ca'm nang 6n luyjn thi dgi hgc 18 chuyfin
H(Sa hpc - NguySn Van H3i
Cty TNHH MTV D W H Khang Vigt
Lot gidi:
Cach 1: Khi hoa tan vao dung dich H 2 S O 4 , Fe(N03)2 se phan li thanh cac ion.
Do vay, truoc het cac em can tinh so mol cac ion nhu sau:
O' •
'
0,2
•••^'J^/•^..-:> <>^:r'^ ^i^mrp
^''^''/'i
CO the hoa tan toi da m gam Fe. Biet trong cac qua trinh tren, san pham khu
A. 5,6.
->
Mol;
0,6 ^
0,8 ^
0,3
0,2
0,6
Mol:
rr,fj{),p« y 4 -
0,1
^
^,
^,
,
vr ^
Mol:
Mol:
Dap an B.
Vi d^ 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4
0,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch
-I X. Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ke't tua Ian nhat.
.
Gia trj nho nhat ciia V la
A. 80.
B.50.
C.60.
.v, t, • 5
njs^aOH =
MjoriX^:'
n + + 2n
H
y
„
"
WO..'-*
"'"^ *
mi
2+
+
4H* + N O ;
> Fe3* + N O + 2H2O
QkHi
- > 0,05 ••idl,0» j „ r t S « . ^ ^ i T -.'ihtvU/
> 3Fe2q % lOH dfegfror* ,;>fefii JfiSv)
Vi d^ 15: Iron 200ml dung djch X gom Ba(OH)2 0,1M va NaOH 0,3M voi
100ml dung dich Y gom Al2(S04)3 0,1M va H2SO4 0,1M, thu dugc a gam ke't
tua. Gia tri cua a la
B. 4,66.
C. 5,82.
D. 5,24.
Trong Y: n 3+ = 0,02 mol; n ^ ^ j - = 0,04 mol; n^+ = 0,02 mol.
+ OH'
0,02 ^
>
0,02
AP* + 3 0 H "
'ol:
0,02
±
Cu
0,02
> Al(OH)3i
0,06
<-
->
.^w
oh
0,02
> AlO;
0,02
,Ba2^ + S04~
;
H2O
,
Al(OH)3 + O H "
/
_> V = 0,08 lit = 80ml
* ^
TrongX: n„ 2+= 0,02 mol; n,,_+ = 0,06 mol; n O H _ = 0,10 m o l . « ^ ' ^ - >
^^"
= 0,04 + 2.0,06 = 0,16 mol
— Dap an A.
>
,
—> mpe = 0,125.56 = 7,0 gam - > Dap an B.
Mol:
> 3Cu^*+ 2NO + H 2 O
;^Mol:
a06
ai6
a04
Cu tan het -> X C O chiia: n^ 2+ = 0,06; n + du = 0,04.
"NaOH
.,
0,05 4-0,4
Fe
+ 2Fe3^
0,075 <- 0,15
H+
Phuong trinh ion thu gpn:
De thu dugc luang ke't tua Idn nhat thi:
0,2
Cac phuong trinh phan ung khi pha trpn:
ncu =0,06 mol; n^+ = 2.0,2.0,5 = 0,20 mol; n ^ ^ , =0,2.0,2 = 0,04 mol.
3Cu + 8H* + 2NO3-
0,1
Laigidi:
Nhan xet: Dung dich chiia muoi NaNOs va H2SO4 loang -> can giai theo
;>
?
0,8
A. 6,22.
D. 40.
Ldi gidi:
phuong trinh ion.
,
Fe
mol
V = 0,4.22,4 = 8,96 lit ^
-> Fe(N03)3 + 5NO + 2H2SO4 + 2H2O
Fe3^ = 0,1 mol; H* = 0,4 mol; N O ; = 0,3 mol va S O ]' = 0,2 mol.
Bao toan electron: 2ncu + lnp^2+ = Sn^o
=0,4
ry, •;
Cac phan ung hoa tan Fe:
Chat oxi hoa: N*-^ + 3e - > N O
z
Lai gidi:
Dung dich X gom cac ion:
0,2 ^ ^ _
Cach 2: Cac chat khu: Cu - 2e ->• Cu^^; Fe^^ - le -> Fe^*.
2.a3+a6
•
FeS2 + 8HNO3 -
> 3¥e^ + N O + 2H2O
0,2
D. 2,8
C.8,4.*
phan ling hoa hoc:
cO/; ,
-> V = (0,2 + 0,2).22,4 = 8,96 lit -> Dap an B.
riNO=
B.7,0.
' 'V',
Cac em luu y, Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi N O :
3Fe2* + 4H* + N O ;
'
> 3Cu2- + ' ^ N O + 4H2O
3Cu + 8H* + 2NO^ •
a3 <- 0,8
san pham thu dugc gom dung dich X va mot chat khi thoat ra. Dung dich X
duy nhat cua N*^ deu la NO. Gia tri ciia m la
Phuong trinh ion rut gon:
Mol:
14: Hoa tan hoan toan 0,1 mol FeS2 trong 200ml dung dich HNO3 4M,
0,02 -> 0,02
> BaS04>l
^
0,02
Vgy: a = 0,02.233 = 4,66 gam - > Dap an B.
+ 2H2O
Cty T N H H M T V D V V H Khang Vi$t
C&m nang fln l u y j n thi dgi hgc 18 chuy6n dg H6a hoc - MguySn Van H5i
V i d v 16: H o a tan hoan toan 9,46 g a m h o n h g p g o m N a , K v a Ba vao nuoc, thu
Al
duQC d u n g d i c h X va 1,792 l i t k h i H2 (dktc). D u n g d i c h Y g o m H C l I M va
AIO2 + - H 2
+ O H - + H2O
H2SO4 0,5M. T r u n g hoa d u n g dich X b o i d u n g dich Y, tong k h o i l u g n g cae
Mol:
m u o l d u o c tao ra la
1
^
Ta c6: n H j = 2,5a = 0,05 m o l - > a = 0,02 m o l .
,
A . 14,72 gam.
B. 16,14 gam.
C. 19,98 gam.
i
D . 17,14 gam.
'2
= 0,08 m o l .
22,4
H
.
( r{ mi'n> tV
K
i
1
-H2
Na+ + O H " +
+ H2O
K+
Ba + 2H2O
i.
K h i cho t u t u Y vao X:
OH- +
Iv
2
+ OH- +
-Hi
1^x4
f1
J
1, t
Mol:
0,01
„,p.r, f t T
<
j\riv,
= 2a m o l ; n ,
so|"
T r u n g hoa X b o i Y: H + + O H "
<-
> Al3+ + 3H2O
dnSrf/;
0,03
0,01 <- 0,01
Mol:
> BaS04>l'
->
0,01
- > m = mA,(OH)3 + mBaS04 = 0-01-78+ a01.233 = 3,11 g a m . :
*.
" H + = " H C l + 2nH2S04 = 2a + 2.a = 4a m o l
cr
> A l ( O H ) 3 4'
0,02
Ba^+ + SO 4-
^
n
- > 0,02
Al(OH)3 + 3 H +
1
M a t khac, n o n g d o H C l gap d o i H2SO4 - > t r o n g cung m o t the tich t h i
^ T r o n g Y: •
;;i •
H+ -
0,02 -> 0,02
Mol:
fin J^j ti )
N/jflnxet: n ^ „ . = 2nH, =0,16 m o l .
,
1,5a
. <
AIO2 + H + + H2O
r.i I
> Ba^^ + 2 0 H " + H2 "M '
OH
0,02
Mol:
P
+ H2O
a
T r o n g 100ml Y: n ^ + = 0,07 m o l ; n g Q 2 - = O'Ol m o l ; n^|_ = 0,05 m o l .
Cac p h a n u n g hoa hoc:
Na
a
T r o n g X: n^^^_ = 0,02 m o l ; n^^2^ = 0,02 m o l ; n^,Q_ = 0,02 m o l
Lai giai:
1,792
a
2
->DapanC.
, ,
,
,
.m
V , , , '
=amol.
9. P H l / O N G P H A P L I E N H E N G U Y E N T 6 - N H 6 M C H L T C
•
H2O
a. Npi dung
'
- > 4a = 0,16 - > a = 0,04 m o l .
So m o l cac n g u y e n t u c6 t r o n g n h o m chiic l u o n t i 1^ thu|in v o i so' m o l n h o m
K h o i l u g n g m u o i t h u d u g c = 9,46 + m^^. + mg^a-
chuc.
V
= 9,46 + a08.35,5 + 0,04.96 = 16,14 gam.
;
b. Cac t n r o n g hg(p t h u a n g gap
—> D a p a n B.
ROH
d u n g djch X v a 1,12 l i t k h i H2 (dktc). D u n g dich Y g o m H C l 0 , 5 M va H2SO4
RCOOH
0,1M. C h o t u t u d e n het 100ml d u n g dich Y vao X, t h u d u g c m g a m ket tiia.
Giatricuamla
^ - a - . , . i* ^ ' . , • •
- , - a^r* •
A . 0,78 gam.
B. 1,56 gam.
" H 2 = 0,05 m o l . Cac p h a n u n g hoa hoc:
Mol:
^9
a
,.WUt,
C. 3,11 gam.'
Laigidi:
^.•vr.,
Ba + 2H2O
•
^OJ)
2a
a
•* S
2
^NaHCOg ^ R C O O N a +
R_NH2
.,
Vi
'fd
Cach t i n h
no=2nH2
> R O N a + ^ H2
CO2 + H2O
' D . 5,44.
J - ' ^ ; .X^CMU.
> Ba2* + 2 0 H " + H2
a
• - i ^ ; •: :
• - r'h-Hn
M o i lien h |
Phan l i n g hoa hpc
V i d\ 17: C h o h o n h o p g o m Ba va A l ( t i le m o l 1:1) v a o n u a c (du), t h u dugc
,
, torn
no=2ncooH
n o = 2nco2
^
) R-NH3CI
n N = riHCi
Dv M A U
V i d v 1 (A-09): K h i d o t chay hoan toan m g a m h o n h g p X g o m h a i ancol no,
d o n chuc, mach h o t h u d u g c V l i t k h i CO2 (6 dktc) v a a g a m H2O. Bieu thu-c
lien h? g i i i a m , a v a V la:
dm nang On luygn thi dji hpc 18 chuy6n
V
A. m = a- — .
5,6
C
.
:
m=2 a - ^ .
22,4
Cty TNHH MTV DWH Khang Vigt
H6a hpc - IMguySn Van HSi
f;,: v'X,-
. ,
'
V
B. m = 2a
11,2
.
So n g u y e n t u C t r u n g b i n h = "'''"^ = 2,5
' ; 'y.-}
D. m = a + - ^ .
'•
^^^^^
,,,)
5,6 ^J,, • ^ - . n : ; j . | i T
,
<
so n g u y e n t u C n h o h o n 2 , 5 a n c o l d o la C 2 H 4 ( O H ) 2 .
"ancol -
n H 2 0 - "CO2
a
nancol= —
-
18
D o X chua cac ancol d a n chiic
- . . . . . ., , ,.^^5^
22,4
no = noH = nx -> no
Bao toan kho'i lucmg: mx = m c + mH + m o
, a
, a
->m=12.
+ 2 . — + 16.(
22,4
18
^8
V
-> D a p an A .
,
,
) - > m=a
22,4^
.
oM
< — • '^'Ht
.r^j,/
-^"^^
' fCf-if:)^rA
V i d u 2: D o t chay hoan toan m g a m h o n h o p Y g o m ba ancol d o n chuc, thuoc
c u n g d a y d o n g d i n g , t h u d u g c 37,4 g a m k h i C O 2 (dktc) v a 27 g a m H 2 O . Gia
triciiamla
P'''rfm
A . 27,1.
l
iO,U
B. 28,6.
C.23,6.
J
,3m +
#
'
n H = 3,0 m o l .
-> nv = n H - o - n c o , = 1'5 - 0,85 = 0,65 m o l .
,
B. V i = V 2 - 2 2 , 4 a .
;
C. V i = V 2 + 22,4a.
D . V i = 2 V 2 - 11,2a.
,/ i,
OH ;
Lbigidi:
V2
22,4
Bao toan n g u y e n to O: no (on) + 2 n o 2 = 2 nQQ^ +
-> 2a -
2V2
^ 2 V i _ 2V2
+ ^=-^ = ^ - ^ + a
22,4
22,4 22,4
Vi =
2^
2V2
22,4
VXH^Q
2V2 - 11,2a
on
nancoi =
n^^Q " "CO2 '
V i d ^ 5 (B-12): Do't chay hoan toan m g a m h o n h g p X g o m h a i ancol, t h u dugc
13,44 l i t k h i C O 2 (dktc) va 15,3 g a m H 2 O . M a t khac, cho m g a m X tac d y n g
- > D a p an C.
V i d y 3: Do't chay hoan toan m o t l u g n g h o n h o p X g o m h a i ancol (no, d a chuc,
m a c h h o , c i i n g so n h o m - O H ) can v u a d i i V l i t k h i O 2 , t h u dugc
5,6 l i t k h i
C O 2 v a 6,3 g a m H 2 O (cac the tich k h i d o 0 dktc). Gia t r j ciia V la
at
B.3,92.
C. 7,28.
63
' r
'
mol; n H 2 0 = 0 , 3 5 mol.
•
''^''^.ymm-Mtifbw
; ,
.
v o i N a (du), t h u d u g c 4,48 l i t k h i H 2 (dktc). Gia t r j cua m la
A. 12,9.
B. 15,3.
.•v^ .
C.12,3.
Lbigidi:
^
• , • • - ( i =c ~J ui r*3Y^;^:-
"^ol •
, ^
,a,Rt; q? i •
D v a t r e n m o i q u a n h | n g u y e n to' - n h o m chuc t h i v o i ancol: no = noH
n o = n o H = 2 n H 2 = 0,4mol.
Meit khac: n c = n c o 2 = 0,6 m o l ; n H = 2nH20= 1/7 m o l .
Theo b a i ra, X c h u a 2 ancol n o -> nx = n ^ j o " "CO2
-
D . 16,9.
^^002 = 0/6 m o l ; n H 2 0 = 0,85 m o l ; nH2 = 0 , 2 m o l .
D . 1,12.
Laigidi:
^C02
A. V i = 2 V 2 + l l , 2 a .
d o n g t h a i bie't ap d u n g bao toan n g u y e n t o oxi.
= 23,6 gam.
yi:
V i , V 2 , a la
Nhan xet: Bai n a y cac e m can n h o v o i ancol n o t h i :
Bao t o a n k h o i l u g n g : mv = mc + mH + m o = 0,85.12 + 3,0.1 + 0,65.16
5,6
Cac k h i d e u d o 6 d i e u kien tieu chuan. Bieu t h u c l i e n he g i i i a cac gia t r i
-> D a p an D .
