NGUYEN TUYENH A
CAC DANG fillN HiNH VA
PHJdNG PHAP GIAI NHANH BAI TAP TRAC NGH||M
I
Ji
1
BIEN SOAN THEO CHUDNG TRINH MCJI
DANH CHO HQC SINH BAN CO BAN VA NANG CAO
ON LUYEN THI TU TAI, DAI HQC VA CAO DANG
rH(i V!EN TINHBiNH THU-V^
NHA XUAT BAN DAI HQC QUOC GIA TP. H OCHI MINH
CAC
DANG
O I E N HINH VA
BAI
PHl/dNG
P H A P GIAI
NHANH
TAP TRAC NGHIEM
L C l I N(&I D A U
H6AHOCI2
Nguyen T u y e n H a
N H A
X U A T
QUdc
DAI H O C
De h6 trg viec hoc tap va on t h i T U T A I - D A I HOC - CAO D A N G
hang nam, toi viet cuon sach n^y theo tinh than giiip cac em trang bi
du kien thiJc de tham diT k i t h i dat ket qua tot nhat.
B A N
G I A TP H O
C H I M I N H
Khu pho 6, phirdng Linh Trung, quan Thu Dutc, TP.HCM
So 3 Cong tradng Quoc te, quan 3, TP ROM
D T : 38 239 172, 38 239 170
F a x : 38 239 172 - E m a i l :
Chiu track nhiem xudt bdn
TS H U Y N H BA L A N
T6 chiic bdn thao vd chiu track nkiem vi tac
quyin
DOAN V A N K H A N U
Bien tap
NGUYEN T H I NGOC H A N
S^a bdn in
T H A N T H I HONG
+ Phan I I : Cac phiicfng phap giai bai tap trac nghiem
+ Phan I I I : Gidi thieu 05 de t h i thuf Tu t a i va 06 de t h i thuf Dai hoc
de cac em thijf siJc minh.
Cac de t h i thuf mang tinh he thong day du cau hoi l i thuyet va hki
tap. Sau do cac em tham khao ddp an de rut kinh nghiem trong k i t h i
sSp tdi nham dat ket qua nhuf mong doi. Chiing toi h i vong cuon sdch
nay la tai lieu tham khao hCJu ich giup cac em ren luyen k i nSng,
nang cao kien thiJc cua minh va dat ket qua tot nhat trong k i t h i
T U T A I - D A I HOC - CAO DANG s^p t d i . Mac du rat co g^ng de bign
soan nhuhg vdi thcfi gian c6 han chac ch^n van cbn nhufng khiem khuyet.
Tac gia rat mong nhan dacfc sir gop y xay diTng cua dong nghi?p, doc gid
gan xa de nhufng Ian sau tai ban se duac hokn thien hdn.
Tac gid
DIEM KHANH
Nguyen Tuyen Hd
TK.02.H(V)
494.2013/CXB/07-25
DHQG.HCM-13
TK.H.453-13(T)
In 1.000 cuon, khd 16 x 24cm.
S6' dang ky ke hogch xuat bin: 494-2013/CXB/07-25/DHQGTPHCM.
Quyet dinh xuSt bhn so: 333/QD-DHQGTPHCM ngay 21 thang 6 nam 2013 cua Nha
xuat bdn DHQGTPHCM.
In t?i Cong ty In Song Nguy6n, n^p lau c h i l u quy I V nSm
+ Phan I : Cac dang bai tap dp dung cong thiJc tinh nhanh
Xin tran trong cam cfn.
Trink bay bia
.
Ve noi dung cuon sach cd 3 phan:
2013.
T
CAC D4NG BAI
A P D^JNG
CACH TiNH NHANH
C a c h tinh nhanh so dong phan ciia:
- Ancol no, dctn chiiCc
- Andehit
dcfn chvCc, no
- Axit cacboxylic
(CnH2„0):
dcfn chiic, no
- Este no, dcfn chtic
- Ete darn chiic, no
(C„H2n02):
(1<
n <
2n-3
(2<
n < 7)
2n-3
(2<
n < 7)
2n-2
(C„H2„02):
(C„H2n+'20):
- Xeton dam chiic, no
- Amin dctn chiic, no
2n-2
(C„H2n+20):
(C„H2„0):
(1
n <
5)
(n-l)(n-2)
2
(2<
n < 6)
(n-2)(n-3)
2
(2<
n < 7)
(n<
5)
2n-l
(CnHzn+sN):
<
6)
Dang 1: KIM LOAI TAG DUNG VOl AXIT
A. H C I , H2SO4 ( l o a n g )
Doi vdi hat loai axit tren thi chi phdn
ling voi nhUng kim loai
diing
trade H trong day hoat d6ng hoa hoc.
K
Ca
Na
Hg
Ag
Au
Mg
Al
Zn
Cr
Fe
Ni
Sn
Pb
Phucfng trinh tong quat: (Vdi M la kim loai)
2M + nH2S04 (loang) -> M 2 ( S 0 4 ) n + nHzl
ta luon c6
2M + 2nHCl ^ 2MC1„+ nHg
n„. = 2. n^^
J
De tinh khoi luong muoi thu diTcfc thi
•k Dung dich H2SO4: m^^^i
sunfat =
nihSn h(7p
kim loai
+ 96 n,i^
* Dung dich HCI: nimudi dorua = nihSn hgp kim loai + 71 nj,^
H
Cu
BAI
TAP
AP
DUNG
B a i 1: Cho l,04g hon hop hai k i m loai tan hoan toan trong dung dich
H2SO4 loang du thodt ra 0,672 Ht k h i H 2 (dktc). Khoi luong hon hop
muoi sunfat khan thu di/oc la:
A. 3,92g
B. l,96g
C.3,52g
D.5,88g
HUdng
2 kim loai +
H2SO4I
dan
gidi
hh muoi sunfat +
B a i 4: Bok tan 1,19 gam hSn hop A gom A l , Zn b&ng dung dich HCl v i T a
du thu difoc dung dich X vk V l i t k h i Y (dktc). Co can dung dich X
di/oc 4,03 gam muoi khan. Gia t r i cua V 1^.
,
A. 0,224 lit.
B. 0,448 lit.
Ap dung cong thiJc: m^ua-i ciorua
H2
4,03-1,19
+ 96 n„^
^
= 1,04 + 0,03.96 = 3,92 (gam)
Chon dap an A.
B a i 2: Hoa tan het 3,53 gam hon hop A gom ba kim loai Mg, A l va Fe
trong dung dich HCl, c6 2,352 l i t k h i hidro thoat ra (dktc) va thu daoc
dung dich D. Co c a n dung dich D, thu difdc m gam h6n hop muoi
k h a n . T r i so ciia m la:
B. 10,985 gam
C. 11,195 gam
Hitdng dan gidi
nn^so, = 0,25.0,5 = 0,125 mol =^ n^,, = 0,125 . 2 = 0,25 mol
n„. = 0,25 mol
=> tong so mol H* = 0,25 + 0,25 = 0,5 mol
thuTc: m m u e i doma = nihSn iwp kim loai
q
+ 7 1 n„^
Chon ddp 6n B.
B a i 3: Hoa t a n hoan to^n 2,17 g a m h o n hap 3 k i m loai A, B, C trong
dung dich HCl da thu diTOc 2,24 l i t k h i H 2 (dktc) va m g a m muoi. Gia
t r i cua m 1^.
B. 5,72 gam.
C. 6,85 gam.
D. 6,48 gam.
Hiidng dan gidi
m„us-iciorua = nihSn h;>p kim lo^i + 71 n,,^ = 2,17
n„,
no
=1^
= 0,175 (mol)
Ta c6:
2H^ ^
Ban dau:
0,5 mol
PU:
0,35 <- 0,175 mol
Sau pif:
0,15 mol
H2
Vi axit dir nen kho'i \\iang rin b^ng tong kho'i lifcfng kim loai va cdc
ion C O trong dung dich.
mr^nkhan
n„ =?^
= 0,l (mol)
22,4
Chon dap an A
= 0,25.1 = 0,25 mol
HHCI
= 3,53 + 0,105 . 71= 10,985 gam
A. 9,27 gam.
r^r^A ^ ^^
V„^ = 0,04 . 22,4 = 0,896 l i t
D. 7,2575 gam
= 0,105(mol)
Ap dung cong
+ 71 n^^
Chon dap an C.
B a i 5: Cho 5,35 gam h6n hop X gom Mg, Fe, A l vao 250ml dung dich Y
gom H 2 S O 4 0,5M va HCl I M thu duoc 3,921it k h i H 2 (dktc) va dung
dich A. Co can dung dich A trong dieu kien khong c6 khong k h i , thu
diTOc m gam chat rSn khan. Gia t r i cua m la
A. 20,900 gam.
B. 26,225 gam.
C. 26,375 gam. D. 28,600 gam.
HU&ng d&n gidi
n H . - | ^
= nihSn hap kim loai
nji^ = ^ ^ ^ p — = 0,04 ( m o l )
C O n g thufC: m m u t f i sunfat = mhSn hap kim lo?i
A. 12,405 gam
D. 1,792 l i t .
Hii&ng dan gidi
n „ . - - | | f = 0,03(mol)
Ap dung
C. 0,896 l i t .
= 5,35 + (0,125.96) + (0,25.35,5) = 26,225 gam
Chon dap an B
+ 0,1
. 71
= 9,27
gam
B a i 6: (DH khoi A.2010): Cho m gam hon hap bot X gom ba k i m loai
Zn, Cr, Sn c6 so mol b^ng nhau tac dung het vdi lirotng diT dung dich
H C l loang, nong t h u dugc dung dich Y
k h i Ha- Co can dung dich
Y t h u dtfdc 8,98 gam muoi k h a n . Neu cho m gam h o n hop X tac
dung hoan toan v d i O2 (du) de tao hon hop 3 oxit t h i the t i c h k h i
(dktc) p h a n ijrng la
A . 2,016 l i t .
B. 0,672 l i t .
C. 1,344 Ht.
Hii&ng dan
D. 1,008 l i t
= 0,4(mol)
" H , =|g
27x + 56y =11
Ta lap va g i a i he phuong t r i n h
- X
gidi
Fe
Chon dap a n B
- Con k h i tac dung O2, Zn tao +2, Cr tao +3, Sn tao +4. (X + O2 t h u
B a i 8: (De KA.2007):
V i cac k i m loai c6 so mol bSng nhau nen cAc muoi c6 so m o l b^ng
A . 1.
B. 6.
C. 2.
Hiidng
136x + 123x + 190x = 8,98 =>x = 0,02;
=> t d n g so m o l H^ = 0,25 + 0,25 = 0,5 m o l
5,32
22,4
Chon dap an D
^
nor,,-/ —
2H*
t h u dirge 8,96 l i t H2 (dktc). P h a n t r a m k h o i lugng cua Fe t r o n g h o n
Ban dau:
0,5 m o l
hop l a .
A. 49,09%.
Pu:
0,475
Sau pir:
0,025mol
B. 50,91%.
Hiiofng dan
C. 40,91%.
D. 59,09%.
gidi
Liiu y: K i m loai tac dung v d i m o t chat nao do ma sinh ra
le mol giufa k i m loai va H2 luon b^ng hod tri kim log,i chia 2
y mol
->
- H2
2
—>
— X
-4
H2
mol
y mol
thi ti
. V
„
= 0,2375(mol)
T a c6:
B a i 7: Cho 11 gam h o n hop A l va Fe tac dung h e t v6i dung dich H C l
Fe
gidi
nHci = 0,25 . 1 = 0,25 m o l => n „ . = 0,25 m o l
=> Vo^ = 0,045.22,4 = 1,008 l i t
2
dan
D . 7.
n„^o, = 0.25 . 0,5 = 0,125 m o l => n,j. = 0,125 . 2 = 0,25 m o l
3
"oj = 2 ' ' ^ 4 ^ ^ ' ' ' ^ 2,25x = 0,045 m o l
mol
0,1
Cho m gam hon hop M g , A l vao 250ml dung dich
nhau va bang x
X
y =
dung dich Y. Coi the tich dung dich khong doi. Dung dich Y c6 p H 1^
- Goi so m o l m o i k i m loai la x (mol) t h i :
Al
0,2
chufa hon hop H C l I M va H2SO4 0,5M t h u dugc 5,32 l i t k h i H2 (dktc) v^
daoc cac oxit: ZnO, CraOa, Sn02)
So do
+ y = 0,4
X =
0,1.56 X 100% = 5 0 , 9 1 %
11
=
3 k i m loai t r e n k h i p h a n i l n g v d i H C l loang nong deu h i oxi hoa
t h ^ n h so oxi hod +2. (X + H C l t h u di/Oc muoi: ZnClz, CrClz, SnClg.)
