\ 2: Hon hgp Y gom FeO, Fe203 va CuO. Hoa tan hoan toan 6,8 gam Y

A.C3H8.

B.C2H6.

C.C3H4.

D.C3H6.

Lot giai:

bang dung djch HCI (du), thu dugc dung dich chua 12,3 gam muoi. Mat

Khi hap thu hoan toan san pham chay gom C O 2 va hoi H 2 O vao dung dich

khac, neu khu hoan toan 6,8 gam Y bang CO (du), thu dugc m gam kim

Ba(OH)2 du:

loai. Gia tri cua m la :
A. 5,2.

C.6,0.

D. 4,8.

^

Led giai:

> 2 mol CI" - >

a mol
^

a= —

Khoi lugng tang 71 - 1 6 = 55 gam.

<—
= 0,1 mol -> no (Y) = 0^1 mol.

- > Dap an A.

^

'

' " "*

V i dvi 3: Hon hgp X gom CuO va Fe203. Hoa tan hoan toan 22 gam X bang
thu dugc 52 gam muoi. Mat khac, neu khu het 22 gam X

bang CO (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao
thanh m gam ket tua. Gia trj cua m la
A. 45,5.

B.37,5.

C. 40,5.

D. 50,0.

Lai giai:
Nhan xet: Khi cho X +

H2SO4

<-

0,15

thi oxit chuyen thanh muoi sunfat va mgt ion

0-2 trong oxit dugc thay the bang mgt ion SO 4 " .

-f^'
= Mat - Dugc
= mBaCOs -

->

^

^

Bao toan khoi lugng: my = m + mo - > m = 6,8 - 0,1.16 = 5,2 gam.

H2SO4,

0,15

mH20=

(r\

(mco2 + n^Hoo) = 19,35gam.

29,55 - 0,15.44 - 19,35 = 3,6 gam ->

nH20

= 0,2 mol.

' *"

Vay hidrocacbon X c6: So C : so'H = nc : nn = 0,15 : 0,4 = 3 :>8.

tang 12,3 - 6,8 = 5,5 gam.

Khi cho Y tac dung voi CO thi cac oxit deu bi khu thanh kim loai.

dung dich

Mol:

Khoi lugng dung dich giam

Nhan xet: Khi cho Y + HCl thi oxit chuyen thanh muoi clorua va mgt ion
0^~ trong oxit dugc thay the bang hai ion CI".
, 1 mol O^-

> BaCOs + H 2 O .

Ba(OH)2 + C O 2
B.5,6.

Dap

an A.

„.,,.n ,,a

, ,

V i dy 5 (B-12): Cho 21 gam hon hgp X gom glyxin va axit axetic tac dung vua
dii vol dung djch KOH, thu dugc dung djch Y chiia 32,4 gam muoi. Cho Y
tac dung voi dung dich HCl du, thu dugc dung djch chua m gam muoi.
Gia trj cua m la
A. 44,65.

B. 50,65.

C. 22,30. ^

D. 33,50.

Ldigiai:

Glyxin ( H 2 N - C H 2 - C O O H ) = a mol va axit axetic ( C H 3 C O O H ) = b mol.
Khi cho X tac dung voi dung dich K O H :
:>
Theobai:

•

•'

•

1 mol - C O O H ^ 1 mo! -COOK -> khoi lugng tang 38 gam.
0,3 mol

-> a + b = 0,3 mol.
Mat khac: 75a + 60b = 21

<-

tang32,4-21 = 11,4 gam

a = 0,2 mol; b = 0,1 mol.

Khi cho Y + HCl d u , thu dugc cac muoi: C I H 3 N - C H 2 - C O O H (0,2 mol) va
KCl (0,3 mol) theo cac phan ung:

23


Cty TNHH MTV DVVH Khang Vigt

elm nang fln luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hi\

H2N-CH2-COOK + 2HC1
'

+ HCl

CH3COOK

V i dv 8: Hon hgp X gom hai amino axit (mach ho, moi amino axit deu chiia

> ClH3N-CH2-CC)OH + K C l

> CHCOOH + KCl

•,l.>m^V

Vay: m = 111,5.0,2 + 74,5.0,3 = 44,65 gam - » Dap an A .
Luu y: Trong bai nay cac em de quen KCl khi tinh kho'i lugng muo'i va chpn
nham phirang an C!
Vi

6 (A-12): Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chat khi a dieu
ki^n thuong) roi dem toan bp san pham chay hap thu het vao binh dvmg

mgt nhom chiic - N H 2 va mgt nhom chuc -COOH). Cho 16,4 gam X tac
dyng vua d u vai dung dich K O H , thu dugc 24 gam muo'i. M | t khac, 16,4
gam X tac dung vua du V lit dung djch HCl 2M. Gia t r i ciia V la
A . 0,1.

lugng phan dung dich giam bot 19,912 gam. Cong thiic phan t u ciia X la

^.

A.CH4.

,

B. C 3 H 4 .

C. C 4 H 1 0 .

^
^

12
2
M9tkhac:mx = mc + m H = — n i r o , + — ^ H - J O =4,64.

'*

mco2 =15,312 gam (0,348 mol); m n j o =4,176 gam (0,232 m o l ) .

> 1 mol -COOK

0,2 mol

<-

-> Dap an C.

y,i . . i ,/ <,
khoi lugng tang 38 gam.

VHCI=0,2 lit.

^

,,,,, , ,

V i dy 9: Cho 12 gam hon hgp X gom glyxin va etylamin tac dyng vua du voi
dung dich HCl, thu dugc dung dich Y chiia 19,3 gam muo'i. Cho Y tac dung
voi dung dich NaOH d u , thu dugc dung dich chua m gam muo'i. Gia trj cua
m la
A . 9,60.

B. 15,45.

C. 21,40.

D. 19,10.

Loigidi:

Dap an B.

.

Glyxin (H2N-CH2-COOH) = a mol va etylamin (C2H5NH2) = b mol.

Lim y: Ci day bai toan khong cho dung dich Ba(OH)2 d u nen neu cac em viet
phuang trinh hap thu: Ba(OH)2 + CO2

Khi cho X tac dung voi dung dich HCl:

> BaCOa + H2O.

0,2 mol thi se bi mac sai lam ngay!

V i di;i 7: Hon hgp Y gom axit axetic, axit acrylic va axit adipic. Cho 16,9 gam Y

^

1 mol -NH2

> 1 mol -NH3CI -> khoi lugng tang 36,5 gam.

0,2 mol

tang 19,3-12 = 7,3 gam

Theobai:

16,9 gam X tac dung vai dung dich NaHCOs d u , thu dugc V lit khi CO2

-> a + b = 0,2 mol.
Mat khac: 75a + 45b = 12 ^ a = 0,1 mol; b = 0,1 mol.
Khi cho Y + NaOH d u , thu dugc cac muoi: H2N-CH2-COONa (0,1 mol) va

(dktc). Gia trj a i a V la

NaCl (0,2 mol).

tac dyng vua dti vai dung djch NaOH, thu dugc 22,4 gam muo'i. Mat khac,

A.

5,60.

q

B. 8,40. -

• - • .• p
C. 4,48.

n

D. 7,84.

Cac phan ung hoa hgc:
CIH3N-CH2-COOH + 2NaOH

\
Khi cho Y tac dyng voi dung djch KOH:

Mol:

> H2N-CH2-COONa + NaCl

0,1

0,1

0,1

1 mol - C O O H -> 1 mol -COONa - > Khoi lugng tang 22 gam.
Theobai:
0,25 mol
tang 22,4-16,9 = 5,5 gam
ncooH =0,25 mol.

Mol:

Khi cho Y tac dyng vai dung djch NaHCOa: nco2 = " c o o H = 0/25 mol

Luu y: Neu quen tinh khoi lugng NaCl cac em se chgn nham phuong an A !

^
^

i'..^ , ,

tang24-16,4 = 7,6 gam

nHci=0,2mol ^

Vay hidrocacbon X c6: So C : so H = nc : nH = 0,348 : 0,464 = 3 : 4

=

{i >

1 mol - C O O H

nHCl=nNH2 = " x ^

^ mco2 + mH20= 39,4-19,912 = 19,488 gam.

va suy ra ncQ2 = TI^^QQ^

f.

nx = ncooH = 0,2 mol. Khi cho X tac dung vai dung djch HCl:

D. C 2 H 4 .

= mBaCOa - (mc02 + ^ H 2 0 ) = 19,912 gam.

-> X la C3H4

D.0,3.
8""?' " " U :

Khi cho X tac dung vai dung dich KOH:
Theobai:

Loigidi:
Kho'i lugng dung dich giam = Mat - Dugc

d-

C.0,2.
Loigidi:

dung dich Ba(OH)2. Sau cac phan ling thu dugc 39,4 gam ket tua va khoi
^j.

B.0,5.

C2H5NH3CI + NaOH

ai

> C2H5NH2 + NaCl
0,1

0,1

-

Vay: m = 97.0,1 + 58,5.0,2 = 21,40 gam ^ Dap an C.

Vco2 = 0/25.22,4 = 5,6 l i t

-> Dap an

A.

25


Cty TNHH MTV DVVH Khang Vi$t

Ca'm nang 6n luygn thi dgi hgc 18 chuySn dg H6a hoc - Nguyin VSn Hii

4. PHL/ONG PHAP BAO T O A N E L E C T R O N
a. Npi dung
'Sh^n

7
4 v si-lk-l

Lai gidi:

ymi t ^'bdn vym

xet: Cac em luu y la dung dich X chi chua hai muoi sunfat

*• ,1 Tong so' mol electron cac chat khu nhuong = Tong so' mol electron cac cha't

phan ung, S nam he't 6 dang goc sunfat.

oxi hoa nhan.
b. Cach ap di^ng

Ta C O cac so do chuyen hoa:

; f-Aji,

Cac em can xac djnh dung va day du cac chat khii va chat oxi hoa ciing nhu

FeS2

,
: • '
Cu2S

..