.-^
D o Y chua cac ancol d o n chuc -> n o = noH = nx -> no = 0,65 m o l .
A. 5,60.'
C.
Do X chua 2 ancol hai chuc -> no = noH = 2nx -> no = 2a -
^ chtia 3 ancol no
,
= 0,325.22,4 = 7,28 l i t .
V i d v 4 (CD-12): D o t chay hoan toan h o n hg-p X g o m hai ancol (no, h a i chuc,
^^^^^
Cac e m can thay rang, k h i d o t chay Y: n n j o > " c O a
' ' ' ' ^ ' ' ' ^ '
Theo bai ra, X chua 2 ancol no -> nx = n H 2 0 ' ^C02 ~ ^ '
D . 37,1.»- m
- ' WtyfAiHi>HOXH-M v '
'^
'
27
^• '
, .
^2.0,25 + 0 , 3 5 - 0 , 1 ^Q 225 m o l ^
aoM
i
37 4
" C O T ^ — ^ = 0,85 m o l - > n c = 0,85 m o l .
44
n H 2 0 = — = 1.5 m o l
i
'*
mach h o ) can v u a du V i l i t k h i O 2 , t h u dugc V 2 l i t k h i C O 2 va a m o l H 2 O .
cfi n
^^^'^
5,6
^
D a p an
22,4
^
V
V
Q s ^ * •* "••'''A
18
no = noH = 2nx -> no = 0,2 m o l .
Bao toan n g u y e n to O: no(OH) + 2 n o 2 = 2 n c o j + nH20
V
•
p o X chiia 2 ancol c u n g so'nhom - O H -> 2 ancol deu hai chuc:
^
Nhan xet: K h i do't chay cac ancol (no, d o n chuc, m^ch ho), ta l u o n c6:
X chua m g t ancol d a chuc c6
Cty TNHH MTV DWH Khang Vijt'
Ca'm nang 6n luy^n thi d
K h i d o t
A. 3,36.
chay 1 m o l X, t h u d u g c 3 m o l H 2 O -> nx = • ^ r i H 2 0 = 0,04 m o l .
„
B. 11,20.
C.5,60..y.g;
•
. nco2 = 0'3mol
^
Lmgidi:
D. 6,72.
•
0'mti^Qj
nc=a3mol.
N h a n thay t r o n g X cac ancol deu c6 so'cacbon = s o ' n h o m c h i i c l "CllVjt * ijj
(
nc = noH = 0,3 m o l .
,r
p M a t khac, k h i cho X tac d u n g v o i Na: n o H = ^^H2 ~* ' ^ H 2 ~ O'^^ mol. s;
- > V = 0,15.22,4 = 3,36 l i t ^ D a p an A .
V i d u 7: H o n h o p X g o m h a i axit cacboxylic d o n chuc. D o t chay h o a n toan 0,1
m o l X can 0,24 m o l O 2 , t h u dugc C O 2 va 0,2 m o l H 2 O . C o n g t h u c h a i axit la
A. H C O O H va C H 3 C O O H .
i
i o , rm
• -f i
C. C H 3 C O O H v a C 2 H 5 C O O H .
,
*
3iwb fefl loarifi
;
dG
N h | n thay X c h u a 2 axit cacboxylic d o n chiic -> chua 2 n g u y e n t u o x i
^
,.;! >
mx = m c + m H + m o -> mc = 3,08-0,24.1 -0,08.16 = 1,56gam.
->• n c o 2 = '^c = 0,13 m o l .
'
-
' > •
N h a n thay, k h i d o t chay 1 m o l m o i chat m e t y l axetat v a etyl f o m a t d e u t h u
dugc n c o 2 = n H 2 0 ' " e n g v o i v i n y l axetat t h i : n^^^ - n^^o = 1D o v^y:
nvinyi axetat =
% nvinyi axetat
r\QQ^ - n H 2 0
v /..
0,13 - 0,12 = 0,01.
= ^ ^ . 1 0 0 % = 2 5 % - ¥ D a p an D .
'
,
"
,bfniJ3»,,,A
V i d\ 9 (A-11): H o n h g p X g o m axit axetic, axit fomic v a axit oxalic. K h i cho m
g a m X tac d u n g v o i N a H C O s ( d u ) t h i t h u d u g c 15,68 l i t k h i C O 2 (dktc). M a t
khac, d o t chay h o a n t o a n m g a m X can 8,96 l i t k h i O 2 (dktc), t h u d u g c 35,2
gam
C O 2
va v m o l
A. 0,2.
H2O.
Gia t r i cua v la
B.0,3.
C.0,6.
D.0,8.
no = 2ncooH = 2nx —> no = 0,2 m o l .
Bao toan n g u y e n to O:
^
no (COOH) + 2 n o , = 2 n c o , +
^
^
-> 0,2 + 2.0,24 = 2 n c o 2 + 0,2
n c o j = 0/24 m o l .
+
So' n g u y e n t u H t r u n g b i n h =
—S2.= ^-^'^ = 4
"X
"HOO
^
,,
,;,fj fin qP;Uv.4™-
• rtljiW
_> Loai B v a C ( v i cac axit
0,1
So' n g u y e n t u C t r u n g b i n h =
= - 5 ^ = 2,4 -> L o a i A ( v i cac axit chua
0,1
so n g u y e n t u C < 2).
- > D a p an D .
^ '••m'r\fl•„,,••
V i d\ 8 (B-11): H o n h g p X g o m v i n y l axetat, m e t y l axetat v a etyl fomat. D o t
chay hoan toan 3,08 g a m X, t h u d u g c 2,16 g a m H 2 O . Phan t r a m so m o l ciia
v i n y l axetat t r o n g X la
A. 75%.
B. 72,08%.
N/ion xet: K h i cho axit cacboxylic tac d u n g v o i N a H C O a ta l u o n c6:
"C02=r»C00H
->
- > no(X) = 2ncooH = 2nco2
"
«
^
r , y '
"^ • ^
^
no(X) = 2. —'-— = 2.0,7= 1,4 m o l .
22,4
Bao toan n g u y e n to O: no(X) + 2 no2 = 2 nco2 '^H20
, chua so'nguyen t u H > 4.
+
no = 2nx = 0,08 m o l .
Bao toan kho'i l u g n g cac n g u y e n to' t r o n g h o n h g p X, ta c6:
->
B. C H 2 = C H C O O H v a C 2 H 5 C O O H .
D. C H 3 C O O H va CH2=CHCOOH.
Cac chat t r o n g X d e u chua 2 n g u y e n t u O
C. 27,92%.
D . 25%.
^ 1 4 +2
^ =2.^
'
'22,4
44
+ nH,o ^
''•••ft ^
2,2 = l , 6 + y
y = 0,6mol.
nang 6n luygn thi d?! hgc 18 chuy6n d j H6a hpc - IMguygn Van Hai
dm
Cty TNHH MTV D W H Khang Vi^t
L o t gidi:
ivl^t khac, cac em l u u y dya q u a n h? nguyen to - nhom chuc:
Nhan xet: Khi cho axit cacboxylic tac dyng voi NaHCOa ta luon c6:
nc02= "COOH
J
' '
j ,3
no(X)=2ncoOHno(x)=0,2mol.
•
n N ( X ) = 0,06
1 344
no(x) = 2ncooH = 2 ticoj
i'r'f, •
Bao toan nguyen to O:
0,12 + 2.
no(x) = 2.
no (x) + 2
HN =
= 2.0,06= 0,12 mol.
22,4
mol.
"HCl =
"NH2 = " H C l ^
0,03 mol . ^ ^
Bao toan khoi lugng:
ncoj + riHjO
=2
<•:'•:
m = mx+ m H c i = 7,66 + 0,06.36,5 = 9,85 gam.
2,016 ^ 2 .4.84
+ "Hjo ^
22,4
44
"Hjo = 0/08 mol.
r ,
,•
OrM^'
^
"
" ,
_> Dap an C.
Vi
mH20= 0,08.18 = 1,44 gam
13: Dun nong m gam hSn hgp X gom cac este voi 350ml dung dich NaOH
2M, thu dugc dung dich Y chiia muol cua mgt axit cacboxylic don chuc va
—> Dap an B.
Vi d y 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc
13,9 gam hon hgp ancol Z. Cho Z tac dyng voi Na du, thu dugc 4,48 lit khi H2
4,48 lit khi H 2 . Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit khi
(dktc). Co can Y, nung nong chat ran thu dugc voi CaO cho deh khi phan ling
C O 2 va 16,2 gam H 2 O . Cac the tich do 6 dktc. Gia tri ciia m la
xay ra hoan toan, thu dugc 4,8 gam mgt chat khi. Gia tri cua m la
A. 14,5.
)
A. 40,6.
B. 15,4.
(
nH2 = 0,2
I
mol.
• •'
C.12,2.
B.26,6.
•
DHJO^O'^
->nH=l,8
•
.
.,1 4
Y
n
' - a w N;'"^-" #
' r i ..*rr * M >
„ i ,
v
Dya tren moi quan h^ nguyen to-nhom chiic thi voi ancol: no (X) = noH
-> no(X)= noH" 2nH2 = 0,4 mol.
• ! ? m-nmi mod
•«»
C. 30,7.
Nhan xet: Cac este tao thanh tu cung mgt axit cacboxylic don chuc.
Khi ancol tac dung voi Na:
«;i,ib fob ,xtM
A,U,A
m = 0,6.12 + 1,8.1 + 0,4.16 = 15,4 gam
-> Dap an B.
.\i .life
= 0,7 mol; n H2 = 0,2 mol.
- O H + Na
-> - O N a + - H 2
Bao toan kho'i lugng: m = m(- + m H + mo
^
D. 34,5.
Loigidi:
rAv>r^ itT,'.' ".sn-.'
Ldfigiai:
, " '
nco2 = 0,6->nc=0,6;
D. 13,8.
Mol:
0,4
iMv^jndaEv
fa;
0,2
<-
noH=2nH2 =0,4 mol.
• M
Vi d y 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic - C O O H va -
M|it khac: n.coo-= " O H = 0,4 mol
"RCOONa=0,4mol.
NH2 trong phan tu), trong do ti 1^ mo : m N = 80:21. De tac dung vua du voi
7,66 gam X can 100 ml dung djch K O H I M . Cho 7,66 gam X tac dyng vtra
->
du voi dung dich HCl, thu dugc dung dich Y. Co can Y thu dugc m gam
Phan ling voi toi xut ( N a O H he't, R C O O N a con du):
muoi khan. Gia tri ciia m la
A. 11,31.
B. 10,58.
C.9,85.
^
I
Mol:
..Q ne cjsGf ^-
nKOH=0,lmol.
Nhan xet:
'» Tu ti 1§ khoi lugng:
y
mo
niN
80
=
21
De thay: ncooH = ^KOH = O,lmol.
no
80/16
=
0,3
M RH-
<-
4,8
-= 16
0,3
) R H + Na2C03
0,3
^
-
.
;_y^,
y[ ^
mwaOH = mRCooNa
-> m = 0,4.82 + 18,4 - 0,4.40 = 30,7 gam.
— Dap an C.
>
0,3
R H la CH4.
Bao toan khoi lugng: m +
10
21/14 _ 3
mol.
RCOONa + NaOH
D. 9,12.
Lai gidi:
nwaOH dir = 0,7 - 0,4 = 0,3
+ mz
'"
Ca'm nang 6n luygn thi dgi hpc 18 chuy§n dg H6a hqc - Nguygn Van Hki
Cty
TN
TV DVVH Khang Vi§t
V i dy 14: Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 11,2
h't khi C 0 2 (dktc) va 12,6 gam H2O. Mat khac, cho m gam X tac dung voi
N a (du), thu dixgc 4,48 h't khi H2 (dktc). Gia tri cua m la
A. 12,4.
;
B. 15,2.
C . 12,6.
m 0 -
D . 13,8.
.^
Laigidi:
V
rico2=^=0,5mol;nH2O=-^=0'7mol.
Ta c6: mc = 0,5.12 = 6,0 gam; m H = 0,7.2 = 1,4 gam. '
H
K h i X tac dung voi N a :
i'-^
Mol:
- O H + Na
0,4
n"^-:- '<
j ,j r t o ^rt, ^ ff,
''
<
•
m c j "* '•onn ni'<»'P!' wl- -
0,2
.„.
- > D a p an D .
., .
^ j,^ ^
CO mm
HIND
AXIT C L O H I D R I C : H C l
a, L i t h u y e t
+ Tfnft
dung voi kim loai, bazo, oxit baza, muoi. V i du:
Fe + 2HC1
n? f
Si
> FeCh + H a t
'^ ^ '''
^
> C a C h + C O a t + H2O
CaC03+ 2HC1
'^^-'^^'^-^
"'^
+ Tin?z fc^""- Tac dung voi cac chat oxi hoa manh: Mn02, KMn04, KCIO3,
M n 0 2 + 4HC1
,.,^„„ „ ;„,,.,,„/,
^
- ^ - ^ M n C h + CI2 +2H2O
, ^-.•cr^^' rfi
> F e C h +H2.
(c) 6 H C l + 2A1
'
.,;.^d..Au.
qfi(}
i
-
i
-^iyti^'^^
B.(a),(b).
'
'-Xji::;
X: jfefb g n i i ^ X i V t .
' - W ! M i:i ,i/:}c:: i
"
iM>f § ' ?. "
Cac phan ling trong do H C l the hi?n tinh oxi hoa la
C . (b), (c), (d).
•
.
,„,f^eH
> 2 K C l + 2MnCl2 + S C h + 8H2O.
(d) 16HCl + 2 K M n a
-
.v,t
,
> 2AICI3 + 3H2.
A.(b),(c).
C O H /
,
> M n C h +CI2 +2H2O.
(a) H C l + M n 0 2
•
'
V i d u l : Cho cac phan ling sau:
(b) 2 H C l + Fe
ry^.^ ^^..^^^^
2KC1 + 2CrCl3 + 3Cl2 +7H2O
K2Cr207 + 14HC1
b. V i d u m a u
"
> 2KC1 + 2MnCl2 +5CI2 + 8H2O
2KMn04 + 16HC1
no = noH = 2nH2 -> no = 0,4 mol -> mo = 0,4.16 = 6,4 gam.
Bao toan kho'i lugng: mx = m ^ + mpi + m o = 6,0 + 1,4 + 6,4 = 13,8
1
K2Cr207. V i d y :
t riD,b :,:n.b : x m t
>-ONa+
CAC A X I T vo
. ,„
s f8 f i
D.(a),(d).