1
3
.
H2
<-
Sau p h a n iJng a x i t diT H *
0,2375 m o l
I
0,025
0,25
= 0,1M
p H = - l o g [H^^ =1
Chon dap a n A
B a i 9: (De KB.2007): Cho 1,67 gam h o n hop 2 k i m l o a i d 2 chu k y ke
tiep nhau thuoc n h o m I I A tac dung het v d i dung dich H C l d i i , t h o a t
r a 0,672 l i t k h i H2 (dktc). H a i k i m loai do l a (Be = 9,Ca = 4 0 , M g = 24,
^ Sr = 87,Ba = 137)
A . M g va Ca.
B. Ca va Sr.
C. Sr va Ba.
D. Be va M g .
Hiictng dan
gidi
B. KIM LOAI T A G D y N G VC3l HNO3, H2SO4 (d|c)
D a t 2 k i m loai 1^ A
A x i t H2SO4 (dSc) + k i m loai = muoi sunfat + san p h a m khuf + H2O
T i le mol gifla k i m loai va h i d r o b^ng ho^ t r i k i m l o a i chia 2 n 6 n t i le
San phdm
m o l la 1: 1
*
= MZ2^ 0,03
22,4
(mol)
Ha
0,03mol
0,03mol
^
= 55,67 ^
Ca va Sr.
H C l I M va H2SO4 0,5M t h u dMc
dung dich B va 4,368 Ht
H2(dktc). P h a n t r S m k h o i laong M g va A l t r o n g X tucfng ufng la
A . 3 7 , 2 1 % M g va 62,79% A l .
B. 62,79% M g va 3 7 , 2 1 % A l .
C. 45,24% M g va 54,76% A l .
D . 54,76% M g va 45,24% A l .
Bai
khd c6 the la NO2 (nhan
muoi nitrat =
BAI TAP VAN
B ^ i nay l a m tifcfng t i f bai
Bai
lOe), NH4NO3 (nhan
H I k i m loai
3e), N2O
(nhan
8e).
+ molgpk.so e n h a n . 6 2
DUNG
1: Cho 1,86 gam h o n hdp A l va M g tac dung v d i dung dich HNO3
n h a t ) bay r a . K h o i liTOng muoi n i t r a t tao r a t r o n g dung dich l a :
A. 40,5 gam.
B. 14,62 gam.
C. 24,16 gam.
HtCclng dan
D. 14,26 gam.
gidi
n^o = ^ ^ - 0 , 0 2 5 m o l
N,o
22,4
3,65M (d = l , 1 9 g / m l ) t h u dugc m o t chat k h i va 1250g dung dich D.
A p dung cong thuTc: m musl nitrat = m kim loai + molspk-so e n h a n . 6 2
= 1,86 + 0,025.8.62 = 14,26 gam
V a y m c6 gia t r i :
B. 61,63 (g)
HUdng
C. 63,65 (g)
dan
D. 63,61 (g)
Chon dap a n D
Bai
gidi
m,,„c, = 1000.1,19 = 1190 ( g ) ; n „ c , = 3,65.1 = 3,65
2: Hoa t a n hoan toan 9,94 gam h o n hop X gom A l , Fe, Cu t r o n g
lugng du dung dich HNO3 t h u dugc 3,584 l i t k h i N O duy n h a t (dktc).
(mol)
T o n g k h o i lugng muoi k h a n tao t h a n h l a :
A. 39,7gam
Zn + 2HC1 ^ ZnCl2 + H2
B. 29,7gam
A p dung D L B T K L , t a c6: nihh.zn.Fe) + "^ddiici = "^udo + " ^ n ,
= 1250 + 2 ( ^ )
C. 39,3gam
Hiidng
Fe + 2HC1 ^ FeCl2 + H2
Chon dap a n C.
le), NO (nhan
loang dtf t h i t h u di/cJc 560 m l l i t k h i N2O (dktc, san p h a m khuf duy
5
1 1 : Hoa t a n m(g) h 6 n hop Zn va Fe can vCra du 11 dung dich H C l
A . 65,63 (g)
n h a n . i .96
Luu y: A l , Fe, Cr khong phan ting vcd HNO3 dSc nguoi va H2SO4 dac nguoi.
10: Cho 3,87gam h 6 n hop X g6m M g va A l v^o 250ml dung dich X
gom
phdm
8e), N2 (nhan
Chon dap a n B
Bai
6
6e)
a x i t HNO3 + k i m loai = muoi n i t r a t + san p h a m khijf + H2O
* m
MA =
"1 kin, loai + molgpk.so
m m u o i sunfat =
San
A
khii c6 thi Id SO2 (nhan 2e), H2S (nhan 8e),S (nhan
- 1190 ^ 63,65(g)
dan
D. 27,7gam
gidi
n„„=||M.o,16n.o.
A p dung cong thUc:
mmuoi
nitrat = m
k i m loai
+ molspk.so e n h a n . 6 2
= 9,94 + 0,16.3.62 = 39,7 gam
Chon dap a n A
B a i 3: Hoa tan 4,97 gam hon hap A l , Cu, Fe trong dung dich HNOr,
loang dii thu dUcfc 1,792 Ut khi NO(dktc). Tong khoi luong muoi khan
tao thanh:
A. 19,85 gam
B. 26,5 gam
C. 39,7 gam
D. 40,2 gam
Bai nay gidi tuang tii bai 2.
B a i 6: (BH khoi B.2008): Cho 2,16 gam M g tAc dung vdi dung dich
HNO3 (du). Sau khi cac phan iJng xay r a hoan toan, thu duoc 0,896 lit
NO (d dktc) va dung dich X. Khoi luong muoi khan thu diTOc khi lam
bay hoi dung dich X la
A. 8,88 gam
B.13,92 gam
C. 6,52 gam
D.13,32 gam
Hii&ng dan gidi
B a i 4: Hoa tan hoan toan 13,68 gam hon hdp X gom A l , Cu, Fe bSng
dung dich HNO3 loang, diX thu diroc 1,568 lit khi N2O (dktc) va dung
n^g = 246 : 24 = 0,09 (moDin^o - 0,896 : 22,4 = 0,04 (mol)
dich chuTa m gam muoi. Gia tri cua m la.
Cdch 1:
A. 48,40 gam.
B. 31,04 gam.
C. 57,08 gam.
D. 62,70 gam.
3Mg + 8HNO3 ^ 3Mg(N03)2 + 2 N 0 + 4H2O
Bai nay gidi tuang tu bai 2.
0,06
B a i 5: (DH khoi A.2010): Nung 2,23 gam hon hop X gom cac kim loai
<<
0,06 <
0,04
4Mg + IOHNO3 > 4Mg(N03)2 + NH4NO3 + 3H2O
0,03
> 0,0075
0,03
>
Fe, A l , Zn, M g trong oxi, sau mot thdi gian thu difdc 2,71 gam hon
hop Y. Hoa tan hoan toan Y vao dung dich HNO3 (da), thu dtfOc 0,672
Khoi liTcfng muoi khan thu diioc khi lam bay hoi dung dich X \k
Ht khi NO (san pham khuf duy nhat, d dktc). So mol HNO3 da phan
m = 0,09. 148 + 0,0075.80 = 13,92 (gam) ^ chon B.
Cdch 2: Diia vao dinh luat bao toan electron
ufng la
A. 0,12.
B. 0,14.
C. 0,16.
D. 0,18.
HiCdng ddn gidi
Mg
2e
0,09
2.0,09
-
Mg^2
N
+5
N
+5
0,09
3e
Tom t^t:
Fe, A l , Zn, Mg + O2 -> hh ; hh + HNO3
'
'
-> muoi +
NO
8e
N - ' NH4NO3)
^ U N O j (pu O trong oxit)
2.0,09 = 3.0,04 + 8x -> X = 0,0075 (mol)
= 0,03 (mol)
B a i 7: Hoa tan 23,4 gam G gom A l , Fe, Cu bang mot lirong vCra du dung
dich H2SO4 d5c, nong, thu di/dc 15,12 lit khi SO2 (dktc) va dung dich
^ H N O j (pu oxi ho4 - khut)
^NO
A. 153,0 gam.
B. 95,8 gam.
C. 88,2 gam.
Hii&ng ddn gidi
Mat khac:
HHNO^ (pu o t ^ n g oxiu =
( cho
n H N o , ,pu ox, ho. - uha, = 3.n^o
( Do
H^O);
N^^ —
+ 3nf^o+ n^o =
i^so,
^
=
15,12
22,4
= 0,675mol
NO)
nimuol simfat =
^T^o(<«iM+
= 2.0,03 + 4.0,03 = 0,18 mol
Chon dap a n D
X
chuTa m gam muoi. Gia tri cua m la.
Ta CO bao toan nguyen to N:
^ ^ ^ i i N o , = 2no,oxit)
-
m = 0,09. 148 + 0,0075. 80 = 13,92 (gam) -> chon B
mo = m , , - m„ = 2,71 - 2,23 = 0,48(g) -> n^ - ^
'^HNOj ~
0,04
Khoi lirong muoi khan thu dtfoc khi lam bay hoi dung dich X la
Bao toan khoi liTOng:
^
3.0,04
8.x
So mol electron dLfOc bao to^n
0.672 (l)«5|^=0,03(mol)
^ ' l a ^
N^2(N0)
-
HI kjn, loai
'^^m
+ molgpk.SO
6
nhan. — .96
2
= 23,4 + 0,675.2. \6 = 88,2 gam
Chon dap an C.
|
D. 75,8 gam.
3Cu + 8H* + 2NO3"
B a i 8: Kok t a n h o ^ n t o ^ n m gam h o n X gom A l , Fe, Cu vko dung dich
H N O 3 d S c nong du, t h u duoc dung dich Y chufa 39,99 gam m u o i va
7,168 Ht k h i N O 2 (dktc). Gia t r i cua m la.
A. 20,15 gam.
B. 30,07 gam.
C. 32,28 gam.
Hiicfng ddn
Dau b a i :
P h a n iJng: 0,06 ^
D . 19,84 gam.
m
=^ m
0,16
0,08
0,16 ^
gidi
B a i 10: (DH
khoi
B-2009):
raM
nitrat
k i m loai
= m
= m
+ molspk.so e nhan.62
k i m loai
" molspk-SO
muoi nitrat
G
nhan.62
gam
A. 0,03 va 0,01
B. 0,06 va 0,02
C. 0,03 va 0,02
D . 0,06 va 0,01
Hiicfng ddn
Au
1. Cho 3,84 gam Cu p h a n iJng v d i 80 m l dung dich H N O 3 I M t h o d t r a
->
0,02
M hoa t r i 2 vifa du vao dung dich c h d a H N O 3 va H2SO4 va dun n o n g ,
thu dtroc 2,94 g a m h o n hap k h i B gom NO2 va SO2. The t i c h cua h 5 n
Quan he giOfa V i va V 2 la
B. V 2 = 2 V i .
C. V 2 = 2 , 5 V i .
Hiictng dan
0,08
0,08 ->
l i t (dktc). K h o i l u a n g m u o i k h a n t h u diroc l a .
A. 6,36g.
B. 7,06g.
HUdng
n„, = 0 , 0 8 m o l
,
3Cu + 8 H " + 2NO3-
D. V2 = l,5Vi.
hop k h i B l a 1,344
gidi
= 0,06 m o l
=0.08 mol
0,06
> A U C I 3 + N O + 2H2O
B a i 11: Hoa t a n 3 gam h o n hgfp A gom k i m loai R hoa t r i 1 va k i m l o a i
0,5 M t h o a t r a V 2 Ht N O . B i e t N O la san p h a m khijf duy
n h a t , cac the t i c h k h i do d cCing dieu k i $ n .