\
Mol:

su bien doi trang thai oxi hoa cua chiing.

Mol:

Viet cac qua trinh oxi hoa (nhuong electron) va qua trinh khu (nhan

Bao toan nguyen to S, ta c6: 2 npeSj + ncu2S = " 5 0 4

J, electron) de xac dinh so'mol electron trao doi roi ap djmg dinh luat bao toan
electron.

,

; ...

V i du 1 (CD-11): Hoa tan hoan toan 13 gam Zn trong dung dich HNOs loang,

du thu dup-c dung dich X chua m gam muoi va 0,448 lit khi Ni (dktc). Gia
tricuamla
B. 37,80 gam.

C. 28,35 gam.

FeS2-15e

Chat oxi hoa:

2N^5

> N2

^

•

3n^o = 15npesj +10ncujS

" N O = 0,2 mol

.oorfv

Y chua m gam muoi. Gia tri cua m la
A. 8,88.
B. 13,92.

nenh,in

C. 6,52.

" M e = - ^ ^ =0,09 mol; nNo= ^^^=0,04 mol.
^
24
22,4

= lOn^^ = 0,2 mol.

Chat oxi hoa:

N^^ + 3e

^ NO;

' mot san pham khu da dugc "gia'u di", do la su tao thanh muoi NH4NO3:

Bao toan electron: 2 n^g = 3 n^o + 8 nNH4N03

> NH4NO3
->

D. 13,32.

-i
'

nNH4N03 =

0,025

Vay: m = mzn(N03)2 + mNH4N03 = 0,2.189 + 0,025.80 = 39,8 gam.
—> Dap an D.
Luu y: 1- Bai toan nay c6 the giai nhanh hon bang each ap dung ngay

„
md'hu'tf

> Mg^2

0,2

'^^WM-

Lot gidi:

Mg - 2e

Mol:

c

V^o = 4,48 lit.

Chatkhu:

2N^5 + se

> 2Cu^2 + 5-6

Bao toan electron:

" Nhu v$y so mol electron trao doi chua bang nhau—> "chua on". O day,
Chat oxi hoa:

^^^
'

phan ung xay ra hoan toan thu dugc 0,896 lit khi NO (6 dktc) va dung dich

"lol.

> Zn*^ —> ne nhuong = 2nzn = 0,4 mol.

+ lOe

^'

Cu2S-10e

> Fe*^ + 25*^

6 day cac em se dugc huong dan giai theo phuong phap bao toan electron.
Zn - 2e

2a

Vi dv 3 (B-08): Cho 2,16 gam Mg tac dung voi dung dich HNO3 (du). Sau khi

Bai toan nay cac em c6 the giai khi viet phuong trinh phan ung.

Chatkhu':

->

a = 0,015.

-^DapanD.

D. 39,80 gam.

Lot gidi:
13
0 448
nzn=—= 0,2 mol; nN2 = i : 7 T =

0,03.2 + a = 0,015.3 + 2a ^

a

> 2CuS04

Cac phan ung khu:

VI D U M A U

A. 18,90 gam.

0,03

> -Fe2(S04)3

2
0,015

sau cac

2N^5 + se

> NH4NO3
"

2.0,09-3.0,04
,
^=0,0075mol.

^

-> m= mMg(N03)2 + mNH4N03 = 0,09.142 + 0,0075.80 = 13,92 gam.
->DapanB.

^

^

^

:

,

Vi d\ 4: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 2,7 gam Al vao 200 ml

phuong trinh bao toan electron: 2 n^g = 3 n-^Q + 8 n[,jH4N03 •

dung dich X chua dong thoi H N O 3 I M va H2SO4 1,5M. Sau khi phan ung

2- Ba kim loai kha manh (Mg, Al, Zn) tac dung voi axit HNO3, c6 the t^io
thanh muoi NH4NO3!
^ .

xay ra hoan toan thu dugc khi NO (san pham khu duy nha't) va dung dich

V i d y 2: Hoa tan hoan toan h6n hop gom 0,03 mol FeS2 va a mol CU2S vao axit
HNO3 (vira dii), thu dugc dung djch X (chi chua hai muoi sunfat) va V lit
khi duy nha't NO (dktc). Gia trj ciia V la
A. 1,12.

B.5,60.

C.2,24.

D. 4,48.

Y (chi gom cac muoi). Khoi lugng muoi c6 trong Y la:
A. 41,25 gam.

B. 53,65 gam.

C. 44,05 gam.

D. 49,65 gam.

Lot gidi:
9 75
n2n=-^—= 0,15 mol; nAl = 0,1 mol.

'
: • •


Ca'm nang On luy$n thi dgi hgc 18 c h u y § n de Hoa hqc - Nguyin Van H&i

Cty TNHH MTV

Nhan xet: K h i axit H N O 3 c6 mat dong thai voi axit H2SO4 loang thi lirgng

Led giai:

trong dung dich la do 2 axit phan li ra -> giai theo phuong trinh ion.
= '^HNOg + 2nH2S04 = 0,2 + 2.0,3 = 0,8 mol

, ,^, ^

Ggisomol:Fe304(amol)vaFeS2(bmol).

+

3Zn

ai5

ifr^f-8

Mol:

Al

8H*

+ 2NO;

0,4-^
+

0,1^

4H*

s?,

+ NO~

0,4^

> AP

+

ai

NO

•

—> H + v a N O j tham gia phan ling het.

+

2H2O

ufifki-^iiO
-
- > Khoi luong muoi trong Y = 9,75 + 2,7 + 0,3.96 = 41,25 gam

Dap an A .

V i d u 5 (B-07): N u n g m gam hot sat trong oxi, thu dugc 3 gam hon hg-p chat
ran X. Hoa tan het X trong dung dich H N O 3 du, thoat ra 0,56 lit N O (san
B.2,22.

> 3Fe*3

FeS2 - 15e

> Fe- + 2S-

"f

nN02= " F e 3 0 4 + 15nFeS2

^, ,^

- ' , ;
^ jo...','-

a + 15b = 0,17 ^

a = 0,02; b = 0,01.^ '

Fe304
Mol:

, 3Fe(N03)3

"^'^^""'^ > 3Fe(OH)3

0,03
> Fe(N03)3 + 2H2SO4

) Fe(OH)3 + 2BaS04
0,5Fe2O3+2BaSO4

•

n N O = — = 0 , 0 2 5 mol.
22,4

1,5 Fe203

0,02

D . 2,32.

Laigidi:
.88;;-',A

Mol: 0,01

->

0,005

0,02

- > m = 160.0,035 + 0,02.233 = 10,24 - > Dap an D .
V i d u 7: C h o 12,45 gam hSn hgp X gom Fe, Mg, Z n vao dung dich H N O 3 du,

Bao toan khoi lugng: m.Q^ = m x - mpg= 3-m.

thu dugc dung dich Y (khong chua N H 4 N O 3 ) va hSn hgp khi Z gom 0,2 mol

Nhan xet: Neu dua theo phuong trinh phan ung, bai giai se rat dai va kho giai.

N O va 0,1 mol N O 2 . C o c^n dung dich Y thu dugc m gam muoi khan. Gia

C a c h 1: 6 day, cac em can s u dung so do phan ung:

tri cua m la

Fe (1) - ^ 1 ° ^ X (2)

A. 31,05.

, Fe^3 (3)

B. 43,45.

C . 55,85.

Bao toan electron: 3.npg = 4nQ^ + 3nj^Q
^

3.E. = 1:21

+ 0,075 -> m = 2,52 gam

:

.^v

C a c h 2: Q u i doi X thanh Fe (a mol) va O (b mol). Ta c6: 56a + 16b = 3.
3e ^

Qua trinh k h u : O + 2e - >

Fe*^

O-^; N*^ + 3e - >

N-

+ 3e
netraod6i

Mg - 2e
> NO;

> Mg^^; Z n - 2e
N-

+

le

dich axit H N O 3 (dac, du), thu dugc 3,808 lit khi N O 2 (dktc) va dung djch X.
Cho X tac dung voi dung djch Ba(OH)2 du, Ipc ket tiia va nung trong khong
khi den kho'i lugng khong doi thu dugc m gam chat ran. G i a tri cua m la
C . 15,86.

>Zn^^.

> NO2.

= 3njsjo + l-nN02 = ^''^"^°^-

go'c N O 3 nam trong muoi dugc tinh n h u sau:

V i d\ 6: H o a tan hoan toan 5,84 gam hon hop gom Fe304 va FeS2 trong dung

B.9,68.

Chat oxi hoa:

> Fe-;

.

Den day, cac em can l u u y la khi cho k i m loai tac dung voi H N O s , so mol

m = 0,045.56 = 2,52 gam.

A. 8,66.

pham khi nen can ap dung dinh luat bao toan electron.

Bao toan electron:

NO

Bao toan electron: 3a = 2b + 0,075 —> a = 0,045; b = 0,03.
->

Nhan xet: D a y la bai toan c6 nhieu chat k h u (3 k i m loai) v a tao ra 2 san

Chat k h u : Fe - 3e

Qua trinh oxi hoa: Fe -

D . 62,05.

Lcn gtat:

—> D a p an A .

7S

^ J-j^n;}».in

Ta CO cac so do chuyen hoa:

FeS2

C . 2,62.

^^

Fe304 - l e

pham k h u duy nha't 6 dktc). G i a tri cua m la:
A . 2,52.

'

Bao toan electron:

:Mig(m

"+g': 4

.U.

Cac phan ung k h u :

> 3Zn2+ + 2 N O + 4H2O

0,1

1

.

Ta c6: 232a + 120b = 5,84.

Phuong trinh ion rut gon:

Mol:

3,808

.-

Trong X:

DVVH Khang Vigt

D . 10,24.

n,

(muoi) = netraod6i

= 0,7mol.