Laigidi:
Nhan xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1
(trong H C l ) xuohgO (khi H2).
,
,
- > Dap an A.
V i dv 2: Hoa tan hoan toan 7,6 gam hSn hgp hot FesOA va C u trong 200ml dung
djch H C l 1,2M (loang). Sau khi cac phan ung xay ra hoan toan, thu dugc
dung dich X (khong chua axit du). Co can X thu du^c m gam muo'i khan.
Gia tri cua m la
A. 10,39.
B. 14,20.
C . 5,16.
D . 11,10.
Lffigidi:
Gpi so mol: Fe304 = a; C u = b.
Theo bai: 232a + 64b = 7,6.
• ?-•<.}(
' ^•
•
^han xet: chi c6 Fe304 phan ung tryc tiep voi axit.
•:
•
'^^j^
f| ,* g .j
'j
X :,,;. i
:
Trong X khong con axit d u nen Fe304 phan ung vua dii voi H C l :
Cac phan ung hoa hpc:
Caim nang On luy$n Ihi dgi hgc 18 chuy6n dg H6a hgc - Nguygn VSn
i * •
Fe304 +
->
0,03 < -
0,24
a = 0,03
Cu
Mol:
> FeCh + 2FeCh + 2H2O
8HC1
Mol:
Cty TNHH MTV DVVH Khang Vijt
Hit
-»
0,03
V i dV 5: Cho 2,13 g a m hon h g p X g o m M g , Cu va A l a d a n g bgt tac d y n g
hoan toan v o i O2 t h u d u g c h o n h g p Y g o m cac oxit c6 k h o i l u g n g 3,33 gam.
0,06
"The tich d u n g d i c h H C l 2 M v u a d u de phan u n g het v o i Y la
- > b = 0,01 m o l .
+
2FeCl3
0,01 - >
> CuCh
•
0,02
^
+ 2FeCl2
0,01
J t a r H Q U i i >J > T i X .
A. 150 m l .
B.SOml.
C. 75 m l .
Lai
0,02
D. 90 m l .
gidi:
m = 0,01.135 + 0,04.162,5+ 0,05.127 = 14,2 gam - > D a p an B.
3: Day g o m cac chat deu tac d u n g d u g c v o i d u n g d i c h H C l loang la
Vi
01 bai nay, cac em rJia't thiet phai ap d u n g bao toan k h o i l u g n g de h'nh khoi
l u g n g oxi tham gia phan ung:
A. KNO3, C a C 0 3 , Fe(OH)3.
^
B. FeS, BaS04, K O H .
C. AgNCte, ( N H 4 ) 2 C 0 3 , CuS.
,
D. NaHCOs, FeS, C u O . '
Lai
mkimiiHii + m o x i = moxit — ^
.^.^^^
= 3,33— 2,13 = 1,2 gam.
moxi
12
n o o = — =0/0375 m o l -> n o = 0,075 mol -> n ^
, , ,
^-^2
gidi:
^
32
Loai A v i KNO3 k h o n g tac d u n g ; loai B, C v i BaS04 va CuS k h o n g tan trong
K h i cho oxit bazo tac d u n g v o i axit tao ra nuoc:
d u n g d i c h H C l loang.
02-
^
D a p an D. Cac p h u a n g t r i n h hoa hoc:
NaHCOa
+ HCl
>
FeS
+
2HC1
>
FeCh
CuO
+
2HC1
>
^''^ ^ '
+ H2O
NaCl
CuCh
•
+
iOnM>i
CO2
''
V i d u 4: H o a tan hoan toan 8,55 gam hon h o p g o m Na, K va Ba vao nuoc, t h u
i
, ^ ; ;,
Suy ra: n ^ + = 0,15 m o l - > n ^ c i ^ 0,15 m o l
- > VHCI = 0,075 l i t = 75ml
A. 14,90.
Dap an C.
B. 13,05.
H2SO4, t i 1$ m o l t u o n g u n g la 2:1. T r u n g hoa d u n g d j c h X b o i d u n g d i c h Y,
C. 7,45.
•<• ,
d u o c d u n g d i c h X va 1,792 l i t k h i H2 (dktc). D u n g djch Y g o m H C l va
t o n g k h o i l u o n g cac muoi dugc
.,fm i .
.
H2O.
d u n g d i c h X. Co can X t h u d u g c m gam chat ran k h a n . Gia trj ciia m la
H2O
+
>
=0,075 m o l .
V i d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml H C l 0,2M, t h u d u g c k h i H2 va
H2S
+
+ 2H^
(0x11)
H K = 0,2 m o l ;
tao ra la
K
D . 20,50.
iiiX
Lai gidi:
= 0,1 m o l ; n H c i = 0 , 1 m o l .
+ HCl
> KCl
+
,,
lH2t
=.
;j
«o.'HBOH n . :
,.
foH
,(>
2
A. 13,81 gam.
B. 11,39 gam.
C. 15,23 gam.
D. 19,07 gam.
n^,
^
Na
= 0,08 m o l . Cac p h a n l i n g hoa hgc:
22,4
+ H2O
)• Na*
+ H2O
>
n Ba + 2H2O
K
'
+ OH"
+
-Hi
0,1
't^»^'-
+
H2O
+ OH- +
iH2
> KOH + iH2t
n^
r<^r
r a n k h a n Y. K i m loai k i e m M la
A.Rb.
B. K.
>
M
H2O
- > 4a = 0,16 - > a = 0,04 m o l .
K h o i l u g n g muoi t h u d u g c = 8,55 + m
+ m
Cl
•*
D.Li.
Jl/ion Ob
Lai gidi:
= a mol.
H* + O H "
..,
D a p an B.
C.Na.
Cac p h a n u n g hoa hgc:
T r u n g hoa X b o i Y:
• -
H C l 0,2M, t h u d u g c k h i h i d r o va d u n g d j c h X. Co c?n X t h u d u g c 14,73 chat
mo\.
i_
0,1
-
V i d y 7: Hoa tan hoan toan 8,97 g a m k i m loai k i e m M vao SOOnil d u n g djch
" H * ° " H C l + 2nH2S04 = 2a + 2.a = 4a m o l
n ^ , . = 2a m o l ;
0,1
• OH£'
- > m = m K c i + m K O H = 74,5.0,1 + 56.0,1 = 13,05 g a m
> Ba2- + 20H- + H2
Nhan xet: n^^. = 2nyi^ =0,Id
> Z'
.
2
Mol: •
o
T r o n g Y:
0,1
' • :
2
8 K
0,1
Luu y: K con d u se tiep tuc p h a n u n g v o i nuoc:
1,792
r,
Mol:
'h-,-.
2- = 8,55 + 0,08.35,5 + 0,04.96
so^
= 15,23 gam —> D a p an C.
+ HCl
> MCI + -H2
2
N/ian xet: Bao toan k h o i l u g n g
m^„.
OH
• •= , I^^/TU' »• .|
<
va M +H2O
^
,
> M O H + -H2
2
m y = m i 4 + m ^ i _ + m^^^.
= 1 4 , 7 3 - 8 , 9 7 - 0 , 1 . 3 5 , 5 = 2,21 g a m n ^ „ . = 0,13 m o l .
UH
73
Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai
Cty TNHH MTV DVVH Khang Vift
8 97
r i M = n H c i + n Q ^ . = 0,23
-> M = - ^ = 3 9 ( K ) ^
' do
D a p an B.
^'^^ ^^"^
^
V i dv 8: Cho m g a m h o n h o p X g o m C u , M g , Fe tac d u n g v o i axit H C l d u , thu
d u g c d u n g d i c h Y, 448ml k h i (dktc) v a 0,64 g a m chat ran. C h o d u n g dich
vao Y t h u d u g c 448ml k h i (dktc). K h o i l u g n g N a C l t r o n g X la
^ 0,585.
B. 1,170.
N a O H d u vao Y, loc ket tiia v a n u n g trong k h o n g k h i t o i k h o i l u o n g khong
doi, t h u dugc 1,2 gam chat ran. Gia t r i ciia m la
A. 0,80.
B. 1,16.
,
D . 0,84.
r•
.
= 0,01.
Ggi so m o l t r o n g X: Fe = x; M g = y - > x + y = n H 2 = 0,02 m o l .
'
So d o p h a n u n g :
Mg
^
-±»£U
> Fe(OH)2 - ^ ^ ^ l F e 2 0 3
MgCh
> Mg(OH)2 —
m F e 2 0 3 + m M g O = 1/2 ^
80x + 40y = 1,2 - »
Luu
CI"
+ 2HC1
+ 2HC1
C. 28,21%.
mNaci
C. 19,2.
D. 9,0.
^-^••«:MfBrtf 6'>
Cac p h u o n g t r i n h p h a n u n g :
Mg
Mol:
Mol:
Loigidi:
i
' •'
> M g C h + H2
K.i i
a
•t^|,\V,
a
> M g C h + CQ2 + H2O
+ 2HC1
b
''"^
b
A p d u n g cong thuc cua p h u o n g phap d u o n g cheo, ta c6:
"H2
= 100 - 1 5 , 7 6 ^ 84,24 gam.
44 - 2 3
2 - 23
=1
• 'iliiA::
, , ,
^
,
— = - -> a = b = 0,1 m o l
b 1
m = m^g+ mMgcos "
bandau.
A
)• M g C h + H2
+ 2HC1
Theo bai: M = 11,5.2 = 23 va a + b = 0,2 m o l .
^ FeCh + H2
Nhan xet: L u g n g H2O t r o n g Y cung c h i n h la l u g n g H2O c6 t r o n g d u n g d i c h H C l
" ^^'^ 8 ^ ^
^
^'
2. A X I T S U N F U R I C : H2SO4
L i thuyet
M a t khac, d o n o n g do H C l bang 20%
mH20 =
4mHci
+ 4 m H c i = 84,24.
,
Bao toan n g u y e n to'Cl:
2nFeci2 + 2nMgCl2
= " H C l ">
Dung d/clz H2SO4/oflng: T i n h axit m a n h
Fe + H2SO4
FeS + H2SO4
31 52
mMgCl2
D a p an B.
Loigidi:
^C02
mMgCl2
>''•'' ^
-Hi
= 0,02.58,5 = 1,17 gam
B. 10,8.
MgCOs
D. 15,76%.
Xet v o i 100 g a m d u n g d i c h Y: -> m^^^ij = 1^,76 gam.
mMgci2 + m H 2 0
"
'
mla
15,76%. N o n g d o p h a n t r a m cua M g C h t r o n g Y la
,
''''
>
> HCl
f. K h i n u n g ngoai k h o n g k h i , Fe(OH)2 chuyen t h a n h Fe(OH)3 v a b i
B. 11,79%.
+ HClt
KHSO4 + H C l t
> ZnCh + Hat
A. 13,2.
d u n g d i c h H C l 20%, t h u d u g c d u n g d i c h Y. N o n g d o cua FeCk t r o n g Y la
^
NaHS04
dugc 4,48 l i t h o n h g p k h i X (dktc). T i k h o i cua X so v o i H2 la 11,5. Gia t r i cvia
V i d u 9: H o a tan hoan toan h o n h g p X g o m Fe v a M g b a n g m o t l u g n g vira d i i
Mg
jCCl + H2SO4
Z n + 2HC1
^
V i du 11: H o a tan het m g a m h o n h g p M g v a MgCOa t r o n g d u n g d i c h H C l , t h u
x = y = 0,01 m o l .
p h a n h i i y t h a n h Fe203.
Fe
—
_» x = 0,02; y = 0,02 ->
^ MgO
'
A. 24,24%.
'
Ta c6: mx = 58,5x + 74,5y = 2,66 va n H 2 = 0,5(x + y) = 0,02.
m = 0,64 + 0,01.56 + 0,01.24 = 1,44 gam.
-> D a p an C.
+ H2SO4
Bao toan n g u y e n to:
FeCh
D . 2,340.
Loigidi:
jsjaCl
Loigidi:
Fe
C. 1,755.
Goi so m o l : N a C l = x; K C l = y. Cac p h u o n g t r i n h p h a n u n g :
C. 1,44.
N^flnxef: m c u = 0,64 g a m n c u
NaC\a K C l v o i H i S 0 4 dac, d u .
jChi thoat ra cho hoa tan vao nuoc t h u d u g c d u n g d i c h Y. Cho b g t Z n d u
2
+ ~n^gClj
1
= ^ " ^ H C l
=11'79 ^ C%(MgCl2) = 11,79% -> D a p an B.
'1,1 , ^
> FeS04 + H 2 t
,j,
> FeS04 + H 2 S t
,
"*" Dwn^djc^ H2SO4 (fflc: T i n h oxi hoa m a n h
N g o a i t i n h axit m a n h , axit sunfuric dac con the h i ^ n t i n h o x i hoa m ^ n h , tac
d i i n g d u g c v o i n h i e u k i m loai, h g p chat,...:
ff;-!;!
»
{'"-l
Cty TNHH MTV DWH Khang ViSt
C^m nang 6n luy$n thi dgi hpc 18 chuySn dg H6a hpc - NguySn VSn HJi
Cu +
— — > CuS04
2H2S04(^flc)
2Fe + 6H2SO4 (dac)
2FeO +
4H2SO4
+SO2 + 2H2O
Gia thie't dung djch ban dau chiia 1 mol H 2 S O 4 (tiic chiia 98 gam
100
:>
/
Fe2(S04)3 + 3SO2 + 6H2O
','
'
_> Khoi lupng dung djch
> Fe2(S04)3 + SO2 + 4 H 2 O
(dac)
2Fe304 + IOH2SO4 (dac)
> 3Fe2(S04)3 + SO2 + IOH2O
L u u y: Cac kim loai Al, Fe, Cr khong tac dung vdi Ji2S04 dac, nguQi.
Dieu che
So do: Quang pirit FeS2 hoac S
H2S04.nS03
> (n+1)
Cac phan ung: S + O2
4FeS2 + I I O 2 —
...
)
SO2 —
^
••
• "ID
•
'
^
,
.
fiB'lii^H
:6i
2S03
+ nSOs
SO3 —
.•..,.H,.-..»,I
> SO2
^ 2Fe203+ 8SO2
2SO2 + O2
H2SO4
H2SO4
'
,-xm:o:.fii ,
> H2S04.nS03
H2S04.nS03+ n H 2 0
(Oleum)
^
> (n+l)H2S04
b. V i dv mau:
Vi dy 1: Cho day cac chat sau: KBr, S,
Si02,
FeO, Cu va Fe203. So'chat trong day
the bi oxi hoa boi dung djch axit H2SO4 (dac, nong) la
CO
A. 4.
B.5.