Dau b a i :
3HC1 + H N O 3
Chon dap a n B
2. Cho 3,84 gam Cu p h a n ling v d i 80 m l dung dich chufa H N O 3 I M va
nHN03
+
0,02 -> 0,06
Vi lit NO.
n c „ = ^
b4
gidi
• Ni/dc CLfdng toan l a t i le 3 : 1 giufa H C l va H N O 3
ThUc h i e n h a i t h i n g h i e m :
TNI:
K h i h o a t a n hoan toan 0,02 m o l A u b a n g
ntfdc cudng toan t h i so m o l H C l p h a n i l n g va so m o l N O (san p h a m
2007)
A. V 2 = V i .
0,04 m o l
khuf duy n h a t ) tao t h a n h I a n liToft l a
Chon dap a n A
H2SO4
->
Nhir vay V 2 = 2 V i . Chon dap a n B.
= 39,99 - 0,32.1.62 = 20,15
B a i 9: (DHKB.
0,04
-> Cu va H"^ p h a n iJng h e t
=> V 2 ti/ong urng v d i 0,04 m o l N O .
H M O = ^ ^ ^ = 0,32mol
"""^
22,4
A p dung cong thiJc:
0,06
>3Cu^* + 2NO^ + 4 H 2 O
^ 1^344 ^
22,4
n^^_ = 0,08 m o l
x + y = 0,06
p h a n iJng h e t
P h a n iJng: 0,03 <- 0,08 -> 0,02 ^
0,02 m o l
=> V i tirang ufng v d i 0,02 m o l N O .
TN2:
ncu = 0,06 m o l ; nn^o, = 0,08 m o l ; nn^so. = 0,04 m o l .
=> T o n g
n^^, = 0,16 m o l ; n ^ ^ . = 0,08 m o l .
•
dan
D . 12,26g.
gidi
oGmol dSt NO2 x m o l , SO2 y m o l
r46x + 64y = 2,94
> 3Cu^^ + 2 N 0 T + 4 H 2 O
C. 10,56g.
f x = 0,05
[ y = 0,01
nimuai = 3 + (0,05 .62 + 0 , 0 1 . - . 2 .96) = 7,06 gam
2
Dap a n B.
B a i 12: Cho 8,3 gam h 5 n hcfp A l va Fe tac dung v d i dung dich H N O 3
loang dtr t h i t h u dircfc 45,5 gam m u o i n i t r a t k h a n . The t i c h k h i N O
(dktc, san p h a m khuf duy n h a t ) t h o a t r a l a :
! A. 4,48 l i t .
B . 6,72 l i t .
C. 2,24 l i t .
D . 3,36 l i t .
Qua t r i n h khii:
Hiicmg dan giai
Ap dung cong thiJc:
^ DNO =
=>
m musi
nitrat
= ni
"-^-"^ ~
so e nhan.62
,0,1
=
0,3
+ molspk-so e nhan.62
N*^ + l e
= 0,2 mol
3.62
0,1
Chon dap an A
B a i 13: Hoa tan hoan toan 1,23 gam hon hop X gom Cu va A l vao dung
dich HNO3 dSc, nong thu dixoc 1,344 l i t k h i NO2 (san pham khuf duy
nhat, d dktc). Phan trSm ve khoi liiong cua Cu trong hon hop X la
C. 68,05%.
D. 29,15%.
Hiidng ddn giai
n NO,
. Cu^^ + 2e
2x
X
Al •
y
Ta c6:
N""
0,2
S"^ + 2e
S*^
0,2
0,1
0,1
24x + 27y = 15
fx = 0,4 mol
2x + 3y = l , 4
[y = 0,2 mol
97 n 9
.%A1 = ^ ^ ^ . 1 0 0 % = 36%.
15
%Mg = 100% - 36% = 64%.
Bai 15: Cho 2,8 gam hon hop bot kim loai bac va dong tac dung v6i dung
Thanh phan phan tram cua bac va dong trong h6n hop Ian liTOt la:
A. 73% ; 27%.
B. 77,14% ; 22,86%
N"^ + 3e ->
C. 50%; 50%.
D. 44% ; 56%
0,18
Tuang tu bai 14.
0,06
Al'^ + 3e
3y
64x + 27y = l,23
|2x + 3 y - 0 , 0 6
Dang 2: OXIT KIM LOAI TAG DUNG VOI AXIT HQ, H2SO4 [bang)
fx = 0,015
Phi/dng trinh tdng quat
[ y = 0,01
MzOn + 2nHCl
%Cu=M15:6i.78,05%
1,23
M20„ + nH2S04
Chon dap an B
2MCln+ nHzO
(loang) ^
dich Y gom H N O 3 va H2SO4 dSc thu diioc 0,1 mol moi k h i SO2, NO,
NO2, N2O. Phan tr5m khoi Itfong cua A l va Mg trong X Ian lacft la
* Doi vdi axit H2SO4 (loang)
Khoi li/ong muoi thu dugc la:
A. 63% va 37%.
B. 36% va 64%.
Doi vdi a x i t H C l
C. 50% va 50%.
D. 46% va 54%.
Khoi luong muoi thu duoc la:
HU&ng dan gidi
HMg = X
mol;
M2(S04)„ + nHzO
Cach tmh nhanh cho trie nghi?m
B a i 14: Hoa tan 15 gam hon hop X gom hai k i m loai Mg va A l vao dung
Dat
0,8
2N*^
dich HNO3 dSc, dtr t h i thu duoc 0,896 l i t khi NO2 duy nhat (d dktc).
= M M = 0,06mol (Dat Cu X mol, A l y mol)
22,4
Cu
0,1
Theo dinh luat bao toan electron:
B. 78,05%.
2N*^ + 2 x 4e
=> Tdng so mol e nhan bkng 1,4 mol.
= 0,2.22,4 = 4,48 l i t
A. 21,95%.
+ 3e -> N*=^
DAI
x
2x
=> Tong so mol e nhudng b i n g (2x + 3y).
+
n,[^so,
-80
•
m,„„a-i ciorua = n i h o n hop oxit kim loai
+ 27,5 n,,^;,
BAl TAP VAN DgNG
= y mol. Ta c6:
Qud t r i n h oxi ho^: Mg -> Mg^* + 2e
m^^ai ^unfat = m h S n hap o x u k i m loai
Al
y
A l ' " + 3e
3y
B a i 1: Cho 50g hon hop hot oxit kim loai gom ZnO, FeO, Fe203, Fe304,
MgO tac dung het vdi 200ml dung dich H C l 4 M (vCra du) thu duoc
dung dich X. Luong muoi c6 trong dung dich X bSng.A. 79,2g
B. 78,4g
C . 72g
D. 72,9g
HUcfng ddn
Ap dung
gidi
C O n g t h u f C : nimudi cloma = nihon h<7p oxit kim loai
+ 27,5
n^jp,
= 50 + (0,2.4.27,5) = 72 gam
Chpn dap an C.
B a i 2: Hoa tan hoan toan 2,81 gam hon hop A gom FezOg, MgO, ZnO
bkng 300ml dung dich H 2 S O 4 0,1M (vCra du). Co can can than dung
dich thu diioc sau phan ufng t h i thu difoc lirong muoi sunfat khan la:
A. 3,81 gam
B. 4,81 gam
C. 5,21 gam
D. 4,8 gam
HU&ng dan
Ap dung
gidi
C O n g thufC: mmum sunfat = nihSn hop oxit kim loai +
J^HaSOi
-^^
Chon dap an C.
B a i 3: De tac dung vCra du vdi 7,68g hon hop gom FeO, Fe304, FegOs can
dung 260 m l dung dich HCl I M . Dung dich thu diTOc cho tac dyng vdi
NaOH du, ket tua thu Auac mang nung trong khong khi den khoi
iLfOng khong doi duoc m gam chat r ^ n . Gia t r i cua m la:
A.6g
B. 7g
C.8g
D.9g
oxit k i m loai + HCl
nnc. =
2H^
0,26
'
Taco:
= 0,03mol .nHci = 0,3.1 = 0,3 mol
2H^
> Hg
0,06
<-0,03
2H^+
0,24
O'-
m = mx-
mo(oxit)
>
A. 87,5 m l
B.125
So do hop thiJc:
2Fe ^
C.62,5 m l
D.175 m l
gidi
- 4 chat rin la Fe203
= Y | ^ = 0,01875 mol
So do hop thufc:
=>
0,13mol
ml
Hii&ng dan
nFe,03
>H20
5 6
=> mpe = 7,68 - 0,13.16 = 5,6 gam => ^fe =
=
= 12 - 0,12.16 = 10,08 gam
B a i 5: Hoa tan hoan toan 2,8 gam hon hop FeO, Fe203 va Fe304 can vCra
du V ml dung dich HCl I M , thu di/oc dung dich X. Cho tii tii dung dich
NaOH dir vao dung dich X thu dugfc ket tua Y. Nung Y trong khong
khi den khoi liiOng khong doi thu dirge 3 gam chat r ^ n . Tinh V?
gidi
> muoi clorua + H 2 O
O'-
H2O
->0,12mol
= 0,26.1 =0,26 mol
+
>
Chon dap an A.
= 2,81 + (0,3.0,1.80) = 5,21 gam
Hii&ng dan
n„^ =
mo,ox,t)
2Fe
Fe203
0,0375 <-
0,01875
= 2,8 - (0,0375.56) = 0,7 gam
no,oxit)=^-0,04375mol
0>l"iol
FeaOg
0,1 -> 0,05 mol
=> mp^^o^ = 160.0,05 = 8 gam
Chon dap an C.
B a i 4: Oxi hoa cham m gam Fe ngoai khong k h i sau mot thcJi gian thu
duoc 12 gam hon hop X g6m(Fe, FeO, FezOg, Fe304). De hoa tan het
X, can vCra du 300 ml dung dich HCl I M , dong thdi giai phdng 0,672
Ht khi (dktc). Tinh m?
A.10,08
B.8,96
C.9,84
D.10,64
2H^
+
0,0875 <-
O'-
>
H2O
0,04375
=> V = 87,5 m l
Chon dap an A.
B a i 6: Hoa tan hoan toan 5,4 gam mot oxit sSt vao dung dich H N O 3 dii
thu dagc 1,456 l i t hon hop NO va N O 2 (dktc va khong con san pham
khuf nao khac). Sau phan ufng khoi lUOng dung dich tang len 2,49 gam
so vdi ban dau. Cong thufc cua oxit s^t va so mol H N O 3 phan ufng la:
A. FeO va 0,74 mol
B. Fe304 va 0,29
mol
C. FeO va 0,29 mol
D. Fe304 va 0,75
mol
Hiidng
mkhi = 5,4
Lap
-
2,49
= 2,91
x + y =
he:
1,456
gam
ddn
(NO
x mol va NO2 y
= 0,065
22,4
a
y = 0,06
+ 3e
O
3a
b
^
N O 3 ' + 3e
^
0,015
NO
<-
J
no
^ MZ5
^ 1
0,075
1
T h u c c h a t p h a n ufng khuf cac o x i t t r e n l a
CO
^
0,075
0,075
+
H2
2e
B. BAI T A P A P
2b
+
le
^
NO2
0,06
<-
0,06
0 ( t r o n g oxit)
>
0 ( t r o n g oxit)
N e n t a l u o n c6:
O 2-
^
>
-
CO2
H2O
= n„^o >
no,t„,„g„,it) = " H ,
nctrongoxit) =
^co
=
^co,
DUNG
B a i 1 : Khijf h o a n t o a n
17,6g
h o n hap
gom
Fe, FeO,
FegOs, c a n 4,48
lit
H 2 ( d k t c ) . K h o i l u o n g s a t t h u di/oc l a :
A . 14,5 g
B . 15,5 g
,
,
^
,
=> {a = b = 0 , 0 7 5 m o l
^
C. 14,4 g
Hitdng
4
no(trongoxit) = "^11,
FeO
mpe = 17,6 -
ddn
D . 16,5
g
gidi
48
= »H,0 =
=
0.2
HlOl
g
3,2 = 14,4
g
C h o n d a p d n C.