^
<»i ('

1

jf'ju'/j

N03
Bao toan khoi lugng:
m = m p e , M g , zn+m^^oa

—> D a p an C .

= ^^'^^ ^

(v

^ ^^'^^

.•••L
29


Cty TNHH MTV DVVH Khang Vijt

C^m nang 6n luy^n thi d

Vi

8 (CD-12): H o a tan hoan toan 8,9 gam hon hg-p gom M g va Z n bang
lugng v u a d u 500ml dung dich HNOa I M . Sau khi cac phan ling ket thiic,
thu dugc 1,008 lit khi N2O (dktc) duy nhat v a dung d|ch X chiia m gam
muoi. G i a tri ciia m la
A. 31,22.

,

B. 34,10.

C . 33,70.

,1
. . . D . 34,32.

'

i»

> Muoi nitrat + N2O
2N*^ + Be

va CO the xay ra qua trinh:

'

*

> N2O
2N*^ + Be

V i dyt 9: D a n luong khi C O di qua hon hgp gom C u O v a F e 2 0 3 nung nong,
sau mpt thoi gian thu dug-c cha't ran X va khi Y . C h o Y hap thy hoan toan
vao dung dich Ba(OH)2 d u , thu dugc 29,55 gam ket hia. Chat ran X phan
ung voi dung dich H N O 3 d u thu dugc V lit khi N O (san pham k h u duy
,s,i nhat, a dktc). G i a tri cua V la

Mol:

0,15

0,15

n,, trao doi = IIXQQ

. M a n ^ o = ^C02 ~ 0/15 mol -> ng trao doi = 0,3 mol.

Bao toan electron: ngtraodoi = 3 n N o ^
V N O =2,24 lit

Dap an A .

T^NO =0,1 mol.

+

K h i cho oxit

+ HNO3:

("""OO

= ng trao doi = 3 n ^ o = 0,09 mol.

1 goc O^^ trong oxit se bi thay the bang 2 go'c
3i'

= 2n , = 0,06 mol.
"

^'^^ " ^'"^^

Bao toan electron: ng(x)=2nQ+3nr^o =0,15 mol - > n ^ ^ . =0,15 mol.
Bao toan N : nnNOs = ^^Q'^

" ^'^^

~^

^'

''

V i d^ 11 (B-12): C h o 29 gam hon hgp A l , C u v a A g tac dung v u a d u v a i 950ml
dung dich H N O 3 1 , 5 M , thu dugc dung dich chua m gam muoi v a 5,6 lit khi
X (dktc) gom N O v a N2O. T i khoi caa X so v a i H2 la 16,4. G i a tri ciia m la
B. 97,20.

C . 98,75.

D . 91,00.

qua: nNo= 0,20 mol; n^^o= 0,05 mol.
Al,Cu,Ag
n
'

Nhan xet: K h i cho hSn hgp oxit tac dung vdi khi C O thi:

K h i cho kim loai + HNOs: T^^Q-

6 bai nay, truoc he't cac em can tim so mol moi khi trong X de thu dugc ke't

6 day, cac em can s u dyng so do phan ung:

> BaCOs + H2O

^ " ^ " ^ > Muoi nitrat + N O

Lai gidi:

Lffigiai:

) Muoi nitrat + N O

+

A. 98,20.

D . 3,36.

" B a C 0 3 = 0,15 mol

C u O , Fe203
> X
k>r>
CO2 + Ba(OH)2

rni-ii ricirt •

Cach 2: Qui doi Y thanh X va O (0,03 mol).

Nhan xet: C a n l u u y M g tac dung vai HNO3 c6 the tao ra muoi NH4NO3.

/J <

:,,0,f rrr'''.':..

—> D a p an D .

Dap anB.

C . 6,72.

v

nQ.2 ^^^.jj= 0,03 mol.

Bao toan nguyen to N : nnwos = " N O "

,^ ^

= 8,9 + (0,36+B.0,005).62 + 0,005.80 = 34,1 gam

B.4,48.

1,;,

D- 0/18. ;,; {

' W

> Y

N O o de tao muoi nitrat, do do: n

a = 0,005 mol.

' ' Bao toan khoi luong: m = mMg, Zn+ m^^^^ + mNH4N03

. j-'

Ltm y: Y chua ca kim loai (con du) va oxit.

> NH4NO3 (a mol)

Bao toan nguyen to N : nHNOs = " N O 3 ^ ^ " ^ ^ 2 0 + 2nNH4N03

A. 2,24.

C . 0,16.

e a c h 1: S u dung so do: X

"

i^jsjQ-(muoi) = ng trao doi = 8 njs420 + 8 nNH4N03 ~ (0/36 + 8a) mol.

f

B. 0,14.

-> n o 2 = 0/015 mol -> n o = 0,03 mol ^

'>•

r

-> 0,5 = 0,36 + 8a + 2.0,045 + 2a

pham k h u duy nhat, a dktc). So mol HNO3 da phan ling la

Bao toan khoi lugng: m o 2 = 2,71 - 2,23 = 0,48 gam

,

K h i cho k i m loai + HNO3:
*

hoan toan Y vao dung dich HNO3 (du), thu dugc 0,672 lit khi N O (san

Lodgidi:

nN20= ^ ^ = 0 , 0 4 5 m o l ; n H N 0 3 = 0/5 mol.

Chat oxi hoa:

trong khi oxi, sau mot thoi gian thu dugc 2,71 gam hon hgp Y . H o a tan

A. 0,12.

Loigidi:

Mg,Zn

V i dV 10 (B-10): N u n g 2,23 gam hon hgp X gom cac kim loai Fe, A l , Z n , Mg

) Muoi nitrat + N O + N2O

Chat oxi hoa:

N^=^ + 3e

va CO the xay ra qua trinh:

> NO;
2N*5

2N*^ + 8e

+ Se

> N2O

> NH4NO3 (a mol) ,

K h i cho kim loai + HNO3:
"NO3

^ ""^'"°

^ ^

^ ^ " N 2 O ^ 8 riNH4N03 = (1 + 8a) mol. ,, ^ j

Bao toan nguyen to N : nHN03

^^oi

^^'^

^ "N2O

2nNH4N03

'

-> 0,95.1,5 = 1 + 8a + 0,2 + 2.0,05 + 2a -> a = 0,0125 mol.

31


elm nang On luyjn thi dgi hgc 18 chuySn dg H6a hpc - Nguyjn Van HSl

Bao toan khoi luqmg: m = m^i, c u , A g + "1,^0^ +

Cty TNHH MTV DWH Khang Vijt

mNH4N03

Xet su trao doi electron a cac giai doan:
(3): Fe - 3 e

= 29 + (1+8.0,0125).62 + 0,0125.80 = 98,2 gam.
Dap an A.

*'

Cu - 2 e

>• Cu*^ ~^

Nhan xet: C a n nhan ra bai toan da "giau d i " san pham NH4NO3.
gom 10,8 gam A l va 2,4 gam Mg, thu dugc 40,9 gam hon h(?p chat ran Y .
Phan tram the tich cua khi CI2 trong X la
B.40%.

C.50%.

,

Laigidi:

n^,= M = a 4 m o l ; n M , = ^ = 0 , l m o l .

27

^ 24

,

.

Mol:

gnyb .ig :f d . i 5

Al -

3e

M g - 2e

> Al*^

Chat oxi hoa:
Mol:

CI2 + 2e

> ICf^

4x

X

> Mg*^

M o l : 0 , 1 - > 0,2 * f'^^'^

O2 + 4e

> lOr

5

\ ]. (w^ - ,

;gnu nerfq 6 u ' > a

PHl/ONGPHAPTRUNGHOADIEN

.

Q si*: q&CJ <
.(OMI) f

-'••^h ^.-.-fi

5^! ( ;v.

V i dv 1: D u n g dich X chua cac ion: Fe^* (0,1 m o l ) , A P " (0,2 mol), C h (x mol) va
S04~ (y m o l ) . C o c^n dung djch X thu dugc 46,9 gam chat ran khan Y. Gia

D . 0,14.

Laigidi:
nNO= ^ ; r — = 0,04 mol.
22,4
;
Bao toan kho'i lugng: TTIQ^ = m y - mx = 12,64 - 56a -1,92 = 10,72 - 56a.
Nhan xet: Neu d y a theo phuong trinh phan ung se rat dai va kho giai.

v

tri cua X va y Ian lugt la
A. 0,2 v a 0,3 .

B. 0,1 v a 0,2.

•

nr,^ipf\n\~n

C . 0,4 v a 0,4.

f:,

D . 0,3 v a 0,3. >

3 orbs

Laigidi:

dktc). G i a tri ciia a la

Nhan xet: D u n g dich da cho chua 4 loai ion, trong do 2 loai ion chvra bie't so

r

mol, do vay can lap dugc 2 phuong trinh dai so de tim so mol cua chung.
+

A p dyng dinh luat trung hoa di?n:
Tong so mol di^n tich duong = Tong so mol di^n tich am

-> 0,1.2
+

+ 0,2.3 = x.l + y.2 ^

X

+ 2y = 0,8 mol.

A p dung bao toan khoi lugng: my =
56.0,1 + 27.0,2 + 35,5x + 96y = 46,9

Cach 1:
"""^^^ ) Fe^^ Cu^^ (3)

lit,;

VIDVMAU

dich HNO3 loang (du), thu dugc 0,896 lit khi N O (san pham k h u duy nhat 6

(2)

,„

ysb n

Dung dich cac chat dien li luon luon trung hoa ve di^n.

thoi gian, thu dugc 12,64 gam chat ran Y . Hoa tan hoan toan Y bang dung

Y

;,

i i O ftoi v M l ' o ^ ' i ' :

.;„,,

b. B i e u t h i i c

V i dv 13: N u n g hon hop X gom a mol Fe va 0,03 mol C u trong khong khi mot

6 day, cac em can s u diing so do phan ung:

.

