CU THANH TOAN
\iibm nddl duh
K I - : ' '.,.>'
CAM NANG ON LUYEN THI DAI HOC
•
•
•
R^n luien ^ nhanh th
ba mi^ BiC - mm - NAM
HOA HOC
^^M|Ml^^lilHRI
H$thdngcacphuongphapgi^ nhanh bai tap hoa hoc X
Wn chon va gifli thieu cac d6 thi thii nam 2013, cac dfi thi thii ciia cac truong THPT,
cac truong Chujen va cac Trung tarn liij^n thi dai hoc uij tm ctia 3 mien B^c - Trung - Nam
cac detf«d& dupe gi^ctitiet,dehfduvatheo cac phuongphapgi^i nhanh X
: i. 4
H
u u A V I I S ' T D A M T K U R urtD T U A N U D U f f H f ! H H f
MINH
cty TNHH MTV DWH Khang Vi$t
Phan 1.
LOI NOI DAU
Cac em hoc sinh than me'n!
Chiing t6i xin trSn trong gidi thiSu vdri cdc em tap sach:
luyen thi Dai hoc 3 mien Bdc - Trung - Nam Hoa hoc
CAC PHCWUfG P H A P G I A I imAIUI
B A I Xp> HQA 119c
1
Cam nang on
Tap sach la tai lieu cap nhat cho cac em cac bo de thi thir Dai hoc, Cao
dang tren ba mi^n Bac - Trung - Nam de cac em tu ren luyen ky nang lam
nhanh bai thi cho minh. Cac de thi deu dugc chiing toi tuyen chon k i cang,
noi dung bam sat chaong trinh thi Dai hoc, Cao dang va theo dung ca'u true
de thi ciia B6 G D - D T . M 6 i bai tap d^u dugc giai chi tiet, de hi^u, theo
nhieu each va dac biet c6 cac phuong phap giai nhanh, d6c dao.
Tap sach g6m hai phan:
- Ph&i I . Cac phuong phap giai nhanh bai tap hoa hoc.
- PhSn I I . Cac b6 d^ thi thit Dai hoc va hudng dan giai chi tie't.
Chiing toi tin tucmg rang tap sach se cap nhat cho cac em day dii cac dang
de thi tuyen sinh Dai hoc, Cao dang theo hudmg ra de thi cua Bg G D - D T ,
trang bi dSy du cho cac em toan bg kie'n thiic hoa hoc Trung hgc phd thong
va quan trgng hon la mang lai cho cac em su tu tin trong cac ky thi sap tdi.
Dd cuon sach hoan thien hon, rat mong nhan dugc su dong gop y kien
chan thanh ciia cac ban dong nghiep va ciia cac em hgc sinh.
X i n tran trgng cam on!
TAG GIA
r
u
,
PhUCSng phap 1: PHLfONG PHAP BAO TOAN KHOI LUONG
I. LITHUYET
- Gia sir CO phan ling:
A+ B
>C + D
Ta c6:
+ nig =
'
Jii i •! • .
i>
1-
+ triy
- Ap dung: Trong mot phan ling, c6 n chat (ke ca cha't phan ling va san phdm), n€u
biet khdi lirong ciia (n - 1) chat thi tinh dugc khoi lirgng ciia cha't con lai.
II. VAN DUNG
^A.
Thi du 1: H6n hop X gom 0,15 mol vinylaxetilen va 0,6 mol H2. Nung nong h6n
hop X (xiic tac Ni) mot thcfi gian, thu dugc h6n hop Y c6 ti kh6'i so voi H, bang
10. DSn h6n hgp Y qua dung dich brom du, sau khi phan ling xay ra hoan toan,
so mol brom tham gia phan ling la
A. 0,1 mol
B. 0,15 mol
C. 0,05 mol
D. 0,2 mol
(Trick dethi thu Dai hoc khdi A, B nam 2013}
Hu&ngddngidi
H6n hgp X c6: S6'lien ket 7t = 0,15.3 = 0,45 (mol)
Kh6ilugng: m^ =0,15.52 + 0,6.2 = 9(gam)
*
Sdmol: nx =0,15 + 0,6 = 0,75(mol)
Soddphanurng: X(C4H4,H2)-^Y
Nhd Sdch Khang Viet xin trdn trgng gioi thieu toi Quy dgc gid va
xin Idng nghe mgi y kien dong gop de cuon sdch ngdy cdng hay han, bo
ich han.Thu xin gui ve:
^ ,1
Cty TNHH Mot thanh vien Djch Vu Van Hoa Khang Viet.