B a i 2: H 6 n h o p A g o m sSt v a o x i t sSt c6 k h o i luofng 2,6 g . Cho k h i C O d i
3nN(Fe(N03)3
+
^N(NO)
+
^NCNO^)
= 3.0,075 + 0,005 -t- 0,06 = 0,29
n„N03
+
mo = 16 . 0,2 = 3,2
Fe(N03)3
njjfHNo,)
+
NO3-
oxit sat la
FeO
B T N T NitO:
D o i v d i t r u d n g h o p c h a t khuf l a C O , H 2 t h i
mol
0,005
56a + 16b = 5,4
Tac6:<^
[ 3 a - 2 b = 0 , 0 6 + 0,015
*
mol).
X = 0,005 m o l
30x + 46y = 2,91
Fe - > Fe
giai
= nj,,HN03)
^
^imo,
=
0.29
qua A d u n n o n g , k h i d i r a sau p h a n ufng di/gfc d a n vao b i n h d i i n g nirdfc v o i
mol
t r o n g diT, t h u diTcfc l O g k e t t u a t r S n g . K h o i Itfcfng sSt t r o n g A l a :
A. 1 g
mol.
B . 1,1 g
C. 1,2 g
Hi£&ng ddn
C h o n d a p a n C.
D . 2,1 g
gidi
Ca(0H)2 + C O 2 -> CaCOsi + H2O
Dang 3: KHLf CAC GXIT KIM LOAI BANG (CO. C. Ha. Al)
10
=
^ 0 0 ,
•
ncaco3
•
J^odrongoxit) =
=
^co
=
—
=
0,1
mol
A. LfTHUYET
t
I . Phxftfng t r i n h p h a n i ^ n g t o n g q u a t :
K h o i li/gfng sit
= n^o^
=0,1
mol
t r o n g h 5 n h a p A l a : mpe = 2,6 -
16 . 0 , 1 = 1 g.
C h o n dap a n A.
B a i 3: Cho V l i t ( d k t c ) k h i Hg d i qua h o t C u O d u n n o n g , t h u duoc 32 g C u .
N e u cho V l i t H 2 d i qua h o t F e O d u n n o n g t h i ItfOng Fe t h u duoc l a :
A. 24g
B. 2 6 g
HUitng
D i e u k i e n : (M Id kim
-
lo<^i diKng
sau Al trong
day
di?n
hod)
K h i khuf o x i t k i m l o a i b S n g cac c h a t khuf C O ( H 2 ) t h i C O ( H 2 ) l a y o x i
ctja o x i t k i m l o a i r a k h o i o x i t .
32
= ncu = npe = —
D4
mpe = 56 . 0,5 = 2 8
C h o n d a n a n C.
g
= 0,5
C. 2 8 g
dan
mol
gidi
D.30g
B a i 7: N u n g n o n g h o n hop X gom PbO v^ FeO v d i m o t l u o n g C vijfa du.
B a i 4: Khuf ho^n toan 32g h o n hop CuO va FezOg b^ng k h i H2, t h a y tao
r a 9 g nude. K h o i lirong h 6 n hop k i m loai t h u duoc l a :
A . 12 g
B. 16g
C. 24 g
D. 26 g.
Hitdng d&n gidi
9
= n„^o
n o d r o n g oxit) =
niO (trong oxit) =
m
kim i o , i
16
. 0,5
=
=
=
"^'^
Sau k h i p h a n ufng xay r a hoan t o a n . t h u d U o c h o n hop chat r ^ n Y va
k h i k h o n g mau Z. Dem can h o n hop rSn Y t h a y k h o i liTOng g i a m 4,8g
so v d i h o n hop X. Cho h o n hop Y tac dung v d i dung dich H C l duf, t h u
di/oc chat k h i A. Sue k h i Z vao dung dieh nude v o i t r o n g d i i di/oc k e t
tua trSng. The t i c h k h i A (dktc) va k h o i liTOng k e t tua t h u d i r o c l a :
A. 6,72 l i t va 15g
B. 3,36 l i t va 30g
C. 6,72 l i t va 30g
D. 3,36 l i t va 15g.
Hitdng dan gidi
8g
= 32 - 8 = 24 g
2PbO + C — ^ 2 P b + C 0 2
Chon dap a n C.
Bai
5:
AI2O3
Cho
khi
CO
qua
ong
d i T n g a (g)
hon
hop
gom
CuO,
Fe.304,
FeO,
nung nong. K h i t h o a t r a duoc cho vao niidc v o l t r o n g du t h a y c6
30g k e t tua t r a n g . Sau p h a n
ufng,
chat rSn t r o n g ong
suf
c6 k h o i luong
202g. K h o i lirong a (g) cua h o n hop cac oxit ban dau l a :
A . 200,8g
B. 216,8g
C. 206,8g
D. 103,4g.
Ca(0H)2 + CO2 ^ CaCOgi + H2O
•
= "c;o
nourongoxit) = ^ c o
=
—
= 0,3
= ^ 0 0 , = 0.3
= 4,8g =>
'
ncaco3 = n c o ,
viia du V l i t k h i CO (d dktc), sau p h a n
ufng
A. FeO va 0,224
B. FeaOg va 0,448
C. Fe304 va 0,448
D. Fe304 va 0,224.
HiCdng dan gidi
= " c o , = 0,02
mol)
npe =
X la Fe304
can
t h u dugc 0,84 gam Fe va
0,02 mol k h i CO2. Cong thufc cua X va gia t r i V I a n luot l a .
V = 0,02.22,4 = 0,448 ( l i t )
= 0,3 (mol)
lb
^ n^j,^ =
n^hv
=
(mol)
1
^ V,,^ = 0,3.22,4 = 6,72 ( l i t )
CO2 + C a ( 0 H ) 2 ^ CaC03 + H2O
Khuf hoan toan m o t oxit s a t X d n h i e t do cao
x
0,015
3
Fe.Oy: - = ^ ^ ^ ^ = ' y
0,02
4
4 8
no - - V
• n„^ = n^hY = 0 , 3 (mol)
"^ol;
Chon dap a n C.
no(irongoxit) = "^co
- 0,3
oxit da b i C lay d i tao CO2.
Fe + 2 H C l ^ Fe + H2
mol
ma = 202 + 0,3.16 = 206,8 g
B a i 6: (CD -2009):
O trong
Pb + 2 H C l - > PbClz + H2
30
"co,
li/Ong
" c o , - - ^ n o =0,15(mol)
• a = m c h a t r^n + ^ l o d r o n g oxit)
=
K h o i li/ong chat r ^ n Y g i a m so v d i h 6 n hop X 1^ k h o i
=> m o
HU&ng dan gidi
nc«co3
2FeO + C — ^ 2 F e + C 0 2
0,84
-b^
56
= 0,15 (mol) =>
mc^co,
= 0,15.100 = 15(g)
Vay dap a n diing la A.
B a i 8: Cho 0,3 m o l Fe^Oy t h a m gia p h a n ufng n h i e t n h o m t h a y tao r a 0,4
m o l A I 2 O 3 . Cong thufc oxit ski l a :
A. FeO
B. Fe203
D. K h o n g xac d i n h
diTOc
C. Fe304
v i k h o n g cho biet so m o l Fe tao r a .
Hiidng dan gidi
= 0,015
(mol)
A l lay di oxi cua Fe^Oy de tao ra
trong
AI2O3
AI2O3.
V i vay so mol nguyen tuf O
va t r o n g FoxOy p h a i bang nhau.
Do do: 0,3 y = 0,4 . 3 = 1,2
Chon dap an C.
=> y = 4 =^ Fe304
B a i 9: Dot chay k h o n g ho^n toan 1 luong sSt da dung h e t 2,24 l i t O2 d
dktc, t h u dtfoc h6n hop A gom cac oxit sat va s^t du. Khuf hoan toan
A bang k h i CO diS, k h i d i r a sau p h a n l i n g duoc dSn vao b i n h difng
n i / d c voi t r o n g d U . K h o i l i i O n g k e t tua t h u dUOc l a :
A. 10 g
B. 20g
C. 30g
D. 40 g
Hiidng ddn giai
B a i 12: Khuf 39,2g m o t h o n hop A gom FegOa va FeO b&ng k h i CO t h u
duoc hon hop B gom FeO va Fe. B t a n vUa du t r o n g 2,5 l i t dung dich
H2SO4 0,2M cho r a 4,48 l i t k h i (dktc). T i n h k h o i luong FezOg va FeO
t r o n g hon hop A.
A. 32g FezOs; 7,2g FeO
B. 16g Fe203; 23,2g FeO
C. 18g FeaOg; 21,2g FeO
D. 20g Fe203; 19,2g FeO.
HUdng dan giai
Ca(0H)2 + CO2 ^ CaCOai + H2O
2 24
n a t r o n g oxit) =
n c o = n^o^ = n^^co^ =
^ . 1 - ,
. [FeoOg : x(mol)
Goi h o n hap A<^ ^ '
^
^
F e O : y(mol)
-2 = 0,2 m o l
mc,co3 = 100-0.2 = 20g
H o n hgp B + H2SO4: FeO + H2SO4
Chon dap An B.
A p dung D L B T N T Fe t r o n g h a i h o n hgp A va B t a c6:
vao a x i t H C l t h i the t i c h k h i H2 (dktc) t h u dUcJc l a :
B. 1,12 l i t
C.3,36 l i t
160x + 72y = 39,2
D.2,24 l i t
J ^ H j = n(hh
kim loai) =
2.24
.
„ ,
= u,i
TCr (1), (2)
,
moi
x = 0,2
,y =
o,i
= 0,lmol
Chon dap a n D.
B a i 11: T h o i m o t luong k h i CO dtf d i qua ong dung hon hap h a i oxit
Fe304 v a CuO nung nong den k h i p h a n ufng xay ra hoan toan t h u dugfc
oxit k i m loai ban dau l a :
C. 4g
D . 4,2g
Hii&ng dan giai
nO(trong oxit) =
H c o = HCQ^
moxitkimio^i = n i k i m io,i +
Chon d a p d n A .
^
m
"caCOa
oxi
(g).
B a i 13: De khuf hoan toan 45 gam h o n hgp gom CuO, FeO, Fe304, Fe va
M g O can dung vCra du 8,4 l i t CO d (dktc). K h o i lugng chat r i n t h u
dugc sau p h a n u"ng l a :
A. 39g
2,32 g hon hop k i m loai. K h i thoat r a duac dua vao b i n h dung dung
d i c h Ca(0H)2 du t h a y c6 5g k e t tua t r i n g . K h o i lugfng h 5 n hap
B. 3,21g
(g);
Vay dap a n dung la A.
VH^= 22,4.0,1 = 2,24 l i t
A. 3,12g
mp,^o^ = 0 , 2 . 1 6 0 = 32
mp^o = 0,1.72 = 7,2
K h i hoa t a n h 6 n h g p k i m loai vao a x i t t h i :
h h k i m loai
(2)
" 2 x + y = 0,5
Hii&ng dan giai
n„ = n
FeS04 + H2O
n„^so^ = 0 , 2 . 2 , 5 = 0,5 (mol);
h t H2 (ot dktc).Neu dem h o n hop k i m loai t h u d i / a c hoa t a n hoan toan
nhhoxit =
(1)
Fe + H2SO4 ^ FeS04 + H2
B a i 10: De khijT hoan toan h 6 n hop FeO va ZnO t h a n h k i r n loai can 2,24
A. 4,48 l i t
160x + 72y = 39,2
= 0,05(mol)
= 2,32 + 0,05.16 = 3,12 ( g )
B. 38g
C. 24g
D . 42g
Hii&ng ddn giai
hai
K h o i lugng chat r ^ n sau p h a n iJng l a : 45 - 1 6 . - ^ ^ = 39g
Chon dap a n A .
B a i 14: Khuf hoan to^n 17,6 gam hon hgp X gom Fe, FeO, Fe203 can
2,24 l i t CO (or dktc). Klio" lugng sSt t h u dugc la
A. 5,6 gam.
B. 6,72 gam.
C. 16,0 gam.
D . 8,0 gam.