Lieu y: De tinh so'mol di?n tich, cac em la'y so mol ion x di^n tich ion do.

= 0,2 mol; y = 0,3 mol
•
Q3
%Vn =
'
.100% = 60% -> Dap an D .
CI2 0,2 + 0,3
^

Fe, C u (1) —

nenhSn=3nNo =0,12.

Tong so'mol di^n tich duong = Tong so mol di^n tich am:

X

C.0,12.

„

a. N p i d u n g

M o l : y —> 2y

B.0,10.

,

'.f.

-^DapanD.

y t -> 32x + 71y + 10,8 + 2,4 = 40,9 -> 32x + 71y = 27,7.

A. 0,08.

' '•'

= 1,34-7a

Bao toan electron: 3a + 2.a03 = 2b + ai2 ^ a = ai4; b = ai8

Bao toan kho'i luong: mx + Tn^i + m j ^ g = my

->

> NO

10,72 - 56a

nenh^n=4no2 =

Bao toan khoi lugng: 56a + 16b = 12,64-0,03.64 = 10,72.

Bao toan electron: 4x + 2y = 1,2 + 0,2 = 1,4 mol.

ImLrx.
iff,: ^
m/

(2) -* (3): N*"' +3e

.

"

Cach 2: Q u i doi Y thanh: Fe (a mol); C u (0,03 mol) va O (b mol).

M&ii) Igol^M m mifl:4 y i[ mil

0,4-^1,2

> IQ-^

= 2ncu =0,06 '"•

Dap an D .

Nhan xet: Day la bai toan hai chat k h u (2 kim loai) va hai chat oxi hoa (2 phi
kim) nen can ap dung dinh luat bao toan electron.
Chat k h u :

(1) _^ (2): O2 +4e

s ^ ; ; n u,

Bao toan electron: 3a + 0,0"6 = 1,34 - 7a + 0,12 -> a = 0,14 mol.

D . 60%.

- •',
X

nenhir6ng

^ , _^

V i dv 12: C h o hon hop khi X gom CI2 va O2 tac dung v u a du vol hon hgp bpt

A. 80%.

nenhirang = 3 npg = 3a

> Fe*^ - >

^

X

= 0,2 mol; y = 0,3 mol.

Dap an A.

m p g 2 + + ^^3+

'^^

n
"^ci" ""^

€0,0

H

"^sol" '

35,5x + 96y = 35,9.
. ^, , ; ,

-'"JOb +

>
33


Cty TNHH MTV DVVH Khang Vigt

dm nang On luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hai

Tirdo—>

Luu y: D i e m m a u cho't d bai nay l a cac e m d u a ra p h u a n g t r i n h d^ii so' cua d j n h
l u ^ t t r u n g hoa d i e n tich.:t.'e'-f

,ji*i„trt,»sj:fr

'^4f^-

V i d y 2 (B-12):, M p t d u n g d i c h g o m : Na* (0,01 m o l ) ; Ca^* (0,02 m o l ) ; H C O ,

2HCOi

(0,02 m o l ) v a i o n X (a m o l ) . I o n X va gia t n cua a la
•••

A. O H - va 0,03.

B. Ch va 0,01.

D . N O ^ v a 0,03.

"• < f J>0 + eC rrfor;

A p d y n g d i n h luat t r u n g hoa di?n:
1.0,01 + 2.0,02 = 1.0,02 + n.a

+

'

n.a = 0,03.

'

'

'

'

fJom «>

irtmrU

Y

., ,('
't;'^

*^

•
' ' '

m = 2(0,02.40 + 0,04.23 + 0,015.60 + 35,5.0,05) = 8,79 g a m

D a p an B.

V i d y 4: M p t coc n u o c chua: a m o l Ca^^ b m o l Mg^^ v a c m o l H C O 3 . Cho t o i
thieu V l i t d u n g d i c h Ca(OH)2 x mol/1 vao coc de l a m g i a m t o n g n o n g d p i o n

rSffr

k i m loai t r o n g coc x u o n g m u c n h o nhat. Bieu t h u c t i n h V theo a, b , x la

ion N O

2a+ b
A V = ^ ^ .

3?

a+b
B.V^ — .

,

A6i-U)MU*I-.

..i-.,-'

> C 0 ^ - + H20.

—> D a p an D .

'-IdU'dfi

Lin gidi:

V i d^ 3 (B-IG): D u n g d i c h X chira cac i o n : C a ^ Na*, H C O ,

Cac p h a n l i n g hoa hpc k h i cho V l i t d u n g d i c h Ca(OH)2 x mol/1 vao coc:
Ca(OH)2 + Ca(HC03)2

v a C I " (0,1 m o l ) .

Mol:

Mol:

fjv \

D . 8,79.

Lai gidi:

kUA^gi
Itfbl'

<-

b

'

A . 3,98.

> MgCOs +CaC03

+ 2 H 2 0 ^ ' "

a+b

•'

Uiidnj-ir.,
"'"niih. ri>\ ;iM'«vJ>'>i>t oifo \—> D a p an B.
, ,
,,
,,,,„,„,,
--

i •

B.6,87.

C. 5,43.

D . 4,78.

Lai gidi:
O b a i nay, d u n g d i c h da cho chua 4 lo?ii i o n , t r o n g d o i o n SO 4 " chua biet so

Nhqn xet: K h i cho Ca(OH)2 d u vao 1/2 d u n g d i c h X, toan b p goc H C O 3 se d i

m o l , d o v a y can lap d u p e 1 p h u o n g t r i n h d a i so de t i m so m o l nay.

* vao ket tua:
OH- + HCO:

> CO,^' + H 2 O .

Ca2> + C O ^ -

> CaCOs

K h i N a O H d u vao 1/2 d u n g d i c h X:
OH- + HCO

> CO

3"

0,03

->

Ca2* + C O 3 "
0,02

3"

+ H2O

0,02

A p d y n g d i n h l u a t t r u n g hoa d i ^ n :

* i ^

T o n g s o ' m o l d i ^ n tich d u o n g = T o n g so'mol dien tich a m
0,01.1 + 0,02.3 = 0,04.1 + x.2^^, x = 0,015 m o l .
+

m'^'^'^'^^^'^'

Bao toan k h o i l u p n g : m = m^^^ + m^^^ + m^^^. + m^^2.
= 39.0,01 + 56.0,02 + 62.0,04 + 96.0,015 = 5,43 gam.

f,
.,.^j,

^

- » D a p an A .

0,03

Lieu y: tai nay cac e m can ap d u n g d j n h luat t r u n g hoa di?n de t i m dupe so

> CaCa
<-

+

„ >^;TOMb rbil
- > c = 0,03 mol.

0,03
0,02

••ttVqed

b

V =

A p d y n g d i n h luat t r i i n g hoa d i ^ n : 2a + b = c + 0,05.

Mol:

U,,J,,W...

v a SO 4~ (x m o l ) . C o can Y t h u d u p e m g a m m u o i k h a n . Gia t r j cua m l a

va CI-(0,05 mol).

Mol:

2CaC03 + 2 H 2 O

V i d v 5: D u n g d i c h Y c6 chvia: K ^ (0,01 m o l ) , Fe^^ (0,02 m o l ) , N O 3 (0,04 m o l )

G p i so m o l t r o n g 1/2 d u n g d j c h X: Ca^* (a m o l ) ; Na* (b m o l ) , H C O 3 (c mol)

Mol:

>

'~ '

a

<-

—>Vx = a + b—>

d u g c m g a m chat ran k h a n . Gia t r j cua m la
C . 9,26.

a

Ca(OH)2 + M g ( H C 0 3 ) 2

t h u d u o c 3 g a m ket tua. M a t khac, neu d u n soi den c^n d u n g d i c h X t h i t h u
B.7,47.

"mtlJl,

+

Cho 1/2 d u n g d i c h X phan l i n g v o i d u n g d i c h N a O H d u , t h u d u g c 2 g a m

;V

•

'

A p d u n g d i n h l u a t t r u n g hoa dien: 2a + 2b = c.

ket tua. C h o 1/2 d u n g d i c h X c o n lai p h a n l i n g v o i d u n g d i c h Ca(OH)2 d u ,

A . 9,21.

^ ^ _ i ± b
D . V=
2x

+
ife graiQ

jt^n

^ , , _ a + 2b
C.V=^ ^ .

X

N h a n thay i o n O H " k h o n g the t o n tai cung i o n H C O 3 t r o n g d u n g d i c h ban

OH- + HCO;

+

CO^" + CO2 + H2O.

D e n day c6 2 p h u a n g an thoa m a n la A v a D . V a y c h p n i o n X la i o n O H " hay

daudocophanung:

'

^

m o l ) ; C O 3 ' (0,015 m o l ) ; C I - = 0,05 m o l .

Lai gidi:
+

—

,

D o v%y, 1/2 cha't r a n k h a n t h u d u p e chua cac i o n : Ca^* (0,02 m o l ) ; N a * (0,04

C. COs^- va 0,03.

Goi dien tich ion X la - n .

, >,r

lim y: K h i d u n soi den can d u n g d i c h X, goc H C O 3 b i p h a n h u y :

.J.K ..•:i.«jSJi':'iie|'

:>''J*!;/i

b = 0,04 m o l .

- > a = 0,02 mol.

m o l cua goc sunfat.

" t,Oi:<; r ' i , „ i
• .

35


Cty TNHH MTV DWH Khang Vi$t
Ca'm nang On luyjn thi dgi hgc 18 chuySn dS H6a hpc - Nguygn Van Hi'i

Ap dung djnh luat trung hoa di^n: 0,01.2 + 0,03.1 = 0,02.2 + n^^. .1

Vi dv 6: Dung djch X c6 chua: Fe^* (0,05 mol), Na* (0,07 mol), CI" (0,03 mol) va

-> n^j_ = 0,01 mol.