' G.3.c|f:
Lai gidi:
S
+
2H2SO4 —
2KHS04 + Br2 + SO2 + 2H2O
^
2FeO
+ 4H2SO4 —
Cu
2H2SO4
+
3SO2 + 2H2O
:
'
SO2 + 2H2O
^
fH',:^::
y:
Khi tac dyng voi
FeaOs +
3H2SO4 —
H2SO4
d^c, FeaOs the hi^n tinh baza:
H2SO410% (loang), thu dug-c dung dich muoi c6 nong dp bang 14,45%. Kim
loai M i a
'
:
B. Fe.
,
'
" "
C. Cu.
' ^
Lcn gidi:
+
H2SO4
(3):
Fe
-3e
> Fe*-^
nenhuong = 3 np^ = 3a
> Cu*^
(1)
(2): O2 +4e
> 2Ct^
(3): S** +2e
> SO2
nenhiKmg=2ncu
^
nenhan =
OsH
» ;
•
=0,03
"^s:
4no2 = ^^^^^^=0,67-7a
=0,06. -
nenh,^n=2nso2
Bao toan electron: 3a + 0,03 = 0,67 - 7a + 0,06 ^ a = 0,07 mol -> Dap an C.
CachZ:
Bao toan khoi lupng: 56a + 16b = 6,32 - 0,015.64 = 5,36.
a = 0,07; b = 0,09
, ,,
.
Dap an C.
Fe304, Fe203 tac dung voi dung dich
H2SO4
dac, nong. So truang hpp xay
ra phan ung oxi hoa - khu la - ? Rf
' A . 6.
.
B.3.
C.4. ,
DJfjfi
Lai gidi:
D. Zn.
Nhan
Phan ling hoa hpc:
MO
->
Vi dy 4 (B-12): Cho cac chat rieng bi^t sau: FeS04, AgNOa, Na2S03, H2S, H I ,
^ Fe2(S04)3 + 3 H 2 O
Vi dy 2: Cho oxit ciia kim loai M (hoa tri 2) tac dung vira dii vai dung dich
A. Mg.
Fe,Cu(l)
> Y (2)
> Fe^3Cu^M3)
Xet su trao doi electron 6 cac giai doan:
Bao toan electron: 3a + 2.0,015 = 2b + 2.0,03
Dap an A.
Luu
:
Qui doi Y thanh: Fe (a mol); Cu (0,015 mol) va O (b mol).
^ Fe2(S04)3 + SO2 + 4 H 2 O
CuS04 +
ne'u dya theo phuong trinh phan ung se rat dai va kho giai.
Cach 1:6 day, cac em can su dyng so do phan ung:
. t y s j . j ,
Nhan xet:
(2)
CO tinh khu (chiia nguyen to' dang 6 muc oxi hoa tha'p).
2KBr + 3H2SO4
Bao toan khoi lupng: mo2 = my " "^x = 6,32 - 56a - 0,96 = 5,36 - 56a.
Cu - 2 e
Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung vai chat
= 98. — = 980 gam.
980 + M + 16
^^^^^•^ -=0,1445 - > M = 56 (Fe)-^ Dap an B.
Vi dy 3: Nung hon hop X gom a mol Fe va 0,015 mol Cu trong khong khi mpt
thoi gian, thu dupe 6,32 gam chat ran Y. Hoa tan hoan toan Y bang dung
dich H2SO4 dac nong (du), thu dupe 0,672 lit khi SO2 (san pham k h u duy
nha't 6 dktc). Gia tri ciia a la
A. 0,04.
B.0,05.
C. 0,07.
D. 0,06.
Lm gtat:
^
0 672
-bVT' ".'^
ncri^=—^
=0,03 mol. , , ,,
, i.
,
(1)
D.2.
H2SO4
H2SO4)
xef: Phan ung oxi hoa-khu xay ra khi H2SO4 tac dung vai chat c6 tinh
khu (chiia nguyen to dang 0 muc oxi hoa tha'p).
> M S O 4 + H2O
*
2FeS04
+ 2H2SO4
Fe2(S04)3 + SO2 + 2 H 2 O
77
im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\
HlS
+ 3H2SO4
'
Cty TNHH IVITV DVVH Khang Vigt
> 4SO2 + 4H2O
8HI + H2SO4 ^'-^
4I2 + H2S + 4H2O
2Fe304 + IOH2SO4
Lcngiai:
,
Gpi so mol: Fe304 = a; Cu = b. Theo bai: 232a + 64b = 5,28.
f-SnO'w! ;oi*:^f '
3Fe2(S04)3 + SO2 + IOH2O
Cac phan ung hoa hpc:
Luu y: ^eiOj, AgNOs xay ra phan ung trao doi, Na2SQj khong tr.c dung:
Fe2(S04)3 + 3H2O
2AgN03 + H2SO4
a = 0,02
Ag2S04i + 2HN03
C. 11,80 gam.
Vi
D. 9,42 gam.
y
^,
^ ^
Na + H2O
+ OH- + - H 2
2
1
> Na^ + O H " + - H 2
Ba + 2H2O
•
*•
trong cung mot the ti'ch thi
CI
. ,
ii ion'ji
H2O
^
V. ' , K
2-
SO^
'
"H2
= 0,10 mol. ''^
Vi
"
; ,1-
,
1 dvi 6: Hoa tan hoan toan 5,28 gam hon hgp bpt Fe304 v a Cu trong 80ml dung
d i c h c h u a m g a m m u o i . Gia tri c u a m l a
A. 19,04.
B. 20,54.
'
C. 17,96.
"
D. 14,50.
Lai gidi:
- > m = 17,96 g a m -> Dap a n C.
' •'fm-E ' o
Cach 2: Ta c6: n M = nH2S04 = " H J = 0,12 m o l .
N e u 1 m o l k i r n loai M
> MSO4 t h i k h o i l u g n g t a n g 96 g a m
0,12
dich H2SO4 I M (loang, vua dii). Sau khi cac phan ung xay r a hoan toan, thu
tang0,12.96.
duoc dung djch X. Co can X thu du(?c m gam muoi khan. Gia trj ciia m la
Vay m = 6,44 + ai2.96 = 17,96.
A. 8,64.
->
B.7,68.
'' '
8: Hoa t a n h o a n toan 6,44 g a m h o n hgp X g o m A l , Fe v a Z n b a n g m o t
Bao t o a n k h o i l u o n g : 6,44 + 0,12.98 = m + 0,12.2
'
= 6,08 + a06.35,5 + 0,03.96 = 11,09 gam.
Dap a n B.
D. 97,80 gam.
Cach 1: Bao toan n g u y e n to H : nH2S04 = " H 2 " 0,12 m o l .
\V
a = 0,03 mol.
.
C. 101,48 gam.
= 3,68 + 98 = 101,68 v a se c h g n n h a m d a p a n A!
^y^:-yyh
Khoi luong muoi thu dug-c = 6,08 + m . + m
Xi
n.'
l u o n g v u a d u d u n g d i c h H2SO4 loang, t h u d u g c 2,688 lit H2 (dktc) v a d u n g
o_ = a m o l ; n _ . = 2amol.
CI
Dap an D.
Luu y: Bai n a y c a c e m d i q u e n t r u k h o i l u o n g k h i H2 b a y r a , v a c h i t i n h : mx
" H a = 2nH2S04 • Goi n H c i = 2a -» nH2S04 = a-
4a = 0,12
'
-> Dap a n C.
Mat khac, nong do HCl gap 2 Ian H2SO4
>
'
Bao t o a n k h o i l u o n g : 3,68 + 98 = mx + 0,10.2 ^ m x = 101,48
iiJ
' '''
SO4
'
-> Khoi luong dung dich H2SO4 = 0,10.98. ^
10 = 98 gam.
Nhan xet: n ^ ^ . = 2nH2 = 0,12 mol.
H ,,.•..,,.',„
• '
Lcngiai:
> Ba2+ + 2 0 H - + H2
,
B. 88,20 gam.
, , .
>
Trung hoa X boi Y: H * + O H "
' '
CuS04 + 2FeS04
Bao toan nguyen to hidro: nH2S04 =
-> Trong Y: n + = 4 a m o l ; n
~
7: Cho 3,68 gam hon hop gom A l , M g va Zn tac dung voi mot luong vua
A. 101,68 gam.
= 0,06 mol.
Li + H2O
" '
Khoi lugng dung dich X la
Cac phan ung hoa hoc:
,,
^ b = 0,01 mol.
du dung dich H2SO410%, thu dxxgc dung dich X va 2,24 lit khi H2 (dktc).
1 344
=
•• '
0,02
->
0,01
0,02
0,01 - > 0,01
m = 0,01.160 + 0,01.400 + 0,04.152 = 11,68 gam
Lbi gidi:
"H2
0,02
Mol:
dung dich X v a 1,344 h't khi H2 (dktc). Dung dich Y gom HCl I M v a H2SO4
B. 11,09 gam.
^
Cu + Fe2(S04)3
i d\ 5: Cho 6,08 gam hon hop gom Li, Na v a Ba vao nuoc (du), thu duoc
A. 10,38 gam.
> FeS04 + Fe2(S04)3 + 4H2O
Fe304 + 4H2SO4
: Mol: 0,02 < - 0,08
0,5M. Trung hoa dung dich X boi dung dich Y, tong khoi lugng cac muoi
dugctaorala
,
:
b uAi
Trong X khong con axit d u nen Fe304 phan ling vua du voi H2SO4:
Dap a n C.
Fe203 + 3H2SO4
ji
jV/ian xet: chi CO Fe304 phan ling true Hep voi axit.
, 1/
; i ;
C. 15,68.
D. 11,68.
Dap a n C.
•
.. ;
,
I- '
„ . , (^„j,,4i<);/:H:H
, ).^-
C^m nang 6n luygn thi dai hgc 18 chuy6n
H6a hpc - Nguygn Van H&\
V i du 9: H o a tan hoan toan 2,81 gam hon hop X gom Fe203, FeO, C u O can
50ml axit H2SO4IM (loang). Khir hoan toan 2,81 gam X bang khi C O (nung
nong) thu dugc m gam kim loai. Gia tri ciia m la
A. 3,24.
B.2,65.
C. 3,06.
D.2,41.
aofl
' • ' L a i g i d i :
•i-,'>r"i
Nhan xet: K h i cho oxit kim loai tac dung voi axit/.ion 0^~ trong oxit se ke't
hqp voi H* trong axit tao thanh H2O theo phuong trinh:
^
, ^ Ijjjy,
02-
+
Mol: 0,025
> H2O
2H^
0,05
^
Vay: m o = 0,025.16 = 0,4 gam
-> m kim:o?i = 2,81
- 0,40
= 2,41
gam
-> Dap
an D.
s^iBi J O A I
flj 4-»f
thu dugc dung dich X va 324,8ml khi SO2 (san pham khu duy nhat, 6 dktc).
Co can dung dich X, thu dugc m gam muo'i sunfat khan. Gia tri ciia m la
•HV
B.5,40.
•• ' '
hoac N H 4 N O 3 ;
2. Al, Fe, Cr khong tan trong HNO3 d$c ngupi. ' ' , " '
Tac dung vai hap chat
3FeO + 10HNO3
> 3Fe(N03)3 + NO + 5H2O
'•'^^
3Fe2* + N O ; + 4H*
> 3Fe3* + N O + 2H2O
"
rl-y:
£n,;,
...
> Fe(N03)3 + NO2 + CO2 + 2H2O *
> Fe(N03)3 + 2H2SO4 + I5NO2 + 7H2O
.ciijj
Dieu che
Vi d\ 10: Hoa tan het 2,088 gam FexOy bang dung dich H2SO4 dac, nong (du),
A. 5,22.
1. Ba kim loai kha manh (Mg, Al, Zn) c6 the khu HNO3 thanh N2O, N2
FeC03 + 4HN03
FeS2 + I8HNO3
S},.0-'K
ia,,j^ u
Litu y:
C. 5,80.
D. 4,84.
So do: N2
N2 + 3H2 <
4NH3 + 5O2
2 N O + O2
> 2NH3
> NO2
.
-4 HNO3
<
,i
'bCl
4y
'""^•'^ > 4 N O + 6H2O
> 2NO2
> 4HN03
4NO2 + O2 + 2H2O
Laigidi:
> NO
> NH3
Nhan xet: Oxit sat FexOy tac dung voi H2SO4 dac, nong -> SO2 thi oxit la FeO
hoac Fe304.
Vi du 1: Cho 3,024 gam mpt kim loai M tan het trong dung dich HNO3 loang,
Cac em luu y rang 1 mol FeO hoac Fe304 deu chua 1 mol Fe''- nen deu c6
kha nang nhuong 1 mol electron, do do:
0 3248
= npe^Oy = 2ns02 -> "FexOy = 2 " - ^ ^ = ^'029 mol
, ns«
-> Mpe o = = ^ = 72 (FeO).
'
'
" ' ^•^
^ '^•
^
. >0r
^
urn
'^
thu dugc 940,8 ml khi NxOy (san pham khu duy nhat, a dktc) c6 ti kho'i doi
voi H2 bang 22. Khi NxOy va kim loai M la
A.NOvaMg.
B. N C h v a A l .
C.N20vaAl.
Lai gidi:
Theo bai: M ^ ^ Q
= 22.2 = 44
NxOy la N2O
Loai A va B . f ,(>« .,, .'
n'f)) •
Bao toan nguyen to'Fe: 2FeO
> Fe2(S04)3
mpe2(so4)3= 0/0145.400 = 5,80 gam-> Dap an C.
;
p'
,'
n,,o=^Jo,042™ol.
22,4
'^o^h/r
Cach 1: Phan ung hoa hgc:
3. AXIT NITRIC: HNO3
8M + 10nHNO3
A. L i thuyet
_v
+ Tinh axit manh: Tac dung voi kim loai, bazo, oxit baza, muoi. Vi du:
MgO + 2HNO3
CaC03 + 2HN03
> Mg(N03)2 + H2O
> Ca(N03)2 + C O z t
+ H2O
+ Tinh oxi hoa manh
+ Tac dung voi kim loai: Axit nitric tac dung dugc voi hau het cac kim loai
(tru Au, Pt), ke ca kim loai c6 tinh khu ye'u nhu Cu, Ag,...
Cu +4HN03(f?flc)
D. N20vaFe.
> Cu(N03)2 + 2N02T +2H2O
3Cu + 8 H N O 3 ( / 0 f l M ^ ^ ) — - > 3Cu(N03)2+ 2 N O T + 4 H 2 O
8
^
> 8M(N03)n + nN20 + 5nH20
0,336
3,024
nfo '
^
Q,;.^^
„
n
-> n = 3 va M = 27 (Al)
Dap an C.