Hii&ng dan
gidi
Hii&ng dan
2 24
K h o i luong chat r a n s a u p h a n ufng l a : 17,6 - 1 6 . - ~ -
K h o i luong nguyen tuf oxi = do g i a m cua chat r ^ n
= 16,0 gam
m = 31,9 - 28,7 = 3,2 g
Chon dap a n C.
B a i 15: De k h i l hoan t o ^ n 30 gam h o n hop CuO, FeO, FezOg, Fe304,
3 2
So m o l nguyen tuf O = ^
= 0,2 mol= so m o l C O
M g O can dung 5,6 l i t k h i CO (6 dktc). K h o i lifong chat r a n s a u p h a n
ufng la
A. 28 gam.
B. 26 gam.
HUdng
C. 22 gam.
dan gidi
16
CO
D. 24 gam.
CO2 + 2e
(ZnO + 2e ^ Zn ^ Zn^* + 2e
5 6
K h o i l i i a n g chat r a n sau p h a n ufng l a : 30 - 1 6 - ^ ^
gidi
F e O + 2e -> F e ^ Fe^* + 2e)
= 26 gam
2W
Chon dap a n B.
B a i 16: D a n tCf t\i V l i t k h i CO (6 dktc) d i qua m o t ong s i l difng lUdng d i i
h o n hop r a n gom CuO, FezOs (d n h i e t do cao). Sau k h i cac p h a n ufng
+ 2e -> H2
Vay so m o l nguyen tuf O = so m o l C O = so m o l H2 = 0,2 m o l
Chon dap a n B.
xay r a hoan t o a n , t h u duoc k h i X. D i n toan bo k h i X d t r e n vao lUOng
C . BAI T A P V A N
diT dung dich Ca(0H)2 t h i tao t h a n h 4 gam k e t tua. Gia t r i cua V la
B a i 1: Cho luong k h i C O d i qua m gam Fe203 dun ndng, t h u di/oc 39,2
gam h 6 n hop gom bon chat r a n la sat k i m loai va ba oxit cua n6,
dong thcfi C O h o n hop k h i t h o a t r a . Cho h 6 n hop k h i nay hap t h u vao
dung dich nudc v o i t r o n g c6 dii, t h i t h u di/Oc 55 gam k e t t u a . T r i so
cua m l a :
A. 1,120 l i t .
B. 0,896 l i t .
C. 0,448 l i t .
Hii&ng dan gidi
nco = nco, = nc«co, = ^
= ^'^^
D. 0,224 l i t .
=> Vco = 0,04. 22,4 = 0,896 l i t
Chon dap a n B.
B a i 17: T h o i m o t luong k h i CO d i qua ong suf ditog m gam h o n hcfp
Fe304 va CuO nung nong t h u diTOc 2,32 gam h o n hap r a n . Toan bo
k h i t h o a t r a cho hap t h u het vao b i n h dung dung dich Ca(0H)2 du t h u
dtfoc 5 gam k e t tua. Gia t r i cua m l a :
A. 3,22 gam.
B. 3,12 gam.
C. 4,0 gam.
Hii&ng dan
m
hSnhop =
2,32
+
16.
=
3,12
C. 64 gam
D. T a t ca deu sai, v i se k h o n g xac d i n h duoc.
B a i 2: Cho luong k h i H2 c6 dif d i qua ong suf c6 chufa 20 gam hon hop A
t r o n g hon hop A l a :
gam
gom
AI2O3, ZnO, FeO va CaO t h i t h u difcfc 28,7 gam hon hgp chat r a n (Y).
Cho toan bo h 6 n hop chat r a n (Y) tac dung v d i dung dich H C l dif t h u
duoc V l i t H2 (dkc). Gia t r i V la
A. 2gam; 18gam
B. 4gam; 16gam
C. 6gam; 14gam
D. 8gam; 12gam.
B a i 3: Cho luong k h i CO (du) d i qua 9,1 gam hon hop gom CuO va AI2O3
nung nong den k h i p h a n ufng hoan t o a n , t h u dUdc 8,3 gam chat r a n .
K h o i luong CuO c6 t r o n g h o n hop ban dau la
A. 0,8 gam.
C. 6,72 l i t .
B. 40 gam
l a i , t h a y k h o i liXOng chat r a n g i a m 3,2 gam. K h o i liicfng m o i cha't
gidi
B a i 18 Cho dong k h i CO du d i qua h o n hap (X) chufa 31,9 gam
B. 4,48 l i t .
A. 48 gam
gom MgO va CuO nung nong. Sau k h i p h a n iJng hoan toan, dem can
D . 4,2 gam.
Chon dap a n B.
A. 5.60 l i t .
DUNG
D. 2,24 l i t .
B. 8,3 gam.
C. 2,0 gam.
D. 4,0 gam.
Fe(N03)2 + AgNOa
0,1
Dang 4: KIM LOAI TAG DUNG VOI DUNG DICH MUOl
Ca'^ N a "
F e ' * Ag*
Mg'^ A l ' " Zn'*
Au'*
__
•
Tinh
K
Ca
Na
F e ' * Ag
Au
Mg
Al
Zn
Cr
oxi hod ion kim log,i
Ni
Fe
Sn
Pb
Fe
——
.
Tinh
- Chieu p h a n ufng: Chat oxi hod m^nh
H
0,1
=
nAgNo,
C r ' * Fe^* N i ' * Sn'* Pb'* Fe^* H * Cu'*
tang
Cu
khvC kim loi^i
gidm
+ Chat khii mg,nh ->
Chat
-
0,3
=
0,2
chufa h o n hop gom AgNOs 0,1M va Cu(N03)2 0,5M. Sau k h i cac p h a n
ufng xay ra hoan t o a n , t h u diTOc dung dich X va m gam chat r a n Y.
Gia t r i ciia m l a .
A. 2,80.
B. 4,08.
nFe
dan
D. 0,64.
gidi
= 0,04 m o l , n^^. = 0,02 m o l , n^^^. = 0,1 mol
Chat khuf m a n h n h a t se tac dung v d i chat oxi hoa m a n h n h a t triTdc:
+ 2Ag*^Fe'*
+
0,02
2Ag
0,02mol
+ Cu'* ^
0,03
manh
C. 2,16.
Hiidng
Fe
Dicing bai tap nay can lUu y den quy tdc a
dif
Cho 2,24 gam hot sSt vao 200 m l dung dich
B a i 2: (DH khoi B.2009):
0,01
Fe'* + Cu
mol
Vay muoi gom c6 Fe(N03)3 va AgNOs d i i
Fe
Ddu a cang Ian khd ndng phdn ling xdy ra cdng
0,5
•
oxi hod yeu + Chat khvi yeu
PT: Cu'* + Fe
Fe(N03)3 + A g
Fe'* + Cu
0,03
0,03 mol
m = 0,02.108 + 0,03.64 = 4,08g. Chon dap an B.
B a i 3: (CD -2009):
Cho m gam M g vao dung dich chufa 0,12 m o l FeCls.
Sau k h i p h a n ufng xay r a hoan toan t h u dtfoc 3,36 gam chat rkn.
Gia
t r i cua m la
A. 5,04
B. 4,32
C. 2,88
HiCdng dan
npe = ^
BAI T A P A P
56
B a i 1 : Cho O,lmol Fe vao 500 ml dung dich AgNOs I M thi dung dich thu
difOc chufa:
gidi
= 0,06 mol
M g + 2FeCl3
DUNG
MgCl2 + 2 F e C l 2
0,06 -> 0,12 ^
0,12
M g + FeClz ^ MgCl2 + Fe
A. AgNOg
B. AgNOs va Fe(N03)2
0,06
C. Fe(N03)3
D. AgNOa va Fe(N03)3
m = (0,06 + 0,06).24 = 2,88 (g). Chon dap a n C.
Hitcfng dan
"AgNo^
=
0.5-1
=
0,5
0.2
0.1
gidi
0,06
<-
B a i 4: (DH khoi A.2010):
0,06 (mol)
Cho 19,3 gam hon hop bot Zn va Cu c6 t i le mol
tuong ijfng la 1: 2 vao dung dich chufa 0,2 mol Fe2(S04)3. Sau k h i cac phan
ufng xay ra hoan toan, t h u duoc m gam k i m loai. Gia t r i cua m la
mol
Fe + 2AgN03 ^ Fe(N03)2 + 2Ag
0.1
D. 2,16.
0.2
A . 6,40g
B. 16,53g
C. 12,00g
D. 12,80g.
Hii&ng dan
Goi
X
^
la so mol cua Zn
65x + 64.2x
X = 0,1 (nzn = 0,1
n^,^,. = 0,2.2 = 0,4
Ta c6:
<-
> 2Fe^^ + Cu^^
B a i 5: (CD - 2009;.- Cho mi gam A l vao 100 ml dung dich gom Cu(N03)2
0,3M va AgNOs 0,3M. Sau k h i cac phan ufng xay ra hoan toan t h i thu
dtfgc m2 gam chat rAn X. Neu cho m2 gam X tac dung vdi lugng du
dung dich HCl t h i thu duac 0,336 l i t k h i (d dktc). Gia t r i cua mi va mg
Ian lUOt la
A. 8,10 va 5,43
B. 1,08 va 5,16
C. 0,54 va 5,16
D. 1,08 va 5,43.
2Ag^
0,1
^
0,2
+
0,03
+
2Ag
0,1
^
0,2
^
+
Ag
0,05
->
Ag
0,03
Cu^^ + 2e ^ Cu
0,03
0,06
0,03
z:> Tong so mol A l = 0,04(mol) mi = 27.0,04 = 1,08 (g)
=> m2 = mAi + m c u + niAg = 0,01.27 + 0,03.108 + 0,03.64 = 5,43 (g)
Chon dap an D
B a i 6: (DH KA.2008): Cho hon hdp bot gom 2,7 gam A l va 5,6 gam Fe
vao 550 ml dung dich AgNOa I M . Sau k h i cac phan ufng xay ra hoan
toan, thu dufcfc m gam chat r ^ n . Gia t r i cua m la (biet thuf t u trong
day the dien hoa: Fe^VFe^' diJng trUdc AgVAg).
C. 59,4
Fe^*
0,05
Khoi liigfng Ag = 0,55.108 = 59,4 gam
Chon dap an C.
B a i 7: Nhiing mot thanh kem va mot thanh sat vao cung mot dung dich
C U S O 4 . Sau mot thdi gian lay hai thanh k i m loai ra thay trong dung
dich con lai c6 nong do mol ZnS04 hang 2,5 Ian nong do mol FeS04.
Mat khac, khoi li/ofng dung dich giam 2,2g. I ^ o i Itrofng dong bam len
thanh kem va thanh sSt Ian lucft la:
B. 64g; 25,6g
C. 32g; 12,8g
CM(ZnS04) = 2,5CM(FeS04) ^
Ap dung dinh luat bao toan e: 3x = 0,03 + 0,06 => x = 0,03 mol.
B. 54,0
^
Fe^*
HiC&ng dan
0,015
0,03
0,3 mol
Ag"
gidi
0,01
A. 64,8
+
A. 12,8g; 32g
Theo bai ra A l con diX: 2A1 + 6HC1 -> 2AICI3 + 3H2
3x
3Ag
Sau cac phan ilng chat ran la Ag c6 so mol 0,3 + 0,2 + 0,05 = 0,55 mol.
Chon dap an A.