SO 4 " . Cho dung dich Ba(OH)2 du vao dung dich X thu dugc ket tiia Y.
Nung Y a nhi^t dg cao ngoai khong khi den khol lugng khong doi thu dugc

+ Bao toan khoi lugng:
Khoi lugng cac chat tan trong Y = ^^^^2+ + ' " N H ^

^, m gam chat tin Z (coi BaS04 khong hi nhi?t phan). Gia tri cua m la
A. 19,91.
B. 18,11.
C. 24,31.
D. 20,31.
Lai nidi:
*
^ Ap dung djnh luat trung hoa dien: 2^ n^,

2. 0,05 +1.0,07 = 0,03.1 + 2n

Mol:

lugng chat tan trong mgt nua dung dich Y.
Vi
° •' 0,07

0,07

. 0,07 ,
Fe^-

y
^

Mol:
0,025
Vay m = 233.0,07 + 160.0,025 = 20,31 gam.

•• •

j,,

ket hia va 0,672 lit khi (dktc).

Ba2*

+ CO3"

Ca-

+ COr

C. 5,35 gam.

CaC03l

A. 1.

o'ijhqil

>NH3t+H20

= 0,02 lit = 20 ml - > Dap an B . ,

B. 12.

C. 13.

D.2.

+ Ap dung dinh luat trung hoa dif n vai dung djch X: ,
1.0,07 = 2.0,02+l.n _ -» n _ = 0,03 mol.
OH

-A

OH

•*• Ap dung djnh luat trung hoa dien vai dung dich Y: ,^

Phan 2 + dung djch BaCh du:

= 0/02 mol.

V=

Votigidi:

^

- > % 2 . = nMg(OH),= ^ = a 0 1 m o l .

~^ "Ba2+ = "BaS04 = ^

,

va NO3 la 0,04. Trgn X va Y dugc 100ml dung dich Z. Gia tri pH ciia Z la

jj

> Mg(OH)2i

BaS04;

^'

mol). Dung djch Y c6 chua CIO 4, NO 3 va H"" (y mol); tong so mol CIO4

o Phan 1 + dung djch NaOH du:

^

jii 0

H ^ ^ - , . . . BV p : ^ ; - :

Vi dv 9 (A-10): Dung dich X c6 chua: Na^ (0,07 mol); SO^" (0,02 mol) va O H " (x

D. 7,05 gam.

Lot gtat:

,

.

> BaCOsi

nNa2C03 = 0,02 mol

NMn xet: Bai nay cac em giai dua theo cac phuong trinh ion.

Ba^^ + sol"

-M.J'jj \t

3

Tong kho'i lugng cac chat tan trong Y la

v a n ^ . = n N H 3 = - r r - = 0,03 mol.
4
'^"^ 22,4

iftAl

: Xn^^^2- = X + y = 0,02 mol.

.y

+ Phan 2 tac dyng voi dung djch BaCl2 du, thu dugc 4,66 gam ket tiia.

NH^ +OH-

D. 70.

Cac phuong trinh phan ung :

+ Phan 1 cho tac dung voi dung dich NaOH du, dun nong, thu dugc 0,58 gam

Mg2^ + 2 0 H "

,

Ap dyng djnh luat trung hoa di?n: 2x + 2y = 0,01.1+ 0,03.1 = 0,04 mol.

bang nhau.

'

C.40.

Ggi so'mol ciia cac ion : Ba^* = X ; Ca^* = y.

7: Chia dung dich Y chua cac ion: Mg2% N H J , SO 4", CI" thanh hai phan

B. 6,11 gam.

,

LaigidU

• •

A. 3,055 gam.

,

8: Dung dich Y c6 chua dong thoi cac ion: Ba^*; Ca^*; CI" (0,01 mol) va
Ion nhat. Gia trj nho nhat ciia V la
A. 30.
B.20.

0,0125 --^

—>• Dap an D.
Vi

,

NO J (0,03 mol). Cho V ml dung djch Na2C03 I M vao Y de thu dugc ket tiia

> F e ( O H ) 3 1 Fe203

)Fe(OH)2

"^cr "^sol" ' •'• '

Lim y: Bai nay cac em de chgn nham dap an A (3,055 gam) do chi tinh khoi

^ n , . = 0,07mol.

> BaS04

^':^^Otti-->

_> Dap an B

- 2^ nj,

"""'^ ) BaS04

^

= 2.(0,01.24 + 0,03.18 + 0,01.35,5 + 0,02.96) = 6,11 gam

Cac so do phan ung:
SO 4 "

,'^f^•v:iJiJr^;,'^^ii:...

f
'
-Aq«Q.^-'
'h '"^nt^h 'qsi^^^IKi! -yin ••{in itff ->;;
jf;|„yi j ^ : . ,

,j

:

^•"ciOi -^l-^NOS =^-V
V =0'04mol.
•
Trgn X vai Y: H* + O H > H2O
-> H^du = 0,01 mol -» IH^]= ^
= 0,1 = 10"^ ->.pH = l
- > Dap an A.

., ,
" + 3.,'
i ? -tit

•

( i n i l ) ;
^

rf.^ o: 37


CtyTI\ih.i
Ca'm nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hgc - Nguygn van H&\

6

1TV DWH Khang Vi^t

PHI/ONGPHAPDI/6NGCHEO

V i dv 10 (CD-07): D u n g dich Z chiia: Cu^* (0,02 mol), K"" (0,03 mol), C h (x mol)
va S O 4 "

(y mol). Tong kho'i lugng cac muo'i tan c6 trong Z la 5,435 gam. Gig

tri cua x va y Ian lugt la
A. 0,02 va 0,05.

V6i

hSrhgp bat ki g6m

trung b m h ( M ) t a s e c 6 t i l e :

B. 0,05 va 0,01.

C . 0,01 va 0,03.

D . 0,03 v a 0,02.

M

LOT gidi:

W

0,02.2 + 0,03.1 = x . l + y.2
X + 2y = 0,07 mol.
B a o t o a n k h o i lugng: m z = m ^ 2 . + m . + m^,. + m

B i l l ;H^i,^>vvi.

^"

"

^'

^

S,,

-

• ,-,4 v ' c ! *

. 1 ; ^ ,r|/

Phan 1 tac dung voi dung dich N a O H du, d u n nong thu dugc 0,672 lit khi
(dktc) v a 1,07 gam ket tiia;

...^

Co can X thu dugc m gam muo'i khan. Gia tri cua m la
B. 7,04 gam.
Lai

C . 8,52 gam.

1*'

Nhan xet: bai nay cac em giai dua theo cac phuong trinh ion.
,

> Fe(OH)3l
.
>NH3T+H20
1'07

->

"Fe3^ = " M O H ) 3 =

.'rin»*Wi

,
• 6':> V fti\b
,

= ^'^^

"^"l-

" N H ^ = ''NHg= ^

>

= 0'03

BaSOU

Ap dung dinh luat trung hoa di?n: 0,01.3 + 0,03.1 = 0,02.2 + n^^^. .1

+

Bao toan khoi lugng: m = m^ 3+ +

*'

mol.

i.:
jA •

^

'

+ "^j>jo~ ""^ "^so^'

an C .

,

J:^.Asxu'^'^^^'^^^: i^^^y •

" -

' D . 73%. n :
'o

Cach 1: Ggi so nguyen h i 29 C u la a v a 2^ C u la b.
Ap dung cong thuc cua phuong phap duong cheo, ta c6:'
65 - 63,54

1,46

73

I 63-63,54

0,54

27

» % § C u = 73%.

a

an D .

Cach 2: N h a n thay A c u = 63,54 <

63 + 65

dong v i 29 C u chiem u u the hon

D a p an D (Cac dap an khac deu < 50%).

'

V i d^;i 2 (CD-07): C h o 4,48 lit khi C O (dktc) t u tir di qua ong s i i nung nong
dung 8 gam mgt oxit sat den khi phan ling xay ra hoan toan, thu dugc hon
hgp khi X c6 ti khoi so voi hidro bang 20. Cong thiic ciia oxit sat v a phan
tram the tich ciia khi C O 2 trong X la
A.FeO;75%.
B. FezOs; 75%.'

C . FeaOs; 65%.

D.Fe304;75%.

Lai

gidi:
4 48
Theo bai: M = 20.2 = 40 v a nx = n c o = T T T = 0'2 m o l . .

. , ,

,

A p dving cong thiic ciia phuong phap duang cheo, ta c6: ' ''
"CO

= 2.(0,01.56 + 0,03.18 + 0,02.62 + 0,02.96) = 8,52 gam
Dap

Lot gidi:

- > % 29 C u > 50%
,

4 66

= 0,02 mol.

AM,

tong so nguyen t u ciia dong vj 2 9 C u la
A. 27%.
B. 50%.
C . 54%.

Dap
t.

nnw

0,672

Phan 2: + dung dich B a C h du: Ba^* + S O 4 "

-> n _

n,

29 C u . N g u y e n t u khoi trung binh cua dong la 63,54. T h a n h phan phan tram

D . 4,26 gam.

gidi:

. Phan 1: + Bung dich N a O H d u :
Fe3* + 3 0 H '
t ' C ^
N H ; +OH*

lAMv

M,

• :';©aji-4):;::.,3-plIiJ's

, ,. ;„)-..i^^:;;

Phan 2 tac dung voi dung djch B a C h du, thu dugc 4,66 gam ket tua.
. A . 3,52 gam.

n>

V i dxf. 1 (CD-07): Trong t u nhien, nguyen to dong c6 hai dong v i la 2 9 C u v a

V i dvi 11 (CD-08): Chia dung d k h X chiia cac ion: Fe^^ S O | ~ , N H 4 , N O 3
-

Y:

,>.'i, •
•

v i o y MAU
,

—> D a p an D .