71- Dinh Tien Hoang, Phirfirng Dakao, Quan 1, TP HCM.
Tel:(08) 39115694 - 39111969 - 39111968 - 39105797
Fax:(08) 39110880
^
Email: Khangvietbookstore
my = m^ = 9(g)
,
= o n Y =9/(10.2) = 0,45(mol)
=>n,,h,ii,. =0,75-0,45 = 0,3(mol) = nH2 (phan umg)
=>n„ (con lai) =0,45-0,3 = 0,15(mol) = nB,^
^,
Dap an diing la B.
Thi du 2: H6n hgp X g6m m6t hidrocacbon o the' khi va H. (ti khd'i hoi ciia X so
v6i H, bang 4,8). Cho X di qua Ni dun nong den phan iJng hoan toan, thu dugc
h6n hgp Y (ti kh6'i hoi ciia Y so v6i CH4 bang 1). C6ng thiic phan tir cua
hidrocacbon la:
||; ||.
,
(,
A. C 3 H 4
B. C2H4
C. C 3 H 6
D. C2H2
|,
(Trich de thi thdDai hoc khdi A, B nam 2013)
3
Ca'm nang fln luy$n thi DH 3 mign BSc - Trmg - Nam mOn H6a hpc - Cu Thanh To^n
Hu&ng ddn gidi
M y =16=> Y g6m ankan C„H2„+2 va
Cty TNHH MTV DWH Khang Vigt
^
Ba(OH)2- Sau cac phan ung thu duoc 39,4 gam ke't tiia va khdi lucmg phSn
du
dung djch giam bdt 19,912 gam. C6ng thurc phan i\i ciia X la:
H2)
Gia sufco 1 mol X (g6m C„H2n+2-2k
A. CH4
C„H2n+2-2k +'^^2
,
'
BandSu:
x
Phan ting:
x
Saupu:
0
^
->
^
(l-x-kx)
'
x (mol)
CO2
x (mol)
H20
ddBa(OH)2
fBaCOj i
Ba(HC03)2
Theo djnh lu^t bao toan khd'i luong => my = m^^ = 9,6(g)
Kettua la BaCO, :nB„co3 =39,4/197 = 0,2(mol)
=>nY =9,6/16 = 0,6(mol)
Taco: mco2 +mH20 =mkt
= > ( l - x - k x ) + x = 0,6=i.kx = 0,4
(l)
=>mco2 +^H20
-^dd{g\)
=39,4-19,912 = 19,488(g)
Mat khac: m ^ = (l4n + 2 - 2k)x + 2(l - x) = 9,6
X + O2 -> CO2 + H2O
^14nx-2kx=7,6
Theo djnh luat bao toan khd'i luong, ta cd: mx +
Tir(l,2)^ — = ^
k
(2)
0,4
'"H20
'
'
/
x + - O2 ->xC02+^H20
Dap an dung la A.
Thidu 3: Nung nong m6t h6n hop g6m CaCO^ va MgO tori khd'i luong khong ddi,
ijjBflfji i r i i
thi so gam cha't rSn con lai chi bang 2/3 so gam hdn hop trudc khi nung. Vay
B. 24,24%
C. 66,67%
fi6b oflc) (}f>'J,b.