X
Fe
mol
Ag^ + le
+
0,3
Fe'"
Al^^ + 3e
Al'^
Dir Ag^= 0,25 - 0,2 = 0,05 mol
0,1 mol
Hi^&ng dan
^
vdi Fe
Khoi luong k i m loai con lai la khoi ItfOng cua Cu: 0,1.64 = 6,4g
Al
3Ag*
gidi
Sau phan ufng vdi A l , Ag* c6n 0,55 - 0,3 = 0,25 mol dung phan ufng
0,1 mol
ncucondu = 0,1
+
0,1
> 2Fe^" + Zn^^
2Fe^^ + Cu
0,2
Al
t h i so mol cua Cu la 2x
= 19,3
mol ; n c u = 0,2 mol)
mol
2Fe'^ + Zn
0,2
Hii&ng dan
gidi
D. 32,4
Zn
+ CUSO4
2,5x
2,5x
gidi
=
'^MF.SO,
ZnS04 + C u i
2,5x
2,5x
+ Cui
Fe
+ CUSO4
-^FeS04
X
<—
<—
X
n^^^o,
D. 25,6g; 64g.
x
—>
X
Do giam khoi lagfng cua dung dich la:
Am = mcu (Mm) - nizn (tan) " " I P e (tan)
o
2,2 = 64(2,5x + x) -65.2,5x- 56x
Ax = 0,4 (mol)
Vay mcu b^m len tiianh Z n = 64g; m c u bAm len t h a n h F e = 25,6g. Chon dap an B.
B a i 8: Nhiing mot thanh graphit dUdc phu mot Idp k i m loai hoa t r i I I
vao dung dich C U S O 4 dtf. Sau phan ufng khoi iMng cua thanh graphit
giam di 0,24g. Cung thanh graphit nay neu diTdc nhiing vao dung dich
AgNOa t h i k h i p h a n uTng xong k h o i luang t h a n h g r a p h i t t a n g
0,52g. K i m loai hoa t r i I I la k i m loai nao sau day:
A. Pb
B. Cd
C. Fe
Hiidng
dan
len
HiCctng dan
H A g N O , (ban dSu, =
gidi
M ' " + Cui
1 mol
Cu + 2AgN03
1 m o l —> g i a m M - 64 (g)
-^^(mol)
M - 64
4-Amgian,
M + 2 Ag^ ->
2 Agi
1 mol
2 mol
0,015 <- 0,03
= 0,24 (g)
Amtang = 0,52
= _ M 2 _ ^ M
216-M
= 112
B a i 9: N g a m m o t t h a n h Cu t r o n g dung dich c6 chiJa 0,04 m o l AgNOa,
sau m o t t h d i gian lay t h a n h k i m loai ra t h a y k h o i li/dng tSng hdn so
vdfi luc dau la 2,28 gam. Coi toan bo k i m loai sinh ra deu b a m het vao
t h a n h Cu. So m o l AgNOs con l a i t r o n g dung dich la
HUdng
C. 0,02.
dan
D . 0,015.
2x
64x
=
M g + CuCl2
(tan)
gam.
C. 2,43 gam.
D . 4,13
gam.
gidi
MgCla + Cu
X
y
y
Ta c6:
2,28
(64 - 24).(x + y) = 0,8
=> 40(x + y) =0,8
=>
da =0,04 - (0,015.2) =0,01 mol
X
+ y = 0,02
K h o i lirong muoi tao t h a n h la 3,23 + (0,02.24) - (0,02.64) = 2,43 gam
Chon dap an C.
Chon dap an A.
B a i 10: N g a m m o t v a t bSng dong c6 k h o i li/cJng 15 gam t r o n g 340 gam
dung dich AgNOs 6%. Sau m o t t h d i gian lay v a t ra t h a y k h o i luong
AgNOa t r o n g dung dich g i a m 25%. K h o i lUcfng cua v a t sau p h a n iJng la
A. 3,24 gam.
B. 1,43 gam.
Hii&ng dan
X = 0,015mol
n^^No^
(b^m) ~
M g + Cu(N03)2 ^ C u ( N 0 3 ) 2 + Cu
2x
108.2X -
A. 1,15 gam.
X
gidi
Cu + 2AgN03 - 4 Cu(N03)2 + 2Ag
X
ban d i u +
B a i 11: Hoa t a n 3,23 gam h o n hgtp gom CuCl2 va Cu(N03)2 vao niTdc
difcfc dung dich X. N h u n g t h a n h k i m loai M g vao dung dich X den k h i
dung dich m a t mau xanh r o i lay t h a n h M g ra, can l a i t h a y tSng t h e m
0,8 gam. K h o i lU(?ng muoi tao ra t r o n g dung dich la
Vay dap a n diing la B: Cd.
B. 0,005.
^v&t
0,03 m o l
Chon dap an C.
(g)
V i cung m o t t h a n h g r a p h i t t h a m gia p h a n iJng n e n :
A . 0,01.
^
= 15 + (108.0,03) - (64.0,015) = 17,28
tSng 2.108 - M = 216 - M (g)
^
_a2^
M-64
> Cu(N03)2 + 2 A g i
n i v a t sau phan ilng =
(mol)
,
mol;
25
n A . N O 3 < « = 0 , 1 2 . — =0,03 mol.
Goi k i m loai c6 hoa t r i I I do la M c6 k h o i lifcfng m(g)
M + Cu^^ ^
340.6
^^Q^QQ = 0.1^
D . Sn
giai
B. 2,28 gam.
C. 17,28 gam.
D . 24,12
gam.
B a i 12: Ngii6i ta phu m o t Idp bac t r e n m o t vat b ^ n g dong c6 k h o i lifOng
8,48 gam bSng each n g a m vat do t r o n g dung dich AgNOa. Sau m o t
t h 6 i gian \zy v a t do ra k h o i dung dich, rufa nhe, l a m kho can dufeJc 10
gam. K h o i luong A g da phu t r e n be mSt cua vat la
A .1,52 gam.
B .2,16 gam.
C. 1,08 gam.
D . 3,2 gam.
Hii&ng dan
Cu + 2AgN03 -> Cu(N03)2
Fe
2x
T a c6: 108.2x - 64x = 10 X
+ Cu
0,01
+ CuS04
T a c6:
^FeS04
+ Cui
X
64x - 56x = 0,12
X
= 0,015 m o l
Z
gam
Chon dap a n B.
Sau k h i k e t thiic cac p h a n ufng, loc bo p h a n dung dich t h u
duac m gam bot r ^ n . T h a n h p h a n % theo k h o i li/ong cua Zn t r o n g h6n
^cuso,
= 0 , 0 1 + 0,015 = 0,025 m o l
_ 0,025
B a i 13: Cho m gam h o n hcjp hot Z n va Fe vao li/cfng d u dung dich
CUSO4.
Mg2"
X
8,48
= 0,01 m o l
m A g = 0,01.2.108 = 2,16
+ Cu^^ ^
0,01
+ 2Ag
2x
X
Mg
gidi
Chon dap a n C.
hop ban dau la
A. 90,27%.
B. 85,30%.
C. 82,20%.
Hiidng
Zn +
CUSO4
dan
D . 12,67%.
Dang 5 . CDs. SOa T A G D U N G VOl D U N G D|CH K I E M
gidi
A. K I M L O A I K I E M : T O M T A T LI T H U Y E T
1. V i t r i t r o n g b a n g t u a n h o a n , c a u h i n h e l e c t r o n
-> ZnS04 + C u i
X
K i m loai k i e m gom: L i t i ( L i ) , N a t r i (Na), K a l i (K), Rubidi (Rb), Xesi
(Cs), F r a n x i ( F r ) .
X
Fe +
CUSO4
^ FeS04 + C u i
y
Thuoc n h o m l A
y
V i m hon hop r ^ n dau b^ng m h o n hdp r ^ n sau
Cau h i n h electron:
N e n : 64x + 64y = 65x + 56y <=> x = 8y
N a ( Z = l l ) l s ' 2 s ' 2 p ' 3 s * hay [Ne]3s'
% Zn = — ^ ^ - ^ y
.100% = 90,27%
65.8y + 56y
K (Z=19) Is22s22p^3s^3p^4s' hay [ A r ] 4 s '
Deu CO 1 electron d Idp ngoai cung
Chon dap An A .
B a i 14: Cho 1,12 gam bot Fe vh 0,24 gam bot M g tac dung v d i 250 m l
dung dich
CUSO4
x M , khuay nhe cho den k h i dung dich m a t
mau
x a n h n h a n t h a y k h o i liicfng k i m loai sau p h a n ufng la 1,88 gam. Gia
t r i cua x la
A . 0,04M.
B. 0,06M.
C. O . I M .
Hii&ng ddn
n i k i m io,i sau phan ling =
1,88
-
1,12
=> Am =0,01.(64-24) = 0,4 gam
n i c b n i a i = 0,52
L i (Z=3) ls'2s* hay [He]2si
-
0,4
=
0,12
-
0,24
D . 0,025M.
gidi
= 0,52
I I . T i n h c h a t ho a h o c
Co t i n h khut m a n h : M ->
+ e
1- T a c d u n g v6i p h i k i m :
T h i du:
4Na + 0 2 - ^ 2 N a 2 0
2Na + C ] 2 ^ 2 N a C l
2. T a c d u n g v d i a x i t ( H C l , H2SO4 l o a n g ) : tao muoi
gam
T h i du:
2Na + 2HC1 ^ 2 N a C l + H 2 t
3. T a c d u n g v d i n\i6c: tao dung dich k i e m va H2
T h i du:
2 N a + 2H2O
2 N a O H + Hgt
H2
C. DANG T O A N CO2 (HOAC SO2) VAO DUNG DICH KIEM NaOH (HOAC
KOH)
I I I . Dieu che:
1. Nguyen tSc: khuf ion kim loai kiem thanh nguyen tur.
C a c pht^ofng trinh xay ra:
2. Phifcfng phap: dien phan nong chay muoi halogen hoSc hidroxit cua
chung.
Thi du: dieu che Na b^ng each dien phan n6ng chay NaCl
PTDP:
B.
MQT
SO
HOP
CUA
CO2 + O H - ^
KIM
LOAI
CO2 + 2 OH" ^
KIEM:
+ Tac dung vdi axit: tao muoi va niTdc
<1
NaOH + HCl ^ NaCl + H2O
• CO2
BAI T A P V A N
2NaOH + CUSO4 ^ Na2S04 + Cu(0H)2i
Na2C03 + COot + H2O
^^^^
NaHCOs + HCl ^ NaCl + C 0 2 t + H2O
?
NaHCOa + NaOH ^ Na2C03 + H2O
I I I . Natri cacbonat - Na2C03
•*
+ Tac dung v d i dung dich axit manh:
Na2C03 + 2HC1 ^
I V . K a l i nitrat: KNO3
Tinh chat: c6 phan ling nhiet phan
2KNO3
= 1
du
2KNO2 + O2
• NaHCOa
• NaaCOa
• NajCOi
• NajCOa
• NaOH dir
dan
gidi
ehi tao muoi NaHCOg.
B a i 2: Hap thu hoan toan 4,48 l i t khi SO2 (d dktc) vao dung dich chufa
16g NaOH thu dugc dung dich X. Khoi liicfng muoi tan thu duac trong
dung dich X la bao nhieu?
A.20,8g
B.18,9g
2NaCl + CO2T + H2O
Muoi cacbonat cua kim loai kiem trong nadc cho moi triTdng kiem
>2
CO2
+
NaOH ^
NaHC03
0,1 moi
->
0,1 moi
^ii^ico, = 0.1-84 = 8,4 gam.
+ Tac dung v d i axit:
+ Tac dung vdi dung dich bazd:
=2
B a i 1: Sue 2,24 l i t k h i CO2 vao 100ml dung dich NaOH I M , t i n h khoi
lUdng muoi thu duac.
2. T i n h litofng tinh:
Thi du:
• NaHCOa
HUdng
2NaHC03 — ^
1 <2
DUNG
1. P h a n v(ng p h a n hiiy:
T h i du:
=1
• NaHCOa
2NaOH + CO2 -> Na2C03 + H2O
n. Natri hidrocacbonat - NaHCOg
T h i du:
CO3'- + H2O
San phani
+ Tac dung vdi dung dich muoi:
T h i du:
HCO3-
^NaOH
+ Tac dung vdfi oxit axit: tao mtioi va niidc
T h i du:
(2)
Ta can lap t le
I. Natri hidroxit - NaOH
T h i du:
(1)
Thiic chat ta dung hai pt sau
) 4Na + 2H2O + O2
CHAT QUAN TRQNG
NaHCOa
CO2 + 2NaOH ^ Na2C03 + H2O
NaOH
2NaCl — ^ ^ ^ ^ ^ 2Na + CI2
4NaOH
CO2 + N a O H
Hii&ng dan
2NaOH+ SO2 ->
0,4
m
0,2
muoi
NazSOg
0,2
= 0,2.126 = 25,2 gam
Chon dap an D.