"t* , ' f

,,

1=0

''^
' N ^ a n u Q :8 i#b >;

x = 0,03 mol; y = 0,02 mol.

thanh hai phan bang nhau.

..,

- K i l o f n ' e O i O ) 'rOVi

64.0,02 + 39.0,03 + 35,5x + 96y = 5,435
35,5x + 96y = 2,985 ^

i »! ^ > I )

AMN

X:

^t.

A p dung dinh luat trung hoa dien:

- M Y

b. So do hoa

do vay can lap dugc 2 phuang trinh dai so de tim so' mol cua chung.
+

,

M-Mx

ny

• O bai nay, dung dich Z chiia 4 loai ion, trong do 2 loai ion chua biet so'mol,

. ^

hai chat X va Y , khi biet dugc gia tri khoi lugng mol

nC02

=

44 -40

1

0,05

28 - 4 0

3

0,15

%Vco2 =75%

chgn B hoac D .
39


Ca'm nang 6n luy$n thi dgi hpc 18 chuyfin dg H6a hpc - Nguygn Van Hi\

8-0,15.16

T a c o : no(oxit)= "cOz = 0'15 m o l - >

ion _^

: n o = 0,1 : 0,15 = 2:3 -> Fe203

Cty TNHH MTV DVVH Khang Vigt

56
D a p an B. \ i ' fin.?'"/-:

V i d u 3: Cho m o t m a n h C u tac d u n g v a i d u n g d i c h H N O s d u , t h u d u o c d u n g
d i c h X v a 0,896 l i t (dktc) h o n h o p k h i Y g o m N O v a NO2. T i k h o i ciia Y so
v o i h i d r o bang 19. Co can X t h u dugc m g a m m u o i khan. Gia t r i cua m la
A . 5,64 gam.

B. 7,52 gam.
' "

C. 9,4 gam.

D . 15,04 gam.

30-38 ~1 ~0,02

"Cu =

"N; _ 2-7,2 = 28-7,2
4

! i:M

:

G g i : n^^^ = 1 m o l ; nH2 = 4 m o l .
i ^

- > mx = 28.1 + 2.4 = 36 g a m

Phan u n g :
—
Ta c6: M Y

m v = mx = 36 gam.
+ 3H2 ^=:±
3x

x
36
=

nNO= 0'02 m o l ; n ^ o j = 0,02 m o l . -«j ijxj |V

5-2x

->

UY = (5 - 2x) m o l

2x
X

x = 0,25 m o l ^

=8

2NH3

H = ^ 1 0 0 % = 25% .

- > D a p an A .
V i dv 6 (A-12): H o n h g p X g o m H2 va C2H4 c6 t i k h o i so v o i H2 la 7,5. D a n X

A p d y n g bao toan electron: 2 ncu = 3 n ^ o + 1 • " N O Z
^

i

A p d y n g cong thuc ciia p h u o n g phap d u o n g cheo v o i X, ta c6:

N2

^ 46-38 _ 1 _ 0,02

nN02

Theo bai: M x = 1,8.4 = 7,2; M Y = 2.4 = 8 .

•

A p d u n g cong thuc cua p h u a n g p h a p d u o n g cheo, ta c6:
HNO

iv ri I ^ n6'>

Phan l i n g xay ra n h u sau:

Lcngidi:

0,896
^.
rr
= 0,04 m o l . Theo bai: My = 19.2 = 38.
22,4

nY =

Lai gidi:

= 0,1 m o l .

3.0,02 + 1.0,02
=V
.
-=0,04 m o l - > ncu(N03)2 = nCu = 0 . 0 4 m o l , :

qua N i n u n g nong, t h u d u g c h o n h g p Y c6 t i k h o i so v a i H2 la 12,5. H i ^ u

suat ciia p h a n u n g h i d r o hoa l a
A . 70%.

m = 0,04.188 = 7,52 g a m - > D a p a n B.

B.80%.

C.60%.

i^-'i D . 50%.

Lai gidi:

V i dy 4: H o a t a n hoan toan m g a m A l bang d u n g djch H N O 3 loang, t h u d u g c

^ '•

d u n g d i c h X ( k h o n g chiia NH4NO3) v a 5,6 l i t (dktc) h o n h g p k h i Y g o m N O

Theo bai: M x = 7,5.2 = 15;

va N2O. T i k h o i h o i ciia Y so v a i h i d r o bang 19,2. So m o l axit H N O a da

M Y =12,5.2 = 25.
S lor~dS[
A p d u n g cong t h i i c ciia p h u a n g phap d u o n g cheo v o i X, ta c6:

t h a m gia p h a n u n g la
A . 1,2.

B. 1,5.

C. 1,7.

D . 2,0.

Loigidi:
56
—
nY = —
= 0,25 m o l . Theo bai: M Y = 19,2.2 = 38,4.
22,4

;

"•

.„ ,;i.,v< — !,(,./«

44 - 38,4 ' 2

nN20= 0,15 m o l ; nMo= 0,10 m o l .

Bao toan electron: 3 n ^ i = 8. n ^ ^ o + 3 n ^ o

Ta c6: M Y -

= 0,5 m o l ^ nAi(N03)3 = ^Al = 0,5 m o l .

'

"

X

= 25 ^
2-x

Bao toan n g u y e n to N : nHNOa = 3 n A i ( N 0 3 ) 3
"

+

2n ^20+

N H 3 la
C. 36%.

X

- > nY = ( 2 - x ) m o l

i"'y •. ' •

D a p an B.
X qua N i n u n g nong, t h u d u g c h o n h g p Y c6 t i k h o i so v a i heli la 5. H i ? u

= 3.0,5 + 2.0,15 + 0,1 = 1,9 m o l ^ D a p a n D .

h g p k h i Y c6 t i k h o i so v o i heli bang 2. H i f u sua't ciia p h a n u n g t o n g h g p

• ' B. 50%.

>

C2H6

x = 0,8 m o l - > H = -.100% = 80%.
1

sua't ciia p h a n l i n g h i d r o hoa la

A . 20%.

B.25%.

D u n n o n g X m o t t h o i gian t r o n g b i n h k i n (c6 Fe l a m xuc tac), t h u d u g c h o n

,r

^

Vi d^ 7 (CD-09): H o n h g p X g o m H2 v a C2H4 c6 t i k h o i so v o i heli la 3,75. D a n
" N O

V i dy 5 (A-10): H o n h g p k h i X g o m N2 v a H2 c6 t i k h o i so v a i heli b ^ n g 1,8.

A . 25%.

+

mv = mx = 30 gam.

+ H2

X

Phan l i n g :

8.0,15 + 3.0,1
"Al =

di

C2H4

0,15
0,10

"H2 = ^

Ggi: nc2H4 = ^

-> mx = 28.1 + 2.1 = 30 gam

rn,

Phan u n g xay ra n h u sau:

A p d u n g cong thuc ciia p h u o n g p h a p d u o n g cheo, ta c6:
"N2O ^ 30 - 38,4' .3

"C2H4 ^ 2 - 15
28- -15
"H2

^_ | _

D . 40%.

C.50%.

D . 40%.

Lai gidi:
Theo bai: M x = 3,75.4 = 15;
M Y =5.4=20.

•

;• ;r .


dm

nang On Iuy0n thi dgl hpc 18 chuyfin ai H6a hqc - Nguyin van Hit

Cty TNHH MTV DWH Khang Vigt

A p d y n g cong t h u c ciia p h u a n g p h a p d u o n g cheo v a i X, ta c6:
"C2H4

2-15

^1

~ 28-15

Goi: ric2H4 = 1 iriol; riH2 = ^ "^ol-

1

mv = mx = 30 gam.
-

Phan l i n g xay ra n h u sau:
C2H4

+

x

H2
x

T a c o : M y = ^ ^ = 20
2-x

1

x

G o i : nc4H4 = l m o l ; nH2 = 4 m o l .
^ ^

N/iflM xet: So m o l k h i g i a m = so m o l H 2 t h a m gia p h a n u n g .
Taco: M y = 2 0 - .

» ' ' J:

" ^ = -=3

- i^i^ 1 a
^

So m o l k h i g i a m = n Y - n x = 6 - 3 = 3 m o l

nv = (2 - x) m o l "v/iH

H = - . 1 0 0 % = 75% - » D a p an D .

x = 0 , 5 m o l ^ H - - . 1 0 0 % = 50%.

V i d u 8: H o n h g p X g o m H 2 va C 2 H 2 c6 t i kho'i so v a i h i d r o la 5. D a n X qua N i
X Hr tich k h i H 2 da t h a m gia p h a n l i n g h i d r o hoa la y qt>rt QQ;
C. 50%.

T r o n g m g t h o n h g p n h i e u chat, c6 the bieu d i e n m o t d a i l u g n g nao d o cua
cac chat t h o n g qua m o t dai l u g n g chung, dai d i ? n cho ca h o n h g p , g g i la dai

^

Lai sidi:

(• {) ft

a. N p i d u n g

'} t/b J V

" D . 85%.

\1

7. P H l / O N G P H A P T R U N G B I N H

n u n g nong, t h u d u g c h o n h g p Y c6 t i kho'i so v a i h i d r o la 10. Phan t r a m the

B. 75%.

^

_^ mx = 52.1 + 2.4 = 60 g a m - > mv = mx = 60 gam.

1

—> D a p an C.

o^,j} A . 60%.

= -

|t

C2H5
->

2 - -12
52 - 1 2

n u b IA

-> mx = 28.1 + 2.1 = 30 g a m

Phan l i n g :

nC4H4_^

;lado':.;riT.

k m n g trung binh.

'

,;

' '

m^M:^'r.:hH4£vim

b. Bieu thuc va van d u n g

Theo bai: M x = 5.2 = 10; M y = 10.2 = 20.

^

.