'^'^
'
Cach 2: Nhan thay de tao dugc khi N2O thi kim loai phai kha m^nh (nhu Mg,
Al, Zn) -> Loai D -> Dap an C.
Vi dv 2: Hoa tan hoan toan hon hgp gom 0,04 mol FeS2 va a mol CU2S vao
axit HlSf03 (vua dii), thu dugc dung dich X (chi chua hai muoi sunfat) va V
lit khi duy nhat NO2 (dktc). Gia tri ciia V la
A. 13,44.
B. 17,92.
C. 20,16.
D. 22,40.
Cty TNHH MTV DWH Khang Vi$t
im nang 6n luy^n thi d?i hpc 18 chuySn dS H6a hpc - Nguyin Van HSi
Lai
Cac phan l i n g k h u :
gidi:
Nhan xet: cac e m l i m y la d u n g dich X chi chiia hai m u o i sunfat —> sau cac
peS_9e
phan l i n g , S n a m he't 6 dang goc sunfat.
FeS2-15e
Ta CO cac so do chuyen hoa:
Bao toan electron:
FeS2
Mol:
0,02
Mol:
a
'
> 2CuS04 ,;<:J3+, sOl-'';; *• ( O M b U '
• Cu2S
->
^aicHS: +
2a
Bao toan n g u y e n to S, ta c6:2 npeSz +
>^ Wlh
>
A'
.
Mol:
/ i n
Mol:
0,005
1
^'^'^Cu2S
V ^ o =17,92 l i t
1' t t- tiVK'.
0,005
0,01
NO
y
t
III
II
U '
I
I
0,02
^.4 > f
y
> *'l) -a /'
(san p h a m khvr d u y nhat) va d u n g d i c h Y. The tich d u n g d i c h N a O H
A. 6 0 m l .
thay axit H N O 3 b a n g axit H2SO4 dac, nong, d u t h i the tich k h i SO2 (dktc)
thu d u g c sau p h a n u n g la bao nhieu?
B. 5,60 lit.
B. 120 m l .
> Fe*3 va
C u - 2e
hoac
+2e
Bao toan electron: n g = n f j 0 2 = 2nso2
0,05 m o l
4: H o a t a n hoan toan 2,08
ncu =
D . 1,12 lit.
64
+
= 0,03 m o l ;
V ; • • •..M ,
> Cu^^
> SO2
D. 180ml.
n^NOa = ^^'^^^^
63
=
.J-
.0.$-/
......
:
0,2 m o l . Phan u n g hoa hpc:
3Cu(N03)2 + 2 N O + 4 H 2 O
8HNO3
Y g o m : Cu(N03)2 = 0,03 m o l ; H N O 3 d u = 0,2 - 0,08 = 0,12 m o l .
> H 2 O va Cu^^ + 2 0 H -
;j
> Cu(OH)2',
= n„+ + 2 n ^ 2+ =0/12 + 0,03.2 = 0,18 m o l
OH
H
Cu
- > VNaOH = 0,18 l i t = 180ml ^ D a p an D .
mi y: Cac e m de q u e n p h a n u n g t r u n g hoa axit ( H * + O H ) va c h i t i n h :
. = 2n^^2+ - 0/06 m o l —> VNaOH = 0,6 l i t = 60ml se chpn n h a m A !
rcitjirf »s" ='
Vso2 = 0,05.22,4 = 1,12 l i t ^
C. 1 5 0 m l .
Loi gidt:
...
Cac p h a n u n g : H * + O H '
Ji*
> NO2
'
gidi:
nN02 = 0/1 m o l .
Fe - 2e
KI;
3Cu
C. 2,24 lit.
Lot
"502 = ^ =
ti
I M can thiet de ket tua he't i o n Cu^* t r o n g d u n g d i c h Y la
->DapanB.
d u n g d j c h H N O 3 dac n o n g t h i t h u d u g c 2,24 l i t k h i m a u n a u (dktc). N e u
Chat oxi hoa: N^^ + i e
0,01
-> -Fe2C)3 + 2BaS04
2
3: H o a tan hoan toan m p t h o n h g p g o m h a i k i m lo?ii Fe v a C u bang
Chatkhu:
it
'I
V i d v 5: C h o 1,92 g a m C u vao 60 gam d u n g d i c h H N O 3 2 1 % , t h u d u p e V m l k h i
Bao toan electron: L n ^ O j = 15r>FgS2
A. 3,36 lit.
+ BaS04
-> m = 160.0,01 + 0,03.233 = 8,59 - > Dap an B.
> 2Cu^2 +
nN02 = 0 ' 8 m o l - >
FeS2
ur
> Fe*3 + 2S^
Cu2S - lOe
*
-> - F e 2 0 3
2
0,01
•>
v
,
FeS
,(fWi
> }A}W\ i r
<
,
Cac phan l i n g k h u :
/i
(Ov^jr!
-
Bao toan n g u y e n to Fe va S:
' • *
i^Cu2S = " S O 4
-> 0,04.2 + a = 0,02.3 + 2a ^ a = 0,02.
FeS2-15e
> Fe^3 + 25"*
nN02 = ^ "FeS +15 npeS2 - > 9a + 15b = 0,24 ^ a = 0,01; b = 0,01.
> -Fe2(S04)3
0,04
> Fe*3 +
D a p an D .
g a m h o n h o p g o m FeS va FeS2 t r o n g d u n g
djch axit H N O 3 (dac, d u ) , t h u d u p e 5,376 l i t k h i NO2 (dktc) va d u n g d i c h X.
V i d\ 6: C h o 19,2 gam k i m loai M (hoa t r i n) tan hoan toan t r o n g d u n g d i c h
H N O 3 , t h u dupe 4,48 l i t k h i N O (san p h a m k h u d u y nhat, 6 dktc). K i m loai
Mia:
A. M g .
D. Fe.
C. C u .
B. A l .
Cho X tac d u n g v a i d u n g d j c h Ba(OH)2 d u , Ipc ket tiia va n u n g t r o n g k h o n g
Lot
khi d e n k h o i l u p n g k h o n g d o i t h u dupe m g a m chat ran. Gia t r i ciia m la
nNo = 0,2 m o l . Phan u n g hoa hpc:
A. 8,43.
3M
gidi:
B. 8,59.
C. 10,19.7
Lai gidi:
nN02=
D- 6,19.
,
> ,
1^=0,24 m o l .
Gpi so m o l : FeS (a mo l ) va FeS2 (b mol). Ta c6: 88a + 120b = 2,08.
+ 4nHN03
N h a n thay: XXM =
> 3M(N03)n + n N O + 2 n H 2 0
3n NO _ 0,6
mol
n
n = 2 va M = 64 (Cu)
MM
=
19 2
u,o
= 32n
n
D a p an C.
83
a'm nang an luy^n thi d^i hoc 18 chuyen
H6a hgc - Mguyen van Hi'i
Cty TNHH MTV D W H Khang Vi?t
/ i d u 7: N u n g 2,94 gam hon hgp X gom cac kim loai A l , Zn, M g trong khi oxi,
sau mot thai gian thu duoc 3,42 gam hSn hap Y. Hoa tan hoan toan Y vao
Chat o x i h o a : N*'* + n e
> X
dung dich H N O s (du), thu duoc 2,016 lit khi NO2 (san pham k h u duy nhat,
Bao toan e l e c t r o n : 2 n M g = n . n x ^
a dktc). So mol H N O 3 da phan ung la
_^ „ = 8
A. 0,20.
B.0,24.
C.0,16.
'
.
no2 = 0,015
mol
no=
0,03
mol
Cachl:Sudungsad6: X
Lm
'^^^^ "^'^^^^^
n ^ . 2 ^^^.j^= 0,03
mol.
A 0,12.
''^
, = 0,06 mol.
8A1 + 3OHNO3
^
'
Bao toan electron: n ^ ^ x j ^ ^ n g + n j ^ Q ^ =0,15 mol —> n
. =0,15
~^
"
va B. Loai C v i C u O tac dung voi c a 3 axit - > dung dich muoi mau xanh.
c ,,
_>
UHNOS
= 10.0,015 + 4.0,01 = 0,19 m o l - > D a p an D .
V
, B .
124,45.
y.'
qua:
ni>jo=0,l m o l ; n N 2 o = 0 , l m o l .
A\,Zn,Cu
> CuS04 + SO2 + 2H2O
Laigidi:
img = 0,15 mol; n x = 0,0375 mol.
D.NO.
^ -
-'•>
-
N^^ + 3e
> NO;
2N^-'*
2N*-^ + 8e
+ 8e
-
"
S'KS
: « o r i f:w 7
^
''^^^
> N2O
> NH4NO3 ( a m o l )
K h i cho k i m loai + HNO3:
" N O 3
1 du 9: Cho 3,6 gam M g tac dung het voi dung dich HNO3 (du), sinh ra 0,84 lit
CNC)2.
D . 112,05.
) M u o i n i t r a t + N O + N2O
vacothexayracaquatrinh:
B.N2O.
C. 99,65.
6 bai nay, t r u a 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
CU + 4 H N 0 3
> Cu(N03)2 + 2NO2 + 2H2O
D u n g dich H2SO4 d ^ c hoa tan C u va c6 khi khong mau, miii xoc thoat ra:
A.N2.
1
t r i cua m la
Chat oxi hoa:
khi X (san p h a m k h u duy nha't, a dktc). K h i X la
3nAi(N03)3 + 2 n N 2 0 + ^^NO
lit h o n h g p k h i X (dktc) g o m N O va N2O. T i k h o i ciia X so v a i H2 la 18,5. Gia
lil
Giai t h i c h : H N O s hoa tan C u v a c6 khi NO2 mau n a u bay ra:
C u + 2H2S04d,ic
'
Laigidi:
Nhan xet: A l , Fe khong tac dung voi H2SO4 va HNO3 dac ngupi - > L o a i A
„,.,'i;.j
•
Dua theo cac p h a n l i n g , n h a n thay: n H N 0 3 = 1 0 n N 2 O + ^^NO
. 120,45.
D. Cu.
Laigidi:
Dap an D . - - n f i ^ - t
uu^.x^
950ml d u n g d i c h HNO3 2 M , t h u d u g c d u n g d j c h chua m g a m m u o i v a 4,48
^'
CCuO.
• '' '
'
'
V i d^ 1 1 : C h o 27,45 g a m h o n h g p g o m A l , Z n va C u tac d u n g vvra d u v o i
trong ba lo bi ma't nhan, ta dung thuo'c thu la
B.Fe.
> A1(N03)3 + N O + 2H2O
'^'
n H N 0 3 = 0,05.3 + 0,015.2 + 0,01 = 0,19 m o l - > D a p an D .
mol.
n du 8: De nhan biet ba axit dac, nguoi: H C l , H2SO4, H N O s dung rieng bi$t
A.Al.
D . 0,19.
* J t l ^ >'
Bao toan n g u y e n to N : n n N O a
' *
3
^ ^'^^
'
3nAi = 8 n N 2 0 + 3 n N o = 8.0,015 + 3.0,01 = 0,15 m o l - > nAi = 0,05 m o l .
-> Dap an B.
Bao toan N : n n N O a " N O 3
IrV
Cach 2: Bao toan electron:
" ^'^'^
:ach 2: Q u i doi Y thanh X va O (0,03 mol).
C. 0,17.
> 8A1(N03)3 + 3N2O + 15H2O
A l + 4HN03
" «li 1
"^^o "
B. 0,15.
each 1: Cac p h a n u n g :
trong oxit se bi thay the bang 2 goc N O 3 de
Bao toan nguyen to N : n H N O s = " N O 3
-
^'""^ - ' "
Laigidi:
4—"-
+ Khi cho kim loai + HNO3: n j ^ ^ _ (muoi) — ngtraodoi— fij^jQ^ — 0,09 mol.
=2n
^>,dn mJ
" -
va 0,01 m o l k h i N O . So m o l axit H N O 3 da t h a m gia p h a n u n g la
^ Muoinitrat + NO2
y: Y chua ca kim loai (con du) va oxitsgS + s O r l
tao muoi nitrat, do do: n
0,0375.n
d u n g d i c h X ( k h o n g chua N H 4 N O 3 ) va h o n h g p k h i g o m 0,015 m o l k h i N 2 O
> Y
+ K h i cho oxit + HNOa: 1 goc
ai5.2 =
D a p an B.
V i d i ? 10: H o a tan hoan toan m g a m A l bang d u n g d i c h H N O 3 loang, t h u d u g c
D . 0,14.
Lotgtat:
Bao toan k h ^ luc^ng: mo^ = 3,42 - 2,94 = 0,48 gam
->
K h i X la N2O ^
^'^^^ '
' " " " i " ' = 3 n f j o + 8 n N 2 0 + 8 n N H 4 N 0 3 = ( I ' l + 8a) mol.
Bao toan n g u y e n to N :
UHNOS ^
" N O 3 ^ "^^^
0,95.2 = 1,1 + 8a + 0,1 + 2.0,1 + 2a
Bao toan k h o i l u g n g : m = m ^ i , zn, C u
^ "N2O + 2nNH4N03
,. ;
a = 0,05 m o l .
' " N O S ^
"^NH4N03
= 27,45 + (1,1 + 8.0,05).62 + 0,05.80 = 124,45 gam.
85
Cty TNHH MTV DVVH Kliang Vi§t
:im nang On luygn thi dgi hgc 18 chuyfin 66 H6a hgc - Nguygn Van HSi
Theo cong t h i i c d u o n g cheo ta c6:
—> Dap an B.
Nhan xet: Can n h a n ra bai toan da " g i a u d i " s a n p h a m NH4NO3.
Vi dyi 12: Cho m a n h C u phan u n g he't v o i d u n g dich H N O a , t h u d u g c 0,896 h't
(dktc) h o n h g p k h i X ( g o m N O va NO2) c6 k h o i l u g n g b a n g 1,52 gam. Co
can d u n g d j c h sau phan l i n g t h u d u g c a gam m u o i k h a n . Gia t r j cua a la
A . 5,64.
B. 7,52.
C. 9,4.
Lai
,
;
+ |A
D a p an B.
d u g c 0,672 l i t N2 bay ra 6 d k t c va d u n g dich X. K h o i l u g n g m u o i k h a n t h u
d u g c k h i c6 can d u n g dich X la
B. 30,1 gam.
;s '
HMg =
C. 55,8 gam.
D . 15,04
Chat oxi hoa:
2N*''
+ lOe
> N2
— ^ nenh?n
2N*^
->
+ 8e
N H 4 N a (a m o l )
a = 0,105 m o l .
^^''^' -
thanh m u o i NH4NO3!