C.23,0g
gidi
D.25,2g.
B a i 3: DSn 10 l i t h 5 n hap k h i gom N2 va C d 2 do or dktc sue vao 2 l i t
dung dich Ca(OH)2 0,02M t h u difgrc I g k e t t u a . T i n h p h a n t r a m theo
the t i c h CO2 t r o n g h 5 n hap k h i .
A.2,24% va 15,68%
B. 2,24%
C. 15,68%
D. 2,24% va 7,84%.
Hitcfng ddn
B a i 5: Cho 6,72 l i t k h i CO2 (dktc) vao 380 m l dd N a O H I M , t h u duac dd "
A. Cho 100 m l dung dich Ba(0H)2 I M vao dung dich A di/ac m gam
giai
k e t tua. Gia t r i m bang:
A. 19,7g
nco^ = n c , c o 3
=Y^ =
B. 15,76g
0,01mol
C. 5 9 , l g
Hiidng
= 0,02.2 = 0,04mol
nca.oH),
m^uo. = 0,1.23 + 0,1.39 + 0,1 .81 + 0,05.80 = 18,3 gam
Chon dap a n D.
Triforng hofp 1:
•
Theo d i n h luat bao toan nguyen tuT (tdng khoi lugng ion trong muoi
bang long khoi lugng muoi)
"N^OH
= M i : ^ . 1 0 0 % = 2,24%
= Hj^^, = n^,,^ = 0,38
dan
mol
gidi
"^n^iom, = n^^^, = 0 , 1
E n ^ j ^ . = 0,38 + 0,2 = 0,58 m o l
nco, =
-^Q
CO.
TrvLUns hc/p 2:
•
- n^^co^ = 2.0,04 - 0,01 = 0,07mol
n^o^ =
^ 0,07.22,4
^
^
g
^
CO2 + O H X
^
D.55,16g
22,4
- 0,3
mol
mol
HCO3-
X
X
CO2 + 2 0 H - -> C O 3 ' - + H 2 O
y
Chon dap a n A
B a i 4: H a p t h u 3,36 l i t SO2 (dktc) vac 0,5 l i t h 5 n hgp gom N a O H 0,2M
Ta c6:
va K O H 0,2M. Co can dung dich sau p h a n ufng t h u duc/c k h o i I'Jcfng
muoi k h a n l a .
A. 9,5gam
B. 13,5g
HU&ng dan
nwaOH
nKOH
= n ^ , = n^j,, = 0,5.0,2 = 0,1 m o l
SO2 + 0 H - - >
X
HSO3
D. 18,3g
y
x + y = 0,3
fx = 0,02
x + 2y = 0,58
[ y = 0,28
B a ' " + CO^-
BaCOsi
0,1
0,1
0,28
K h o i lirang k e t t u a l a : m B ^ c O j = 0,1.197 = 19,7 gam
Chon dap a n A .
B a i 6: D o t chay hoan toan 0,1 m o l etan r o i hap t h u t o a n bo san p h a m
chay vao b i n h chufa 300 m l dd N a O H I M . K h o i liiong m u o i t h u difgrc
sau phan iJng?
A. 8,4g va 10,6g
B. 84g va 106g
C. 0,84g va l , 0 6 g
D. 4,2g va 5,3g.
X
Hiidng
SO2 + 2 0 H - - ^ S0^-+ H 2 O
y
gidi
= n^^. = n^„_ - 0,5.0,2 = 0,1 mol
Z n ^ ^ . = 0,1 + 0,1 = 0,2 m o l
X
C. 12,6g
2y
2y
y
Taco:
C2H6 ^ 2CO2
0,1 - > 0,2mol
x + y = 0,15
fx = 0,1
" x + 2y = 0,2
|y = 0,05
CO2 + N a O H -> NaHCOa
X
X
X
ddn
gidi
b) Tdc dung vai axit
COa + 2 N a O H ^ NazCOg + H2O
y
2y
y
Ta c6: <
fx = 0,1
=> <
x + 2y = 0,3
,.y = 0 , l
V4y
-mN,HC03
H2 t\i do.
rx + y - 0 , 2
= 0.1-84 = 8,4
M + 2H^
gam
M^* + H2t
- D o i v d i axit c6 t i n h oxi hod m a n h nhiT HNOs, H2SO4.
gam
-> N"'(N02), N*^(NO), N-^(NH4N03)...
Chon dap a n A.
S^*^ ^ S^'(S02), S-=^(H2S), S°(S).
D. DANG TOAN CO2
(HOAC SO2)
VAO
DUNG DjCH Ca(0H)2 h o g c
Ba(0H)2. KIM LOAI PHAN NHOM CHINH NHOM II
TAT
MSO4 + H z t
M + H2SO4
mNa,co, - 0,1.106 = 10,6
T6M
trong dung d i c h a x i t ( H C l , H2SO4 loang) t h a n h
_ D e d a n g khilf ion
LITHUYET
4 M + IOHNO3
c) Tdc dung vai
T r o n g H2O,
4M(N03)2 + NH4NO3 + 3H2O
H2O
B e k h o n g p h a n urng, M g khuf c h a m , c a c k i m loai con l a i
khuf m a n h .
1. Vj tri trong bang
thong t u a n hoan, tinh chat v^t If
M + 2H2O ^
a) Vi tri
M(0H)2 +
Ca(0H)2 +
C a + 2H2O
K i m loai p h a n n h o m I I gom:
B e r i (Be); Magie (Mg); Canxi
(Ca);
T r o n g cac chu k i cac nguyen to nay diing l i e n sau k h i loai k i e m .
b) Tinh chdt vat li
- M g day c a c k i m loai hoat dong y e u hcfn r a k h o i dung d i c h muoi
M g + CUSO4 ^
- C a c k i m loai con l a i t a c dung vdi H2O trong dung dich
D i e n p h a n n o n g c h a y muoi halogenua cua chiing
- La k i m loai m e m (mem hon n h o m )
MX2
- K h o i li/Ong r i e n g tiidng doi nho
X : halogen
Jig" phan n6ng chay
^ M
+
X2
hoc
Cac nguyen to p h a n n h o m c h i n h n h o m I I c6:
- 2 electron hoa t r i (s^)
- Co ban k i n h nguyen tuf \6n
- L a nhOfng chat khuf m a n h M - 2e -> M^^
T r o n g cac hap chat c^c nguyen to nay c6 so oxi hoa +2.
a) Tdc dung vai phi
kim
2 M 0 ( M la nguyen tut k i m loai)
2CaO
1. C a n x i oxit: CaO
C a x i oxit l a oxit b a z a
- T a c dung m a n h l i e t v d i H2O tao b a z a
C a O + H2O ^
Ca(0H)2
C a O + 2HC1 ^ CaCl2 + H2O
- T a c dung vdi oxit a x i t tao muoi tuang iJng
C a O + CO2
- V d i CI2: M + CI2 -> MCI2
M g + CI2 ^
MQT SO HOP CHAT QUAN TRONG CUA CANXI
- T a c d u n g vdi n h i e u a x i t tao muoi tuang ufng
- V d i oxi k h i dot nong:
2Ca + 0 2 ^
MgS04 + C u i
3. D i e u che
- N h i e t do n o n g chay n h i e t do soi thap
2 M + O2
H2T
d) Tdc dung vai dung dich muoi
S t r o n t i (Sr); B a r i (Ba) va Radi (Ra)
2. T i n h chat hoa
H2t
MgCl2
CaC03
- C a n x i oxit dirge dieu che bkng phijang phap nhiet p h a n muoi cacbonat.
C a C 0 3 — ^ C a O + CO2
5. Nifofc ciJtng
2. C a n x i h i d r o x i t : C a ( O H ) 2
1. Nx^oTc cufng
L a c h a t r S n it t a n t r o n g H2O
Nadc curng la nUdc c6 chura n h i e u ion Ca^*, Mg^*. Nifdc k h o n g chiJa
hoSc chura i t nhOfng ion t r e n , goi la niTdc m e m .
D u n g d i c h C a ( 0 H ) 2 c6 t i n h bazO y e u h o n N a O H
- T a c dung v d i a x i t v a oxit a x i t tao m u o i tUofng ting
Ca(0H)2 + 2HC1 ^
C a ( 0 H ) 2 + CO2 ^
2. P h a n l o a i nvCdc
C a C l a + 2H2O
Nude curng chia t h a n h 3 loai
C a C O a i + H2Q
1. Nifde ciJng tarn t h d i : la nirdc curng c6 chura ion HCO3"
C a ( 0 H ) 2 + 2CO2 ^ C a ( H C 0 3 ) 2
, CaCOH),
N e u t i le m o l
CO2
2. NiXdc curng v i n h cijfu: la niJdc curng c6 chufa ion CI" hoSe SO^
1 ^
< — tao muoi a x i t
2 •
3. Nifdc curng toan p h a n : L a niidc cufng c6 chiJa dong thcfi cac ion
HCO3", C r hoSe SO^-
N e u t i le m o l ^^^^^^2 < ^ ^.^Q jjjy5'j t r u n g t i n h
CO2
N e u t i le m o l 9^^^k
CO2
trong k h o a n g
i < ^^^^^
2
3. T a c h a i c i i a nitofc ciifng
< 1
tao d6ng
- Xa phong k h o n g t a n
- V a i soi mau muc n a t
t h d i 2 muoi
- Nau thurc a n lau c h i n , g i a m m u i v i
- T a c dung v d i dung d i c h muoi
- Tao chat c&n t r o n g n o i h a i l a m l a n g p h i n h i e n lieu...
Ca(0H)2 + NaaCOs
C a ' ^ + CO^- ^
3. C a n x i c a c b o n a t
cufng
CaCOai + 2 N a O H
4. C a c h l a m m e m ni^cfc
CaCOai
Nguyen tSc: L a m g i a m n o n g do cac ion Ca^^ va Mg^* t r o n g nufdc bang
each chuyen nhij'ng ion t i i do nay vao t h a n h p h a n chat k h o n g t a n .
CaCOa
C a n x i cacbonat l a c h a t r S n m a u t r a n g k h o n g t a n trong
H2O
C a C O s l a muoi c u a a x i t y e u v a k h o n g b e n
CaCOa + 2 H C 1
CaCl2 + H2O + C O z t
CaCOa + 2CH3COOH -> Ca(CH3COO)2 + H2O + C O a t
ot n h i e t do t h a p CaCOa t a n d a n trong H2O c6 CO2
CaCOa + H2O + CO2 -> C a ( H C 0 3 ) 2
4. C a n x i s u n f a t : C a S 0 4
C a S 0 4 con goi l a t h a c h cao, m ^ u t r ^ n g , i t t a n t r o n g H2O
C a S 0 4 . 2 H 2 0 : t h a c h cao song
2 C a S 0 4 . H 2 0 : t h a c h cao n u n g n h o lufa
C a S 0 4 : t h a c h cao k h a n
T h a c h cao n u n g duoc dung de due tiiOng.dung trong y hoc de bo bot...
Phirong phap: Phuong phap hoa hoc va phuong phap trao ddi ion.
a) Phuang
phap
hod
hoc
* D o i v d i ntrdc cufng t a m t h d i . Dun nong trUdc k h i dung
Ca(HC0a)2 — ^
C a C O a i + H2O + C 0 2 t
Loc bo chat k h o n g t a n , dirge nifdc m e m
- D u n g Ca(0H)2 v t o du de t r u n g hoa
Ca(HC03)2 + Ca(0H)2
2 C a C 0 a i + 2H2O
Loc bo chat k h o n g t a n dtrgc nude m e m
* D o i v d i nude cufng v i n h cufu va ni/dc cufng t o a n p h a n
Dung dung dich Na2C0a, Na3P04
C a S 0 4 + Na2C03 -> C a C 0 3 i + Na2S04
Ca(HC03)2 + Na2C03 ^ CaCOgi + 2NaHC03
Ca?* + CO3
^ CaC03>l
Mg'* + CO3 ^"
MgCOa
b) Phuang phdp trao doi ion
Cho nutdfc cufng di qua chat trao doi ion (ionit) chat nay se hap thu
cac ion Ca^^ va Mg^* the vao do la ion N a \a duoc nude mem.