A p d u n g cong t h u c cua p h u o n g phap d u o n g cheo vod X, ta c6:
"C2H2 ^ 2 - 1 0
nH2

26-10

X = -i^^^
= -

^

' ^,Dm+-M„

- > mx = 26.1 + 2.2 = 30 g a m

mv = mx = 30 gam. ) mrfj

Cac p h a n u n g xay ra n h u sau:
C2H2
C2H4

C2H4

H2
2H2

(X - d a i l u g n g d a n g xet; n - s o ' m o l ; i - so t h u t u chat).

G g i : nc2H2 = l m o l ; n n j = 2 m o l .

.0

+

Khoi luang phan t u trung binh: M = ^ ^ ^ ^

+

So' n g u y e n t u cacbon t r u n g b i n h : C =

-

C2H6
_

Nhan xet: So m o l k h i g i a m = so m o l H 2 t h a m gia p h a n u n g .
Taco: M Y = 2 0 - >

"CO2

+

n Y = — = 1,5

20

:e£.o :.,.,„

5,o

So n g u y e n t u h i d r o t r u n g b i n h : ^ H = — —

VIDUMAU

So m o l k h i g i a m = n Y - n x = 3 - 1 , 5 = 1,5 m o l
H = — . 1 0 0 % = 75%

2nH,o

V i d u 1 (CD-11): D e hoa tan hoan toan 6,4 gam h o n h g p X g o m k i r n loai R (chi c6

D a p an B.

hoa t r i 2) va oxit ciia no can v u a du 400ml d u n g dich H C l I M . K i m l o ^ i R la
V i d y 9: H o n h g p X g o m h i d r o va v i n y l a x e t i l e n c6 t i k h o i so v a i h i d r o la 6.

A . Ba.

Phan t r a m the t i c h k h i H 2 da t h a m gia p h a n u n g h i d r o hoa la
A . 60%.

B.85%.

C.50%.

D . 75%.

Lcn gtat:
Theo bai: M x = 6.2 = 12; M y = 10.2 = 20.

B. Be.
:i.

D a n X qua N i n u n g nong, t h u d u g c h o n h g p Y c6 t i k h o i so v a i h i d r o la 10.

C. M g .
Laigiai:

M x = — = 3 2 R < 3 2 < R O

163

-A.

Cac p h u o n g t r i n h p h a n u n g :
R + 2HC1
> RCI2 + H2
N h a n thay: nx = - j U H C I = 0,2 m o l .

D . Ca.

'^'^' •
R O + 2HC1

> RCI2 + H2O

- > R < 3 2 < R + 16

A p d u n g cong t h u c cua p h u o n g p h a p d u o n g cheo v a i X, ta c6:
16
R = 24 ( M g ) - > D a p an C.
An


Ca'm nang 6n luygn thi d^i hpc 18 chuySn 6i H6a hgc - Nguygn Van Hai

Cty TNHH MTV DWH Khang Vi§t

V i d\ 2: Cho 1,67 gam hon hgip gom hai kim loai (6 2 chu ky lien tiep thupc
nhom IIA) tac dung het voi dung dich HCl (du), thoat ra 0,672 lit khi H2
(dktc). Hai k i m loai do la
A. B e v a M g .

B.MgvaCa.

CSrvaBa.

Lot gidi:
Gpi cong thiic chung ciia hai kim loai la M , hoa tri chung la n.
Ta c6: n^ = 2n^^^ = 0,1 mol. Bao toan electron: n^

_
t

> MCI2 + H2

! >fr ^ ,

Nhan thay: nj;^ = n ^ j = 0,03 mol.

Vi
if^l

dxjL 3

= 0,1

kim loai kiem M tac dung he't voi dung dich HCl (du), sinh ra 0,448 lit khi (a

i.

dktc). K i m loai M la
A.Na.

kiem tho Y tac dung he't vai lugng d u dung dich HCl loang, thu dupe 5,6 lit
khi H2 (dktc). K i m loai X, Y la
C. Natri va magie. D. Kali va canxi.

Lcri gidi:

B. K.

Cach 1: Cac phuong trinh phan ung:
2X + 2HC1

> 2XC1 + H2

Mol: a

0,5a

Taco : a5a + b = 0,25

Y + 2HC1

> YCI2 + H2

Mol: b

b

-> a + 2b = 0,5.

Mat khac: a + b < a + 2b = 0,5 < 2(a+b) -» 0,25 < a+b < 0,5
7 1 - 7 1
_
.••ulnAv,,.^,,'. „
-> — < M < —
> U< M <28.
0,5
0,25
Mx hoac M Y > M > 14 -> Loai B v i hai kim loai deu < 14.

nco2

= 0,02 mol. Cac phuong trinh phan ung:

M2CO3 + 2HC1

_^ M x = — = 95 ^
0,02

^ M + 61< 95 < 2M + 60

17,5 < M <34

M
Mat khac: 1 < n < 2

14,2 < M < 28,4. Lap luan nhu tren -» Dap an C.

W

^ M la Na ^ Dap an A.

X tac dung voi axit H2SO4 loang, d u thi the tich khi hidro sinh ra chua deh
1,12 lit (dktc). K i m loai X la
A.Ba.

B.Ca.

C. Sr.
Loigidi:

D. M g .

«) ;

,

Cac phuong trinh phan ung:
> XCI2 + H2

M = AL= 56,67
0,03
.
Mat khac: M x >

V i A\y 4 (CD-12): Hoa tan hoan toan 1,1 gam hSn hpp gom mpt k i m loai kiem X

k'mS.Dw

dung dich HCl du, sinh ra 0,672 lit khi H2 (dktc). Mat khac, khi cho 1,9 gam

Nhan thay: nx + nzn = n n , = 0,03 mol

= 0,5 ^ M = 14,2n.

^ .

V i d^ 6: Cho 1,7 gam hon hop gom kim loai X (nhom IIA) va Zn tac dung voi

Cach 2: Gpi cong thiic chung aia hai kim loai la M , hoa tri chung la n.

n.nM = 0 , 5 m o l - > - ^

^j., j ^ -

M H C O 3 < 95 < M2CO3.

X + 2HC1

Ta c6: n^ = 2nj^^ = 0,5 mol. Bao toan electron: ng (j^j = ng= 0,5 mol.

,|,,/

> 2MC1 + CO2 + H2O

-> Dap an C.

^

D.Li.

^oigidi:

Lap luan: Mx hoac M Y < M < 28 -» Loai A va D v i hai k i m loai deu > 28.
b'j

C.Rb.

M H C O 3 + HCl
> MCI + CO2 + H2O
Nh^n thay : nx = n c o 2 = 0,02 mol.

.V,; v ; . ' ^ N I

nH2 = a 2 5 m o l .

Loai B, C, D.

V i du 5 (B-08): Cho 1,9 gam hon hpp X gom muoi cacbonat va hidrocacbonat ciia

Hai kim loai la Ca va Sr -> Dap an D.

B. Liti va beri.

= lln.

Dap an A.

(A-10): Cho 7,1 gam hon hop gom mot kim loai kiem X va mpt kim loai

A. Kali va bari.

^ M

M a t k h a c : l < n < 2 -> l l < M < 2 2 . V a y M x < M < 2 2

nH2=

-> ^ ^ ^ ^ " ^ 55,67

" " e " 0,1 mol.

D. Cava Sr.
_> n.nM=0/l m o l - ^

^ ^ = 0 , 0 3 m o l . ,V
^^'^
_
Phuang trinh phan ung: M + 2HC1

** -

0,05

Zn + 2HC1

> ZnCh + H2

M x < 56,67 < Mzn -> X l a M g h o a c C a . .

,
et-' •

38 -> X la Ca -> Dap an B.

0 = d + B,

m

V i dii 7: Dot chay hoan toan 6,72 lit (dktc) hon hpp M gom hai hidrocacbon X

va mpt k i m loai kiem tho Y (Mx < MY) trong dung dich HCl du, thu dupe

va Y (Mx < MY), thu dupe 11,2 lit khi CO2 (dktc) va 10,8 gam H2O. Cong thiic

1,12 lit khi H2 (dktc). K i m loai X la

phan t u cua X la
.C2H4.

A. L i .

B.Na.

C. Rb.

D. K.

B.C2H2.

C.C2H6.

•Q™D.CH4.

'
4";


Ca'm nang On luygn thi dgi h9C 18 chuygn

H6a hpc - Mguyin Van Hii

Cty TNHH MlV DWH Khang Vigt

Lai giai:

V i dV 9 (B-08): D o t chay hoan toan 1 l i t h o n h o p k h i g o m C2H2 v a hidrocacbon

11,2

"^€02 " ; r r 7 " 0,5 m o l ^ nc = 0,5 m o l .

22,4
10 8
"H20= - T ^ =
lo

X s i n h ra 2 l i t k h i CO2 v a 2 l i t h o i H2O (cac the tich k h i va h o i d o 6 ciing

> : •! .M, /

dieu k i e n nhi^t d g , ap suat). Cong thuc phan t u ciia X la
A.C2H6.

m o l - > n H = 2.0,6 = 1,2 m o l .

Xheobai:

M =- ^ = 2 4 . ViMx
'

0,3

' m qit* *

->DapanD.

.t

i*-')ti>f. yfc

v o i H2 bang 11,25. D o t chay hoan toan 4,48 l i t X, t h u d u g c 6,72 l i t C O 2 (cac
the tich k h i d o 6 dktc). Cong t h u c ciia ankan va anken Ian l u g t la

+

t h u d u g c h o n h g p ancol Y, t r o n g d o t o n g k h o i l u g n g cac ancol bac m o t gap
13/6 Ian k h o i l u g n g ancol bac hai. Phan t r a m k h o i l u g n g ciia ancol bac m o t
(c6 so cacbon I a n h o n ) t r o n g Y la

C ; ; H , - + — O2
n

Lit:

r-f
16.0,15 + 42.0,05
,
-> M x =
—

= 22,5 - > Thoa m a n ^ D a p an C.
'

Ta c6: nx = a + b = 0,2 va nco2 = a + 4b = 0 , 3 - > a = — ; b =
3
Mx =

2 — '- ^
0,2
L o ^ i p h i r o n g an D .
46

'^

—
3

'

2n

D . 7,89%,

'

—

^

nC02

+ nH20

2

6

9n

—
- > 9 n = 21 ^
Taco: K = - ^
3

— 7
n =
3

> H a i anken l a C2H4 (a m o l ) v a CsHe (b m o l ) .