V i dv 15: H o n h g p X g o m M g , A l va Z n . Cho m gam X tac d u n g VvJ-i d u n g dich
H C l d u , t h u d u g c 36,45 g a m m u o i clorua. M a t khac, hoa tan he't m gam X
A . 5,8.
.
B.8,7.
C.11,6.
D . 14,5.
•
, V-\T->
0,4 - >
2nMg = 0 , 7 m o l .
,
= 10n[vj2 = 0 , 3 m o l .
ao toan k h o i l u g n g , ta c6: m + 35,5a = 36,45; m + 62a = 55."
Dap an C.
Lu-u y: N e n ap d u n g ngay bao toan electron: 2 n ^ g = 3 n ^ o + 8 nNH4N03 •
i du 14: H o a tan hoan toan 12,42 gam A l bang d u n g dich H N O a loang (du),
t h u d u g c d u n g d i c h X va 1,344 l i t (a dktc) h o n h g p k h i Y g o m N2O v a N2.
T i k h o i cua Y so v o i h i d r o la 18. Co can X, t h u d u g c m g a m chat ran khan.
Gia t r i cua m la
A . 38,34.
B. 34,08.
C. 106,38.
Lai gidi:
-
Ta c6: nAi = 0,46 m o l ; nv = 0,06 m o l ; M y = 18.2 = 36
H3PO4 + 2 N a O H
H3PO4 + 3 N a O H
mol 1:1
mol
: :\ vnU W ' l
-> NaH2P04 + H2O
1:2 -> Na2HP04 + 2H2O
mol 1:3
/ T i r ',-,->•") :'t[ ill, :
> Na3P04 + 3H2O
a=1
: NaH2P04
1< a < 2 : N a H 2 P 0 4 + N a 2 H P 0 4
Neu:
nH3P04
'en
0-..,-
2 'Y tV :•'::„
i b. V I D y M A U
^ V i du 1 (B-08): Cho 0,1 m o l P2O5 vao d u n g d j c h chua 0,35 m o l K O H . Sau k h i
D . 97,98.
-
^ , ,
- > D a p an C.
H3PO4 + N a O H
^'''^''l;:,
Vay: m = mMg(N03)2 + mNH4N03 = 0,35.148 + 0,05.80 = 55,8 gam
'
- > a = 0,7 m o l ^ m = 36,45 - 35,5.0,7 = 11,6 g a m
"••'^***W;'yi'«^
.1
1
a05
'•••T
ra \
a. L i thuyet
'-•••"'Mol:'.-''
; :
m o t l u g n g i o n k i m loai giong nhau t h i : n^|_ = " N O 3 " ^ ("^^O • * • ^
Chat oxi hoa:
NH4NO3
'' •'••''^
Lim y: Ba k i m loai kha m a n h ( M g , A l , Zn) tac d u n g v o i axit HNO3, c6 the tao
4. A X I T P H O T P H O R I C : H3PO4
>
-
- > D a p an C.
san p h a m k h u da dugc "giau d i " , do la su tao thanh m u o i NH4NO3:
+ se
-> N2
Nhan xet: D o goc Or va N O 3 deu c6 dien tich 1 - nen k l i i ket h g p v o i cung
N h u vay so m o l electron trao doi chua bang nhau —> "chua o n " . 6 day, mot
2N*s
(
+ lOe
2N^5
Lai gidi:
T u y nhien, giai theo p h u o n g phap bao toan electron se nhanh hem.
- > nenhucmg =
1 }
> N2O
i *
Bai toan nay cac e m c6 the giai k h i vie't p h u o n g t r i n h phan l i n g .
Mg^2
+ Se
cua m la
'''
0,35 m o l ; txj^^ = • ^ ^ = 0,03 m o l .
M g - 2e
> AP
trong d u n g d i c h HNO3 d u , t h u d u g c 55 gam m u o i n i t r a t k i m loai. Gia t r i
gam.
Lai gidi:
Chatkhu:
2N*5
fI
mol
V $ y : m = mAi(N03)3 + mNH4N03 = a46.213 + 0,105.80 = 106,38gam.
,f
du 13: H o a tan hoan toan 8,4 gam M g bSng d u n g d i c h HNO3 v u a d u , t h u
A . 51,8 g a m .
C h a t k h u : A l - 3e
- > 3.0,46 = 8.0,03 + 10.a03 + 8a
"^•1\::^.,iriyP4^-^,'..j>fy'{^
Bao toan n g u y e n to C u : ncu(N03)2 = " c u = 0,04 m o l .
a = 0,04.188 = 7,52 gam
n N 2 0 = nN2 = 0,03
Baotoanelectron:3nAi = 8nN2O+10nN2+8nNH4N03
2ncu = Sn^o + " N O Z = 0,02.3 + 0.02 = 0,08 m o l - > ncu = 0,04 m o l .
^
1
va CO the x a y r a qua trinh:
T r u o c he't cac e m can t i m so m o l m o i k h i : n ^ o = nN02 ^ ^'^^
' "••'••H'
44 -36
nN2
1
Chat oxi hoa:
D . 15,04.
gidi:
Bao toan electron: • '''V:/'-
28 - 3 6
nN20
phan u n g xay ra h o a n toan, d u n g dich t h u d u g c c6 cac chat la
iorM.
j
A . K3PO4 va K2HPO4.
B. K2HPO4, KH2PO4.
IA nuu
C.K3P04vaKOH.
D.
'•Q-;
H3PO4
va
KH2PO4.
87
Cim nang On luy^n thi dgi hoc 18 chuy6n dg H(5a hpc - Nguygn Van
Lai
Nhanxet:
HSi
gidi:
V i dv 4= Cho 21,3 gam P2O5 vao d u n g djch c6 chua a g a m N a O H , t h u d u g c
> 2H3PO4
P2O5 + 3 H 2 O
Cty TNHH MTV D W H Khang Vi?t
^
,
'
d u n g d i c h c6 chua 28,4 gam Na2HP04 va b g a m Na3P04. Gia t r i ciia a, b Ian
nH3P04 = 2ni^05 = 0'2 m o l .
->
l u g t la
A . 2 8 ; 8,2.
1 75 ^ t a o 2 m u o i : K H ^ P a v a K 2 H P O 4 .
J}mH_ ^
nH3P04
0,2
—> Dap an B.
'' '
~ -M - '•n4i;
*
;finri:i ©up jvi v?,?: dvlt 65
Lieu y: Cac e m can chuyen P2OS thanh H3PO4 va xet t i I f m o l n h u tren.
V i dv 2 (B-09): Cho 100ml d u n g djch K O H 1,5M vao 200ml d u n g d i c h H3PO4
0,5M, t h u d u g c d u n g d i c h X. Co can d u n g d i c h X t h u d u g c m g a m m u o i
C. 2 8 ; 16,4.
D . 20 ; 8,2.
Lot gidi:
2;i 3
28 4
'"b
= - ^ = 0,15 m o l ; nN32HP04 = 7 ^ = 0'2 m o l . „ ^ ,,.
digji:.^
• r^^hnvl •
*
B. 2 0 ; 16,4.
P2OS + 3H2O
^^^.^^
nH3P04 = 2np205 = 0 , 3 m o l .
> 2H3PO4
Bao toan n g u y e n to P: nH3P04 = nNa2HP04 + nNa3P04
0,3 = 0,2 +
—
khan. Gia t r i ciia m la
A . 15,5 gam.
B. 18,2 gam.
V,.-
^. "KOH =
C. 12,8 g a m .
- > b = 16,4 g a m .
D . 16,4 gam.
So do p h a n u n g : H3PO4 + K O H
Lai gidi:
.
•.
m o l ; nH3P04 = 0,10 .
> M u o i + H2O.
Bao toan k h o i l u g n g : 0,3.98 + a = 28,4 + b
^
. ,
a = 28 gam
D a p an C.
V i d u 5: Cho 0,05 m o l P2O5 vao d u n g d i c h chua 0,4 m o l N a O H . Sau k h i phan
Cach 1 : NMn xet
= •
= 1,5 - > tao 2 m u o i : KH2P04va K2HPO4.
-
u n g xay ra hoan toan, d u n g d i c h t h u d u g c c6 cac chat la
"H3PO4
H3PO4
Mol:
a
->
H3PO4
Mol:
b
A.Na3P04vaNa2HP04.
+ KOH
a
a
+ 2KOH
->
f
->
Ta c6: a + b = 0,1; a + 2b = 0,15
b
''^'r •!'>
i,t:.,dMte-..
•
U:
- > a = b = 0,05 m o l .
- > m = 0,05.136 + 0,05.174 = 15,5 gam
+
KOH
> Muoi
' ^ ^ l ' "H-^(H3P04) = ^ ' ^ ^ " ^ " l ^
'^H20 = 0,15 m o l .
t h u d u g c d u n g d i c h Y c6 chua 14,95 gam h o n h g p m u o i . Gia t r i cua a la
B. 1,00.
C.0,50.
So do p h a n u n g : H3PO4 + K O H
• '
LUYEN
H C l 7,3% t h u d u g c d u n g d i c h Y va 0,672 l i t k h i H2 (dktc). N o n g dp % ciia
H C l t r o n g Y la
B.2,18%.
C. 1,83%.
D . 1,78%.
Bai 2: H o a tan hoan toan h 6 n h g p X g o m Fe v a Z n b a n g m g t l u g n g v u a d i i
d u n g d i c h H C l C%, t h u d u g c d u n g dich Y. N o n g do p h a n t r a m ciia FeCh va
Z n C h t r o n g Y Ian l u g t la 8,05% va 8,63%. Gia t r i cua C la
A . 5.
O H ^ O = " K O H = 0,2.
B.15.
C.20.
D . 10.
Bai 3: H o a t a n h o a n toan 7,8 gam h6n h g p g o m M g va A l b a n g d u n g d i c h H C l
> Muoi + H2O.
Bao toan k h o i l u g n g : 0,la.98 + 0,2.56 = 14,95 + 0,2.18
a = 0 , 7 5 - > D a p an A .
D . 0,80.
c6 m u o i axit - > K O H het.
^
5. B A I T A P 6 N
A . 2,43%.
gidi:
NMn xet: V i Y chua h o n h o p m u o i
"H+(H3P04) =°'3^
0,1
Bai 1: H o a tan het 1,04 g a m hon h o p X g o m M g , Fe bang 40 g a m d u n g d i c h
- > D a p an A .
V i dv 3: Cho 100ml d u n g d i c h H3PO4 a mol/1 vao 100ml d u n g d i c h K O H 2 M
' ' O H - ( K O H ) =°'2;
,.
Luu y: Cac e m can c h u y e n P2O5 thanh H3PO4 va xet t i I f m o l n h u tren.
1J
Lai
r;
—> Dap an C.
+ H2O
m = 0,1.98 + 0,15.56 - 0,15.18 = 15,5 gam
A . 0,75.
, i Sij, v«x ;^fi!j ut;.a(,j 'MO ul.-' >.
nH3PO4 = 2np2O5 = 0,l mol.
nH3P04
~^ "^H3P04 + r " K O H = n ^ + rflH20
^
> 2H3PO4
'
4,
Cdc em luu y:
"OH-(KOH) =
N h a n x e t : P2O5 + 3H2O
->
,,f,„:
Lai gidi:
i]±JaOH. ^ M = 4 > 3 ^ NaOH d u va tao 1 m u o i Na3HP04.
D a p an A .
Cach 2: A p d u n g bao toan k h o i l u g n g cho so do phan l i n g :
H3PO4
D . H3PO4 va NaH2P04.
, 0 ,
>K2HP04+2H20
2b
B. Na2HP04, NaH2P04.
C. Na3P04 va N a O H .
> KH2PO4 + H 2 O
^
d u . Sau p h a n u n g t h u d u g c d u n g djch c6 k h o i l u g n g tang t h e m 7,0 gam so
D
v o i ban d a u . K h o i l u g n g ciia A l trong h o n h g p ban d a u la
A . 5,40 gam.
B. 2,70 gam.
C. 1,35 gam.
D . 4,05
gam.
Ca'm nang 6n luygn thi d^i hgc 18 chuyfin d6 H6a hgc - NguySn van Hai
Cty TNHH MTV DVVH Khang Vi§t
Bai 4: Dot nong 2,80 gam hon hgp X gom Cu, Zn va M g trong khi oxi d u , thu
dugc 4,08 gam hon hgp oxit Y. De hoa tan het Y can toi thieu V m l dung
djch H2SO4IM. Gia tri ciia V la
A. 80.
r?S C I B . 60.
C. 100.
g
D. 120.
^e;/Bai 5: Hoa tan hoan toan m gam hon hgp gom Mg, A l , Z n trong dung dich
H2SO4 loang, d u thu dugc 0,672 lit khi H2 (dktc) va 3,92 gam hon hgp muoi
sunfat. Gia tri cua m la
A. 2,48.
B. 1,84.
C. 1,04.
D. 0,98.
Bai 6: Cho 6,45 gam hon hgp hai kim loai X va Y (deu c6 hoa tri 2) tac dung voi
dung dich H2SO4 loang du. Sau khi phan ung ket thuc, thu dugc 1,12 lit khi
(dktc) va 3,2 gam kim loai. Lugng kim loai nay phan ung vua du vai 3,55
gam khi CI2 khi dot nong. Cac kim loai X va Y Ian lugt la
A. Zn va M g .
B. Zn va Cu.
C. M g va Cu.
D. M g va Pb.
Bai 7: Nung hon hgp X gom 0,28 mol Fe va a mol Cu trong khong khi mgt thoi
gian, thu dugc 25,28 gam chat ran Y. Hoa tan hoan toan Y bang dung dich
H N O 3 loang (du), thu dugc 1,792 lit khi N O (san pham k h u duy nhat 6
dktc). Gia tri cua a la
A. 0,03.
B.0,10.
C.0,06.
'
D. 0,08.
^
Bai 8: Cho 10,8 gam hon hgp gom Fe304 va Cu vao dung dich H2SO4 loang du.
Sau khi cac phan ling xay ra hoan toan, t h u dugc d u n g dich X chiia m
gam muo'i va chat ran Y. Hoa tan het Y trong dung djch H N O 3 loang,
sinh ra 0,448 lit khi N O (san pham khu duy nhat, a dktc). Gia tri ciia m la
A. 16,80.
B. 18,32.
C. 13,92.
D . 18,48.
Bai 9: Hoa tan hoan toan m gam oxit FexOy bang H2SO4 dac nong, thu dugc
dung dich chua 4 gam mgt loai muoi sat duy nhat va 0,224 lit SO2 (dktc).