D^ng bai tap CO2 vao dung dich Ca(OH)2 hoqc Ba(OH)2 thi:
Dang 1: Biet nc^o,,,,^ .ncaco,
"co,
TrUang hap 1: rico, =
ntettua
TrUdng hop 2:
2nb«,„
HCO,=
= 0,02 mol
- ntettua = 2.0,03 - 0,02 = 0,04 mol
Chon dap an A.
B a i 2 : (DH A - 2008): Hap thu hoan toan 4,48 l i t k h i CO2 (d dktc) vao
500 m l dung dich hon hop gom NaOH 0,1M va Ba(0H)2 0,2M, sinh ra
m gam ket tua. Gia t r i cua m la.
A. 19,70 g.
B. 17,73 g.
C. 9,85 g.
D. 11,82 g.
Hiidng dan gidi
^Ca(0H)2
Dang
2:
^^^'^
T H I : n^^o^ - nc^co^
nNaoii = " N a * = " o H " " ^ ' ^ ^
T H 2 : n^o^ - 2.nc^,o„)^ -''^c^co^
I n^„. = 0,05 + 0,2 =0,25 mol;
B i e t n^o^,
nc^oH)^
t i m nc^co^
CO2 + OH" ^
X
o
W 2
^^CatOlllj
CO2
T H I : nco, = nc.co,
TH2
: ncacos
- •^^cMom,
~
"co^
^
BAI T A P V A N
B. 0,02 mol va 0,05 mol
C. 0,01 mol va 0,03 mol
D. 0,03 mol va 0,04 mol.
HiC&ng dan gidi
= 0,02
mol
C O 3 ' " + H2O
y
x + y = 0,2
rx = 0,15
" x + 2y = 0,25
[ y = 0,05
0,05
0,05
Chon dap an C.
A. 0,02 mol va 0,04 mol
=
HCO3"
Khoi laong ket tua la: ra^^co, =0,05 .197 = 9,85 gam
^CMOMU
B a i 1. Hap thu toan bo x mol CO2 vho dung dich chiJa 0,03 mol Ca(0H)2
diroc 2 gam ket tua. gia t r i x?
nr,ro
2y
0,1
DUNG
2
M | = 0,2 mol
Ba'" + CO3'- ^ BaCOai
_ ,
"Ca(OIl)o ~ •
Neu n^o, ^ ric^co, =^
n^o,
X
+ 20H- ^
y
Dang 3: Biet nc^co^. "00, t i m
nCaC03
X
"^u^iom, = ^ 3 , . . = 0 , 1 mol
Bai 3: (BH A - 2009): Cho 0,448 l i t k h i CO2 (d dktc) hap thu het vao
100 ml dung dich chuTa hon hop NaOH 0,06M va Ba(0H)2 0,12M, thu
duoc m gam ket tua. Gia t r i cua m la.
A. 1,182 g.
B. 3,940 g.
C. 1,970 g.
D. 2,364 g.
Hit&ng dan gidi
nNaOH
= n^^. = nQ„_ =0,006 mol;
I n ^ j j =0,006 + 0,012.2 = 0,03 mol;
HB^IOH),
nco, =
= n^^,, = 0,012mol
= 0,02mol
C02 + O f f
^
X
X
X
2y
x + y = 0,02
[ y = 0,01
Ba^^ + CO^-
BaCOai
0,012
0,01
0,01
A.3,136 l i t
B. 1,344 l i t
C. 1,344 l i t hoSc 3,136 l i t
D. 3,36 l i t hoSc 1,12 l i t .
Hit6ng
K h o i laang k e t t u a l a : mg^co, = 0 , 0 1 .197 = 1,97 gam
Ca(0H)2 + CO2
Chon dap a n C.
2,5 l i t dung dich B a ( 0 H ) 2 nong do a mol/1, t h u difdc 15,76 gam k e t
gidi
CaCOsi + H2O
Ca(0H)2 + 2CO2 ^
tua. Gia t r i cua a l a .
Ca(HC03)2
0,08 < - 0,04
B. 0,048 M .
C. 0,06 M .
D. 0,04 M .
Ca(HC03)2 —
Hiiafng d&n gidi
^
0,04
2,688 ^ , , „ „ , _
_ 1 5 ^
= 0,12mol; ng^co, =
= 0,08mol
nco, > neaco,
dan
0,06 < - 0,06
B a i 4: (DH A - 2007;.- H a p t h u hoan t o a n 2,688 l i t k h i CO2 (or dktc) vao
=
= 2.0,15 - 0,1 = 0,2 m o l
B a i 6: T h o i V l i t (dktc) CO2 vao 100 m l dung dich Ca(0H)2 I M , t h u diroc
6 gam k e t tiia. Loc bo k e t t u a lay dung dich t h u ducfc dun n o n g l a i c6
4gam k e t t i i a nufa. Gia t r i V l a :
fx = 0,01
' x + 2y = 0,03 ^
n^f>
ntettua
Chon dap a n D .
y
A . 0,032 M .
-
Vco, = 0,2.22,4 = 4,48 l i t
CO2 + 2 0 H - - > COa^" + H 2 O
y
n^^^ = 2nb,,o
TriCdng hap 2:
HCO3"
CaCOai + H2O + C 0 2 t
<-
0,04
I n c o , =0,08 + 0,06 = 0,14 m o l => Vco,= 0,14 .22,4 = 3,136 l i t
Chon dap a n A.
chi xay r a t r i / d n g hap 2
B a i 7: Hap thu toan bo 0,3 mol CO2 vao dung dich chijfa 0,25 mol Ca(0H)2.
Khoi lifofng dung dich sau phan uTng tang hay giam bao nhieu gam?
C„^,„„,
= - ^ = 0,04M
C. G i a m 16,8gam
D G i a m 6,8gam.
gidi
Be tinh khoi lixang chat sau phan i2ng tang hay giam
B a i 5: D a n V h t CO2 (dkc) vao 300ml dung dich Ca(0H)2 0,5 M . Sau
p h a n l i n g duac lOg k e t tua. Gia t r i cua V bSng:
B. 3,36 l i t
Hildng
TriCang hap 1:
B. T a n g 20gam
Hii&ng dan
Chon dap a n D.
A . 2,24 l i t
A. T a n g 13,2gam
C. 4,48 l i t
d&n gidi
n^o^ = nkettoa = 0,1 m o l
Vco = 0,1 .22,4 = 2,24 l i t
a lay khoi
liigng
vao trie khoi lilgng ra khdi dung dich niu > 0 la tang n^u < 0 la giam
Ta
c6:
n^o^
> n^^^Q^^^^
D. Ca A , C deu dung
J^COj -
2nbaza ~
nen
chi
xay
ra
TH2
^kgt tua
nc,co, = 2nc„oH,, - ^co, = 2.0,25 - 0,3 = 0,2mol
Am = mco, - ixic^co, = 0,3.44 - 0,2.100 = - 6 , 8 gam
Chon dap a n D.
B a i 8: Cho 5,6 h t h o n hdp X gom Ng va CO2 (dktc) d i cham qua 5 K t
dung dich Ca(0H)2 0,02M de p h a n ilng xay ra hoan toan t h u dUdc 5
gam k e t tua. T i n h t i k h o i h o i cua h 6 n hcfp X so v d i H2.
A. 18,8
B. 1,88
C. 37,6
Hiidng
= n^
Truang
hap 1:
dan
= O.lmol
n^o^ =
gidi
100
=3^
""'"^
2
j^>,15.44.0,1.28 ^
0,25
= 15,6
= 18,8
Chon dap a n A.
B a i 9: Sue CO2 vao 200 m l h 5 n hgfp dung dich gom K O H I M va Ba(0H)2
0,75M. Sau k h i k h i h i hap t h u hoan toan t h a y tao 23,6 gam k e t t u a .
C. 2,24 l i t
HU&ng dan
^on-
Nguyen
hap 1:
khi chuyen
hap 2:
= 0,2 m o l ;
Pt:
_ 23,6
-^BaCOj
~
CO2 + O H " -> HCO3"
X
CO2
+
y
X
20H-
2y
^
0 0 3 ' - +
=> X
(1)
(2)
2YCI3 + 3H2O +3CO2T
Theo 2 phuong t r i n h (1),(2) t a t h a y
•^muo'i clorua ~
^
Pt:
h h rauoi cacbonat
H2O
197
= 0,12mol
A C O 3 + H2SO4
-'^cOo
B2CO3 + H2SO4
^ | y = 0,12
"co, =0,12 + 0,26 = 0,38 mol
(loang) ^
(loang) ^
A S O 4 + CO2T + H 2 O
A 2 S O 4 + CO2T + H 2 O
Cach t i n h n h a n h cho m u o i sunfatv.
nirauo'i
sunfat
—
rnmuo'i
cacbonat
^CO^
BAl TAP VAN DUNG
B a i 1: Cho 115g hon hcrp gom A C O 3 , B 2 C O 3 , R 2 C O 3 tac dung het vdi dung
dich H C I thay thoat r a 0,4481 CO2 (dktc). K h o i liicfng muoi clorua tao r a
trong dung dich la:
A. 115,22g
B.151,22g
C. 116,22g
HUcfng dan
fx = 0,26
Vay Vco^ =0,38.22,4 = 8,512 l i t
Chon dap a n D.
XCO3 + 2HC1 -> X C I 2 + H2O + CO2T
Y2(C03)3 + 6HC1
y
y = 0,12
" x + 2y = 0,5
chdt.
Cach t i n h n h a n h dung cho tr&c n g h i e m .
ri^^mn, = ^ B a ^ * = ^' 1^"^°^
n^o^ = nkettta = 0,12 m o l
X
tii chat nay sang chdt khac Ung vdi 1 mol
lugng
.(71-60)
M u o i cacbonat tac dung v d i dung dich H 2 S O 4 loang
Vco.= 0,12 .22,4 = 2,688 l i t
Truang
D . 44,8
tdc cua phuang phdp nay Id dua vdo sU tang, gidm khoi
D. Ca A va B dung
gidi
I n ,' ,O, i, r. =0,2 +' 0,15.2 = 0,5 m o l ;'
Truang
C. 224
n c f = 2n^^,_ (ma M^,. = 35,5; M^^,. = 60)
B. 2,688 l i t
-
B.44,8 hoSc 224
Dang 6. MUQI CACBONAT TAG DUNG VO! DUNG D|CH
HCI, H 2 S 0 4 loang
T i n h Vco^ da dung d (dktc).
-
D . 16,745g
tuf cac b a i t r e n .
TMng
= 0,05 mol
hap 2: n^^^ = 2nb^,„ - nkeuaa = 2.0,1 - 0,05 = 0,15 m o l
n KOH
C. 9,85g
G i a i tiicfng t i i b a i t r e n
A. 44,8 hoSc 89,6
Khac dap an l o a i
A. 8,512 l i t
B. 14,775g
duac 0,2g k e t tua. Gia t r i V m l l a :
= 0,05mol
^
0,05.44 + 0,2.28
= 31,2 =^ d , / , , = ^
M =
0,25
'
"'"^
2
Truang
A. 23,64g
B a i 11: T h o i V m l (dktc) CO2 vao 300 m l dung dich Ca(0H)2 0,02M, t h u
5
n CRCOO
nkettua
D. 21
a i 10: Sue 4,48 h t (dktc) CO2 vao 100ml hon hop dung dich gom K O H
I M va Ba(OH)2 0,75M. Sau k h i k h i h i hap t h u hoan t o a n t h a y tao m
gam k e t tua. T i n h m .
D . 161,22g
gidi
C a c h g i a i 1:
A C O 3 + 2HC1 -
ACI2 + H^O + CO2 t
B 2 C O 3 + 2HC1
> 2BC1 + H2O + CO2 t
R 2 C O , + 2HC1 ^
2RC1 + H.,0 + CO,
t