2a + 3 b ^ 7 ^ ^ ^ ^ b - > Dat a = 2 m o l ; b = 1 m o l .
a+b
3
_
,
. . ^ „;>

Phan u n g hoa hoc:
CH2=CH2 + H2O — ^ ^ ^ ^

Mol:
^20,33 ^ 22,5

C. 31,58%.

Gpi cong thuc c h u n g ciia 2 anken la C - H j - .

, . f O ^ . H . . . ,
':,,tM\k
^••"'^ it-Mi'^r:; h':6

,

B. 10,88%.

Lai giai:

Ne'ulaC:CH4 =amolvaC3H5 = bmol.

N e u la D : C H 4 = a m o l va C4H8 = b m o l .

j j ,. j ,

day d o n g d i n g can v u a d u 21 l i t O2 (cac the tich k h i d o t r o n g c u n g d i e u

A . 46,43%.

r

— = 2 —> H i d r o c a c b o n X c6 so n g u y e n

ki?n nhi^t do, ap suat). H i d r a t hoa hoan toan X t r o n g d i e u k i | n thich h g p

L o a i B.

r

; -i

V i dy 10: D o t chay hoan toan 6 l i t h o n h g p X g o m 2 anken ke tiep n h a u t r o n g

Ta c6: nx = a + b = 0,2 va n c o j = a + 3b = 0,3 - » a = 0,15; b = 0,05.

+

:

D a p an A .

I,
mH = 4,5 - 0,3.12 = 0,9 g a m - > nH = 0,9 m o l
'
'^ ^ So n g u y e n t u H t r u n g b i n h t r o n g X = ^ = ^ = 4,5
"X
0'2
;p|;y;
v

-

= 4 _> H i d r o c a c b o n X c6 so

t u C bang 2 ( v i h o n h g p chua C2H2 so n g u y e n t u C b a n g 2).

J, Bao toan k h o i l i r g n g : mx = mc + mH

,

"^*::i*>,^i:^-'O^H:,,>«:''-'''

+ So n g u y e n t u C t r u n g b i n h = —

Ta c6: mx = M x . n x = 22,5.0,2 = 4,5 gam.

.u • •

' ,:

Loai B va C.

^ ^ ' V B. C2H6 v a C2H4.

Lai giai:
4 48
6 72
= -zr--= 0,2 m o l ; nco2 = ^r—= 0,3 m o l - » nc = 0,3 m o l .
22,4
^22,4
, ,

,.
M =•,,(') >n' m • vsrfi nv,r\v
Theo bai: M x =11,25.2 = 22,5
X chua m o t hidrocacbon c6 p h a n tvr k h o i

->LoaiA.
Thu50:50

D.C3H8.

j.,,

n g u y e n t u H I o n h o n 4 ( v i h o n h g p chua C2H2 so n g u y e n t u H n h o h o n 4).

C. C H 4 v a C3H6.

n h o h o n 22,5 - > D o la metan (CH4)

C.CH4.

+ So n g u y e n t u H t r u n g b i n h = —^^^2-= ^

V i d\ 8 (B-10): H o n h o p k h i X g o m m o t ankan va m o t anken. T i k h o i cua X so

A.CH4vaC2H4.

B.C2H4.

Lai giai:

Bao toan k h o i l u g n g : mM = mc + m n = 0,5.12 + 1,2.1 = 7,2 g a m .
^

,

2

CH3-CH=CH2
+ H2O
C
Mol:

iul^ltXi

u'.

CH3CH2OH

2
> CH3CH(OH)-CH3

i
i , :

s


nang On luygn thi dji hqc 18 chuySn dg H6a hpc - Nguygn Van Hii

dm

C H 3 - C H = C H 2 + H2O

"
Mol:

)

1-x

Cty TIMHH MTV DVVH Khang Vi$t

Loigiai:

CH3CH2CH2OH

. . . 46.2 + 6 0 ( 1 - x )
13
no
1
Theobai:
-=
> x = 0,8 m o l .
60x
6
%mcH3CH2CH20H =

Cach 2: De thay: %

|;

w

.1 1% 1 A
" '

" 60,53%.

J^lhan xet: ncoz

^ "H2O

este la C - H j - O j .

,

,

,

„
= 2n

,

+ " H2O

^

_

'&'f<:.ihtii
••
= / .„ . , p r ^

' .•>
nco2 _ 0,55 _ , 7 c
n =
T—- - A/3

M a t khac, theo bai: % mcH3CH(OH)CH3 =

,^

^ac este deu no, d a n chii-c -> C o n g t h u c cua 2

^,
. ^
„
Bao toan n g u y e n t o 0x1: 2n ggte + 2n

^ ^ 7 ^ - ^ 0 0 % = 7,89% ^ D a p an D .

mcHgCHzOH =

= O'^S m o l ; n H 2 0 = 0/55 m o l . , ,

= 0,625 m o l ; n

1-x

fj

-ft

•

s;V!

mB<*(

d.e m:, rb'r:

-100% = 31,58%.
M a t khac, k h i cho X tac d u n g v o i K O H , t h u duQfc m u o i ciia m g t axit h i h i co

^

% mcH3CH2CH20H = 100 - 60,53 - 31,58 = 7,89%

—> D a p an D .

'

,

,

va hon h o p ancol Y la d o n g dang ke tiep —> 2 este t r o n g X h o n k e m n h a u 1
.,, .

nguyen t u cacbon.

V i dv 11: H o n h o p X g o m hai axit cacboxylic d o n chiic. Do't chay hoan toan
0,1 m o l X can 0,26 m o l O2, t h u d u o c CO2 v a 0,2 m o l H 2 O . C o n g t h i i c hai
axit la
A. C H 3 C O O H va C H 2 = C H C O O H .

X-

C. H C O O H va C 2 H 5 C O O H .

' " *

- > C o n g thuc cau tao: H C O O C H 3 (a m o l ) va HCOOC2H5 (b m o l ) . ,5.

- > 2a + 3b = 0,55 - > a = 0,05 m o l ; b = 0,15 m o l .

HCOOC2H5 + N a O H —
Bao toan n g u y e n to O, ta c6: 2nx + 2 no2 = 2 nco2 + "H2O

^

,

" f•*

^

HCOONa + CH3OH

H C O O N a + C2H5OH

- > m = 0,05.32 + 0,15.46 = 8 , 5 g a m

,

51^*

Phan l i n g hoa hpc:
HCOOCH3 + N a O H —

'

^ 1 1 1 0 2 6 ^

, .j. j

a +b

B. C H 2 = C H C O O H va C H 2 = C ( C H 3 ) C O O H .

_^

-

Ta co: a + b = 0,2 m o l . M a t khac: n = 2,75 = ^ ^ — ^

'

D. C H 3 C O O H v a C 2 H 5 C O O H .

—> C o n g t h i i c p h a n t i i 2 este: C2H4O2 va CsHeCh.

.

'"
£ fv A i&oJ
, fy 3 j § o J

—> D a p an A .
V i dvi 13 (A-10): Cho h o n X g o m ancol metylic v a hai axit cacboxylic (no, d o n

Nhqn xet:

n ^ j o < ^C02

^'^P ^ ^^^^

-»LoaiCvaD.

chiic, ke tiep n h a u t r o n g day d o n g dang) tac diang het v o i Na, giai p h o n g ra

^^^^ "^9* ^^^^ k h o n g no
nP

^

«*) 1

-> So n g u y e n t u C t r u n g b i n h =—£22.= 2,6 -> X c h i i a m g t axit c6 so n g u y e n

nx
ttr C n h o h a n 2,6 — > L o a i B (Hai axit deu chua so'cacbon > 2,6) .
-> D a p an A .

6,72 l i t k h i H2 (dktc). N e u d u n n o n g X (co H2SO4 dac xiic tac) t h i cac chat
t r o n g h o n hgp p h a n l i n g vtra d i i v o i nhau tao t h a n h 25 g a m h o n hgp este
(gia thie't p h a n l i n g este hoa dat h i f u suat 100%). H a i axit t r o n g X la
A. C 3 H 7 C O O H va C4H9COOH.

B. C H 3 C O O H va C2H5COOH.

C. C2H5COOH va C3H7COOH.

D. H C O O H va C H 3 C O O H .

Laigidi:

V i d\ 12: D o t chay hoan toan h o n h o p X g o m hai este can d u n g 14 l i t k h i O2,
t h u d u g c 12,32 l i t k h i C O 2 v a 9,9 g a m H2O. N e u cho m g a m X tac d u n g

nH,=

vira d u v o i d u n g d i c h K O H , c6 can d u n g d i c h sau p h a n u n g t h i t h u dxxgc

'

22,4

_
4

m u o i k h a n cua mQt axit h u u co va m gam h o n hgip ancol Y la d o n g d 5 n g ke

G p i cong t h u c c h u n g ciia 2 axit la R C O O H .

tiep. Gia t r i ciia m la

D a t so m o l : C H 3 O H = a; R C O O H = b. Ta co: a + b = 2 n H j = 0,6 m o l .

A . 8,5.
48

—=a3mol.

;«

x

C.7,1.

D . 10,1.

Phan u n g este hoa:

RCOOH

v, ,.

+ C H a O H ; e = ± RCOOCH3 + H2O

49