Cong thuc ciia oxit sat va gia tri m Ian lugt la
A. FeOval,44.
B. Fe304 va 4,64. C. Fe203 va 3,20. D. Fe304 va 2,32.
Bai 10: Hoa tan hoan toan 4,4 gam hon hgp X gom Fe, Cu, Ag trong dung dich
H2SO4 dac, nong thu dugc 2,24 lit khi SO2 (san pham khu duy nhat, a dktc)
va dung dich Y. Co can Y thu dugc m gam muoi khan. Gia tri ciia m la
A. 14,2.
B. 9,2.
C.9,3.
^
D. 14,0.
Bai 11: Nhiet phan hoan toan 6,32 gam KMn04, toan bg lugng khi O2 sinh ra
cho phan ung het vai 3,6 gam hon hgp X gom Fe va Cu, thu dugc hon hgp
Y. Hoa tan het Y trong dung dich H2SO4 dac (du), thu dugc 0,784 lit SO2
(dktc). Thanh phan % khoi lugng Fe trong X la
A. 46,67%.
B. 53,33%.
C. 31,11%.
D. 69,89%.
Bai 12: Cho 0,015 mol mgt loai chat oleum vao nude, thu dugc 200ml dung dich
X. De trung hoa X can dung 200ml dung dich NaOH 0,3M. Oleum do la
A. H2SO4.2SO3.
B. H2SO4.3SO3.
C. H2SO4.4SO3.
D. H2SO4.SO3. 4
Bai 13: Cho m gam hon hgp X gom A l , Cu vao dung dich HCl (du), sau khi
ket thiic phan ung sinh ra 3,36 lit khi (a dktc). Neu cho m gam hon hgp X
tren vao mot lugng d u axit nitric (dac, ngugi), sau khi ket thiic phan ling
sinh ra 6,72 lit khi NO2 (san pham khu duy nhat, a dktc). Gia tri cua m la
A. 11.5-
'yimn
12,3.
C. 10,5.
D. 15,6.
Bai 14: Hoa tan hoan toan 1,23 gam hon hgp X gom Cu va A l vao dung dich
HNO3 dac, nong thu dugc 1,344 lit khi NO2 (san pham k h u duy nhat, 6
dktc) va dung dich Y. Sue khi NH3 (du) vao Y, thu dugc m gam ket tiia.
phan tram khoi lugng cua Cu trong hon hgp X va gia tri cua m Ian lugt la
A. 21,95% va 2,25.
B. 78,05% va 2,25.
C. 21,95% va 0,78.
D. 78,05% va 0,78.
Bai 15: Hoa tan het 20,6 gam hon hgp gom Cu, Fe, Zn trong dung dich HNO3
du, thu dugc dung dich Z (khong chua NH4NO3) va hon hgp khi gom 0,15
mol N O va 0,25 mol NO2. Co can dung dich Z thu dugc khoi lugng muoi
khan la
" ' *
; ^ * ^ bv.X'-^
;iosTJ'4io .
A. 45,4 gam
B. 64,0 gam.
C. 51,6 gam.
D. 57,8 gam.
Bai 16: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 9,6 gam Cu vao 200ml
dung dich X chua dong thoi H N O 3 2M va H2SO4IM. Sau khi phan ung xay
ra hoan toan thu dugc khi N O (san pham khu duy nhat) va dung djch Y.
Kho'i lugng muo'i CO trong Y la:
A. 38,55 gam.
B. 50,95 gam.
C. 63,35 gam.
D. 43,50 gam.
Bai 17: Cho 3,2 gam bgt Cu tac dung voi 100ml dung dich X hon hgp gom
HNO3 I M va H2SO4 0,1M. Sau khi cac phan ling xay ra hoan toan, sinh ra
V lit khi N O (san pham khu duy nhat, 6 dktc). Gia tri ciia V la
A. 0,784.
B. 0,896.
C. 0,672.
D. 1,344.
Bai 18: Hoa tan hoan toan 0,06 mol FezOs va 0,1 mol FeS2 trong 260ml dung
dich H N O 3 4M, san pham thu dugc gom dung dich X va mot chat khi
thoat ra. Dung dich X c6 the hoa tan toi da m gam Cu. Biet trong cac qua
trinh tren, san pham k h u duy nhat ciia N*"* deu la NO. Gia tri ciia m la
A. 9,60.
B. 16,64.
C. 11,52.
D. 14,03
Bai 19: Cho 12,45 gam hon hgp X gom Fe, Mg, Zn vao dung dich H N O 3 du, thu
dugc dung dich khong chua NH4NO3 va hon hgp khi gom 0,2 mol N O va
0,1 mol NO2. Dot chay 12,45 gam hon hgp X trong khi clo d u thu dugc bao
nhieu gam muo'i clorua?
A. 37,3..
B.23,1.
)U!Ji
C. 30,2.
D o j D. 16,0.
Bai 20: Cho 3,84 gam Cu phan ung voi 80ml dung dich chua H N O 3 I M va
H2SO4 0,5M, sau phan ung thu dugc V m l khi NO (dktc). Gia tri ciia V la
A. 896.
B. 560.
C. 448.
D. 336.
Q1
Cty TNHH MTV DVVH Khang Vigt
dm
nang On luygn thi dai hpc 18 chuySn dg H6a hpc - Nguygn van Hai
Bai 21: Cho m gam hot Cu kim loai vao 200ml dung djch HNOa 2M, c6 khi NO
thoat ra. De hoa tan het hot Cu, can them tiep 100ml dung dich HCl 0,8M,
dong thoi Cling c6 khi NO thoat ra. Gia tri cua m la
A. 9,60 gam.
B. 3,84 gam.
C. 10,24 gam.
D. 11,52 gam.
Bai 22:Hoa tan hoan toan 0,01 mol FeS2 trong 42 gam dung dich HNOs 37,5%
theo phan ung: FeS2 + HNO3
> Fe(N03)3 + Isr02 + H 2 S O 4 + H2O
Cho V ml dung dich NaOH 2M vao dung dich sau phan ung de thu dugc
lugng ket tiia Ion nhat. Gia trj nho nhat cua V la
A. 40ml.
B. 70ml.
C. 80ml.
' D. 100ml. '''^
, H
'
.'.•••A
6. H U ' 6 N G D A N - L C J I G I A I
:i :
40. 7,3
1 ^ = u,uo mol;; n H , I =
= 0,08 m u i 11 H
= 0,03 mol.
Bail: nHni =
Hr
36,5
"2
22,4
Ggi so'mol: Mg = x va Fe = y mol.
Ta CO so do: Mg
) H2; Fe " ^ ^ ^ > H2
.
nH2 = X + y = 0,03 mol -> nnci = '^^^2 ^ ^'^^
Bai 3:
2A1
Mg
i
hS^
^'^^
+
> H2O
2H*
,
0,16
1
Vay: nH2S04 = 2 " H ^ =0,08 mol ^
,,;,;.3««B(-^
i ;V
n no
V^d H2SO4 = ^
^ , 4 r • <•
«
= 0,08 lit = 80ml
>') n^oJ c: '
Bai 5:
H2SO4
N/ifln xet: Bao toan nguyen to'H:
Mol:
Kim loai + H2SO4
Ta c6:
0,03
> Hi
<-
mkimiogi
;« ,
0,03
'^muiBi':"
<,>|
)• Muoi + H2
+ maxit =
mmuoi
+
s
<
m^^^
-> m = 0,03.2 + 3,92 - 0,03.98 = 1,04 gam
,
^ i^^^--A
" "'
Bao toan khoi lugng cho so do:
Zn + 2HC1
> ZnCh + H2
Xet vai 100 gam dung dich Y:
' '
Ggi so mol: Al = x; Mg = y:
••?«
^
Ta c6: 27x + 24y = 7,8; l,5x + y = 0,4
x = 0,2 mol; y = 0,1 mol.
'' ^ ' '
-> mAi = 0,2.27 = 5,4 gam -> Dap an A.
]• ' •
=
Bai 4: Bao toan kho'i lugng ta c6: mo2 = 4,08 - 2,80 = 1,28 gam •
1 78
Vay: no = - — = 0,08 mol -> n 2- = 0,08 mol.
*
> ^(S"' • •
16
Khi cho oxit kim loai tac dung voi axit, ion 0^~ trong oxit se ket hgp voi H*
trong axit tao thanh H2O theo phuong trinh:
:-m.ms nbiji .1
"1
> FeCh + H2
"
.
Dap an A.
mv = 40 + 1,04 - 0,03.2 = 40,98 gam
n 02 36 5
^ C%Hci= ' ^ ^ .100% =1,78% -> Dap an D.
40,98
^
Bai 2: Phan ung hoa hgc:
gam va niz^ciz = 8,63 gam
'
-> mH2 = 0'8 gam -> nH2 = 0,4 mol.
02-
Ik
^
:
Mol: 0,08
t
mFeCl2 =
^
Khoi lugng dung dich tang them = Dugc - Mat -> 7 = 7,8 - mH2
XfVay: nHci di, = a08 - 0,06 = a02 mol.
Fe + 2HC1
> 2A1C13 + 3H2
> MgCh + H 2
+ 6HC1
+ 2HC1
Dap an C.
' '*. ; ' '
-> mH20 (Y) = 100 - 8,05 - 8,63 = 83,32 gam.
Nhan xet: Khi cho X va Y vao dung dich H2SO4 loang du tha'y con lai 3,2 gam
Nhan xet: Lugng H2O trong Y ciing chinh la lugng H2O c6 trong dung dich
kim loai khong tan -> c6 mgt kim loai dung sau H trong day dien hoa (gia
su do la Y) -> mv = 3,2 gam; mx = 6,45 - 3,2 = 3,25 gam.
HCl ban dau.
Mat khac, do nong do HCl bang C%
C
m HCl _
mH20
100-C
X + H2SO4
Y
Bao toan nguyen to'CI:
Dap an D.
2.8,06
2.8,63
83,32.C
-1^^1^=36,5(100-0
'
3 25
'
-> nv = nH-, = 0,05 m o l - » Mx = —— = 65 ^ XlaZn.
83,32.C
FeCl2 +2nz,
2npea2 +2nznc.2 =nHCl ^
> XSO4 + H2
^
^=
^„
+
CI2 —!—> YCI2
i
, ,
.
,
,,
32
-> nY= n a 2 = 0'05mol -> Mx= -rhrz = 64: - ^ Y l a C u .
-> Dap an B.
• 93
Cty TIMHH MTV DVVH Khang Vi$t
dm nang 6n luygn thi d^i hpc 18 chuyen 6i H6a hgc - NguySrr Van Hai
1 792
Bao toan n g u y e n to Fe:
O'O^ m o l .
Bai 7: 11^0=
- > X.0,01 = 0,01
= m y - m x = 25,28 - 0.28.56 - 64a = 9,6 - 64a. >
So do phan u n g : -j^yr:^],,)
'^'^M'''!^^ :P".
> Y(2)
¥e,Cn{l)
> Fe-^ Cu-^S)
Xet su trao d o i eleclion o cac giai doan:
U ) ^ ( 3 ) : Fo - . V
Cu
;
£3
> Cu*^ ^
-2e
( 1 ) ^ . ( 2 ) : O . +4e
i
^ 20-^-> nenhjn = 4no2 = •^^^•^^= 1,2 - 8a ^ ; Y | V
8
^2) ^ ( 3 ) : N * * +3e
> N O n e n h j n = 3nNo = 0,24.
ir-;:! i
•
Bao toan electron: 3.0,28 + 2.a = 2b + 0,24 - > a = 0,06; b = 0,36.
.
•
': ' •
.
a
a
Mol:
a
-> 2FeS04 + CuS04
2a
+ 8HNO3
ai
+
2nH20
^
0,2 r
i
0,2
<-
'
.
?
> i n = 14 g a m - > D a p an D .
^
r a r h 2: K h i cho k i m loai tac d u n g v o i axit F12S04 (dac, nong), so m o l go'c sunfat
Tu-(l)->(3):
Chatkhu:
Chat oxi hoa:
u^^m *
a
2- = " S O T = —— = 0,1 m o l .
22,4
J,
,5,^
•
K2Mn04 + M n 0 2 + O 2
.1'
^ ..•..'•^•.^•^.•'^.fr.:
<-
0,03
Vay: m = 152.3.0,03 +160.0,03 = 18,48 gam
4
=
"SO2 =
> Vv*^
>20 2
mol.
Nhan xei: O x i t sk FexOy tac d u n g v o i H 2 S O 4 dac, n o n g
v^iiiiBi
•? ^
> ' .''ft'
Bai 12: G Q I cong t h i i c o l e u m la H 2 S 0 4 . n S 0 3 .
,.:
Mol:
S O 2 t h i o x i t la FeO
hoac Fe304.
T r o n g 1 m o l FeO hoac Fe304 d e u chua 1 m o l Fe*^ nen d e u c6 k h a nang
= npe^Oy = 2 n s o 2 n p e ^ O y = 2.0,01= 0,02 m o l
^*^'''
.
H2S04.nS03 + n H 2 0
n h u o n g 1 m o l electron, d o do:
> Cu^^
> SO2
•
D a p an A
D a p an D .
= 0'°!
Cu - 2e
+ 2e
Mat khac: 56x + 64y = 3,6 ^ x = 0,03; y = 0,03 - > % m F e = 46,67%.
0,02
0 224
Fe - 3e
O 2 + 4e
> Fe^^ (3) •''^^^^
3nFe+2ncu=4no2 + 2 n s o 2 - ^ 3x + 2y = ai5.
> 3Cu(N03)2 + 2 N O + 4 H 2 O
<<-
0,03
> Y (2)
* ' " ' ' £•
Bao toan electron:
232a + 64(a + 0,03) = 10,8 g a m ^ a = 0,03 m o l .
Bai 9: n F e 2 ( S 0 4 ) 3 =
nS02
i5ao toan k h o i h m n g . 4,4 + 98.0,2 = m + 0,1.64 + 18.0,2
mol:
Fe(amol),Cu(bmol)(l)
a
Chat ran Y la C u d u . Phan u n g hoa tan C u bang H N O 3 loang:
3Cu
'
6 day, cac e m can s u d u n g so do phan u n g :
Fe2(S04)3 + FeS04 + 4 H 2 O
4H2SO4
a
'
giai.
Fe2(S04)3 + C u
Mol:
, .f
f
Nhqn xet: Bai nay neu d u a theo p h u o n g trinh phan u n g se rat dai d o n g va kho
Phan l i n g hoa hoc:
Mol:
'" > R2(S04)„ +
+ 2n}Lb04{dac)
: '
"°2^i-i§^'^'°^"'°^'"S°2=^=0'035mol.
• f
Bai 8:
Fe304 +
t
^jj-h_L: G Q I cong t h i i c r h u n g cua cac k i m loai la R .
Bai 11: 2 K M n 0 4
Bao toan k h o i l u o n g : 64a + 16b = 8,58 - 0,28.56 = 9,6.
,, ..
*
D o vay: m = 4,4 + 0,1.96 = 14 g a m - > D a p an D .
Q u i d o i Y t h a n h : Fe = 0,28 m o l ; C u = a m o l va O = b m o l .
-»Dap anC.
\'
SO4
Cach 2:
'
,,
D a p an A .
tao m u o i duQ-c t i n h n h u sau: n
viy/i
t
,i
2 24
• Bao toan electron: 0,84 + 2a = 1,2 - 8a + 0,24 - > a = 0,06 m o l .
'
-> Dap an C.
m = 72.0,02 = 1,44 g a m
2 24
Bai 10: n s o j = ^
=
2R
nenhirong=2ncy = 2a
x.0,01
x = 1 -> o x i t sat la FeO.
^
— > le'-* -> nenhirong=3npp =0,84
-
> xFe2(S04)3
0,02
Cachl:
Bao toan k h o i lucmg:
2FexOy
0,015
H2SO4
Mol:
,
+
0,015(n+l)
2NaOH
» t f u it
> (n+l)H2S04
0,015(n+l)
,,
> Na2S04 + 2 H 2 O
0,03(n+l)
-> 0,03.(n+l) = 0,2.0,3-> n = l ^ O l e u m la H 2 S 0 4 . S 0 3 ^
Bai 13: n^^=^^=
,
0,15 m o l ; n^o2 = g
= 0,3 m o l .
D a p an D .
^^^
^^
^J^^
95
naiKj on luyeii ilii liai iioc 18 chuy6n d6 H6a hpc - Nguyjn VSn H i i
Cam
Cty T N I I H M T V D V V H Khang Vi§t
Nhan xet: K h i cho X + H C l ( d u ) , chi c6 A l phan u n g (Cu d u n g sau H ) .
= nHNOa + 2nH2S04 = 0,4 + 2.a2
K h i cho X + H N O 3 (dac, n g u p i ) , chi c6 C u phan u n g ( A l b i t h u d o n g hoa,
khongtan).
,.
> 2A1C13 + 3H2
Cu + 4HN03
> Cu(N03)2 +
n ^ i = ^n^^
= 0,1 m o l .
Phuong t r i n h i o n r i i t gpn:
3 Z n + 8H^ + 2NO3
^
Mol:
^
^ ncu =
2NO2 + 2H2O
m = 0,1.27 + 0,15.64 = 12,3 gam
Dap an B.
- " N O Z
Mol:
^'
G o i n c u = X m o l ; n A i = y m o l . Ta c6: 64x + 27y = 1,23 gam.
> Cu(N03)2
> Cu(OH)2i
Al
+
""''^
^
= «!
-(/ufc «6wm OJ;v
So do p h a n u n g :
Cu
> A1(N03)3
> [Cu(NH3)4](OH)2
> Al(OH)34'
Bao toan electron: 2ncu + 3nAi = n N o , - > 2x + 3y = 0,06 m o l .
,
'-^'^
S<,?/'
- > X = 0,015 m o l ; y = 0,01 m o l - > % m c u = ^ ^ ^ ^ ^ . 1 0 0 % = 78,05%.
1,23
Vay: m = mAi(OH)3 = 0,01.78 = 0,78 gam - > D a p an D .
Chat oxi hoa:
N^^ + 3e
Bao toan electron:
netraodoi
^O;
Zn-2e
N*^ + l e
N03
Bao toan k h o l l u p n g : m = m ^ ^ pg
8H*
0,045
0,12
0,7.62 = 64 gam.
^
\^
/
n2n= 0,15 m o l ; n ^ u ^ 0,15 m o l .
t r o n g d u n g d j c h la d o 2 axit p h a n l i ra —> g i a i theo p h u a n g t r i n h i o n .
^.^^^ ^^^^
-
^
. , ,.
^ ^
giai theo p h u o n g t r i n h i o n .
+ 2NO3
->
, .
;A
->
0,03
BailS:
Phan l i n g hoa hpc:
'.fOU'"
•/••••e'
0,1
Mol:
.5.', , ;
* 'tia
\
+ y >
.. . .
^
0,2
> 2Fe(Na)3 + 3H2O ,
0,24
+
+ 4H2O
.OI^S
.a nil qftCl '
'
Q,\2
,
< ,
>
= 0,4 m o l ; N O ^ = 0,66 m o l v a SO ] ' = 0,2 m o l .
Cac p h a n u n g hoa tan Fe:
3Cu
^j^^;
; •-t-r
> Fe(N03)3 + 5 N O + 2H2SO4 + 2H2O
0,8
0,06
,
^
0,03
,
FeS2 + 8HNO3
Mol:
-
> 3Cu2- + 2NO
V = a03.22,4 = 0,672 l i t D a p a n C.
Fe3* = 0,22 m o l ;
'
Nhan xet: K h i axit H N O s c6 mat d o n g t h o i v o i axit H2SO4 loang t h i l u g n g H *
96
,
<
>
Fe203 + 6HNO3
- > Dap an B.
Bai 16:
"
n' r >
D u n g d i c h X g o m cac i o n :
" ^ M O - ^ "^^'^
, 1 , : ^ =.««„,m * : I n m ££.0 - ,,->?i
,
^
+
M o l : 0,1
> NO2.
N O 3 n a m trong m u o i dugc t i n h n h u sau:
' ''
o) O ' ?;
= 0,05 m o l .
3Cu
— 3njyjQ + 1. n[sjQ2 = 0,7 m o l .
(muoi) = n„ trao doi = 0,7 m o l .
+ 4H2O
[ =
nHN03 + 2nH2S04 = 0 4 + 2.0,01 = 0,12 m o l , ...^
Trong X: \ ^^J.
^ Q^Q^ ^^^^
.i.,££.Kl,0 = V «
NO3
SO|
^Zn*2
D e n day, cac e m can l u u y la k h i cho k i m loai tac d u n g v o i HNOs, so m o l goc
n
0,1
H* t r o n g d u n g d j c h la d o 2 axit p h a n l i r a
("f), tf)
k h i nen can ap d u n g d i n h luat bao toan electron.
> Fe«;
->
0,4
> 3Cu2* + 2 N O
3 2
nr^, =
f;
Nhan xet: D a y la bai toan c6 nhieu chat k h u (3 k i m loai) va tgo ra 2 san p h a m
> Cu"2. p e - 3 e
->
0,15
+ 2NO;
Dap anB.
Mol:
,
Cu-2e
8H*
oii^d.'
ai
Phuang trinh ion riit ggn: t
Cdc em can lieu y: Cu(OH)2 tan t r o n g d u n g d i c h N H s d u .
Chatkhu:
+
->
N/jflM xet: K h i axit HNO3 c6 mat d o n g t h a i v o i axit H2SO4 loang t h i lug-ng
'^^
' ^^• ^
•''•^
BailS:
0,4
_> K h o i l u p n g m u o i t r o n g Y = 9,75 + 9,6 + 0,2.96 + 0,2.62 = 50,95 gam.
3nAi = 2 nH2 ^ n ^ i = 0,1 m o l ; 2ncu = nN02 ^ ncu = 0,15 m o l . '
D a p an B.
^
,p ^ - 3
i:;n< nin^''
' ,
"yi.H~'. /rai J I C L . '..i/iih
^,
.
+ 4H2O,
>2>Zn^* + 2 N O
H"^ t h a m gia p h a n u n g het. '
' "
m = 0,1.27 + 0,15.64 = 12,3 g a m
0,15
3Cu
= 0,15 m o l .
m-Uoi +
Cach 2: A p d u n g bao toan electron:
+
_ = 0,4 m o l ; n^ o_ = 0,2 m o l .
NO3
SO4
••'..'Hi
Cach 1 : Giai theo p h u o n g t r i n h
2A1 + 6HC1
Trong X: j
= 0,S m o l
^,
8H* + 2 N O ;
0,15
<-
Cu
+ 2Fe3*
IQ <, ;
)• 3Cu2^ + 2 N O + 4H2O
i V j / ' V
0,4
> Cu^*
+ 2Fe2*
M o l : 0,11 <-0,22
m c u = 0,26.64 = 16,64 g a m - > DapanB.^^^
^,.rid
,
ao:u.,.j, i:;^! ::M;;t:
'
^
.•''i^*''.^'^MM'^^' - '.''^'^
o-iH'W]JM-!<>.:/
97
Cty TNHH M T V DVVH Khang Vijt
Cim nang On luyQn thi dai hgc 18 chuySn dg H6a hgc - Mydygn van Hi\
Bai 19:
N/ifln A:e^: K h i X t a c d u n g v o i m o i c h a t o x i h o a HNO3 v a C h , cac k i m l o a i d e u
T i X H CHAT CUA CAC HIttROXIT
duQic d u a l e n m u c o x i h o a c a o n h a t —> So m o l e l e c t r o n t r a o d o i t r o n g hai
^ HIDROXIT KIM LOAI KIEM, KIEM THO
t r u o n g h g p la b a n g n h a u .
Chat o x i h o a :
N^s + 3e
Chat o x i h o a :
C h + 2e
N*^ + i e • — - > NOz.
^ NO;
+
•
- > D a p an A .
'^^^^^ •
Bai20:ncu = a06mol.
CO2
^
,
d u n g d i c h . K h i do, l u g n g
'^^
Neu:
trong
^
>• 3Cu2* + 2 N O + 4 H 2 O
+ 8H^ + 2 N O ;
->
- > V = 0,04.22,4.1000 = 896ml ^
+
. - ,v
•> ^ i - o u =
'
T a c o : n „ + = 0 , 2 . 2 + 0,1.0,8 = 0,48 m o l ; n , , ^ _ = 0,2.2 = 0,4 m o l .
H
NO3
a
••• T ( : 3 n
> 3Cu2^ + 2 N 0 + 4 H 2 O
,
A
U ^
+ 2NaOH
8
ji^f^
2NaCl
CO2
+ 2H2O
+ H2O
> NaA102
,
^
+ 2H2O
.JBnocfofe) ;o:
/ . X ) :(«:>* r i m i i
^ ' > 2 N a O H + C h t + HzT
^^
>'
tn r..... .x^ry •\r,m<'f-.0 -
> CaCOsi
+ Ca(OH)2
Neu:
j^Y/bib^-,:
,I'-VT; '•('..!}
'•
*
= 2nca(OH)2
= a
: Ca(HC03)2
^>-n^ jiSt-S) t
1< a < 2 : CaCOs + Ca(HC03)2
a
>2
y.;
+H2O
> Ca(HC03)2
nco2
,()r/f?Py 4. ^r\>=,'4
.^^^j,
^OD"
> 2NaA102
a < 1
8
D a p an D .
, *
' M
' ! - !
^ = 0,18 m o l - > m = 0,18.64 = 11,52 g a m
:Na2C03
D i e u che
T i n h so m o l : n
3 0 48
=—
>2
> Fe(OH)3 i + 3 N a N 0 3
Hidroxit kim loai kiem tho -
n e n t i n h so m o l C u t h e o H ^
.
:
rt-'.
: NaHC03
Tac d u n g v o i h g p chat l u a n g t i n h
2 C O 2 + Ca(OH)2
Ta c6: ncu
j rififi^J rmt fcrb rr!t;;<
1< a < 2 : N a H C O s + NazCOg
Al(OH)3 + N a O H
• ;X gncn
N h | n xet: Theo b a i , ta coi C u t a n v u a d u t r o n g h o n h g p HNO3 v a H C l . *
<
;v
Tac d u n g v o i d u n g d i c h m u o i :
AI2O3
1
0,04
D a p an A .
+
+
Nhan thay ^
^
Fe(N03)3 + 3 N a O H
Bai 21:
3 C u + 8H* + 2N03-
= a
nco2
la d o 2 a x i t p h a n l i ra nen c a n g i a i t h e o p h u o n g
0,06 < - 0 , 1 6 ^ 0,04
; ,
>K
o»U.; Mt^O ^ i O . ^ ' . u o g ;< r r .
> NaHCOs
•
a < 1
Ta c6: n ^ + = nHNOg + 2nH2S04 = 0,U m o l ; n^^_ = n^NO^ = 0,08 m o l .
Mol:
>Na2C03 + H 2 O
+ NaOH
>
udl
uimrDi
trinh ion.
3Cu
^ ''•'.yt
'
.
'
N^flM xef: Svr c6 m a t a x i t H2SO4 l o a n g se " d o n g g o p " t h e m l u ^ n g
i.
•'••^'•^g.^jjj
Tac d u n g v o l o x i t axit
C02 + 2 N a O H
= mx + m Q j = 12,45 + 0,35.71 = 37,3 g a m .
'
' ' ; j;,,^,
Hidroxit kim loai kiem
> 2C\-
mmu6i
^^^^
^^^
j,.Uthuyet
Bao t o a n e l e c t r o n : 3 n N o + n N 0 2 = 2 n c i 2 ^ 2 n c i 2 = 0,2.3 + 0,1 = 0,7 m o l
- ) • nci2 = 0,35 m o l - >
^ ^
M£t,0 ^ ^ K ' ,
:CaC03
Bai 22:
b . V i d v mau
„
,
42.0,375 „
,
Ta co: nHNOg = — — — = 0,25 m o l .
•'«
V i dvi 1: Cho cac chat: N a H C 0 3 , CO, A l ( O H ) 3 , Fe(OH)3, H F , CI2, N H 4 C I . So chat
DO
FeS2 + I 8 H N O 3
Mol:
> Fe(N03)3 + 15N02 + 2H2SO4 + 7 H 2 O
0 , 0 1 ^ 0,18
0,01
D u n g d i c h sau p h a n u n g gom:
Fe(N03)3 = 0,01 m o l ; HNa
7
'
'
'
08
,
,
B.5.
' ^'-^^
'1.0
C.3.
Lbi
dm.
, :
D.6.
giai:
Nhan xet: Day la cau hoi tong hqip, lien quan den k i e n thuc l o p 10 (HF, CI2),
l o p 11 (CO, N H 4 C I ) va l o p 12 (NaHC03, Al(OH)3, Fe(OH)3).
" N a O H ^ 3"Fe(N03)3+r>HN03 + 2 n H 2 S 0 4 = 0,14 m o l
= 0,07 l i t = 70ml
'
= 0,07 m o l ; H2SO4 = 0,02 m o l .
tac dung dugc voi dung dich N a O H loang 6 dieu kif n thuong Ik
A . 4.
0,02
De t h u d u g c k e t t i i a I o n nha't:
VNaOH
^
D a p an B.
Cac chalt xay ra phan ung:
NaHC03 + NaOH
•,
> Na2C03 + H 2 O
.^^j^j^ ^ i , . v ( ...^