4
phdn tram theo khd'i luong ciia CaCO, trong hdn hop ban ddu la
A. 75,76%
= mco2
=> mo2 = 19,488 - 4,64 = 14,848(g) => no2 = 0,464(mol)
= 21 =>n = 3;k = 2 ( C , H 4 )
\
D. C3H4
So dd cac phan ling xay ra:
0 (mol)
kx
, ,
(Trich de thi thADai hoc khdi A, B ndm 2013)
Hu&ng ddn gidi
CnH2n+2
1-x
C. C2H4
B. C4H,o
= ^ m x = 4,8.2 = 9.6(g)
/
4,64
D. 33,33%
12x + y
(Trich de thi thuDai hpc khoi A, B nam 2013)
> 0,464
1
Hu&ng ddn gidi
4,64.
Gia six ban d^u c6 100 gam hdn hop CaC03,MgO
Ta c6:
4
•;<
= 0,464
12x + y
=> Khd'i luong hdn hop con lai sau khi nung la: 100.2/3 = 200/3(g)
Cap nghiem thoa man: x = 3;y = 4 = > X 1^ C3H4
C a C O , — ^ C a O + C02 t
p?, ,
(lib)
Dap an dung la D.
Theo djnh luat bao toan khd'i luong:
mco2 = 100-200/3 = 100/3(gam)^nco2 = 100/132(mol)
^ ncaC03 = 100/132(mol) = nco2
, ^
mcaco., = 10000/132(g)
vay %CaC03 = 10000/132 = 75,76%
Dap an diing la A.
Thi du 4: Ddt chay hoan toan 4,64 gam mdt hidrocacbon X (chat khi d di^u kien
thudng) rdi dem toan b6 san ph^m chay ha'p thu het vao binh dung dung dich
|
Thi du 5: Hdn hop X gdm 0,15 mol CH^C—CH=CH, va 0,6 mol hidro. Nung
nong hdn hop X (xiic tac Ni) mot thdi gian, thu dugc hdn hop Y cd ti khd'i so vdi
khi hidro bang 10. DSn hdn hop Y qua dung dich brom du, sau khi phan ung xay
ra hoan toan, khd'i luong brom tham gia phan ufng la
A. Ogam
B. 24 gam
C. 8 gam
D. 16gam
•
(Trich de thi thi Dai hoc khoi A, B nam 2013)
Hu&ng ddn gidi
Xet hdn hop X c6: n^^^) =0,15.3 = 0,45(mol)
,j
/,
»,)»
,J,H,h nk qpO
Cty TMHH M I V JVVH Khang Vigt
C^m nang On luy?n thi BH 3 mign B&c - Trung - Nam mfln H6a hpc - Cu Thanh ToAn
m x = 0,15.52 +0,6.2 = 9 ( g a m ) ; n x = 0 , 7 5 ( m o l )
Sodo:
X-^Y
I K>:i'IJA.;^ ri;;!o j{r»,/>
Theo djnh luat bao toan khdi lucfng ta c6: m y =
^
Thi du 7: Dot chay hoan toan x gam h6n hop g6m hai axit cacboxylic hai chiic,
ut.c .
= 9(gam)
j^^"'
'
= > n Y = 9 / ( l 0 . 2 ) = 0,45(mol)
=> nf,h
giin,
^ n ,
=> m
= 0,75 - 0 , 4 5 = 0 , 3 ( m o l ) = n^^^ (phan ufng)
c.-^
= 0 , 4 5 - 0 , 3 = 0,15(mol) = n B ^
T,,.'
"
mach ho va deu c6 mot lien ket d6i C=C trong phan tir, thu dugc V lit k h i C O 2
(dktc) va y mol H , 0 . Biau thurc lien he giua cac gia tri x, y va V la
B.V =
C . V = ^ ( x + 30y)
D. V = | ( x + 62y)
.
(Trich de tuyen sinh Dai hoc khdi A)
Huong dan gidi
= 0,15.160 = 24 (gam)
A x i t cacboxylic hai churc, mach ho, c6 m6t lien ket d6i C=C c6 CTPT dang:
Dap an diing la B.
C„H2„_2
Chii y ; V I H2,Br2 trung hoa lien k^'t n nen:
< n'
(COOH)^
l,5y
n^ ( c o n l a i ) =11^^^ (phan ling).
hay
C„,2H2n04
<-
v6i 600 m l dung dich NaOH 1,15M, thu duoc dung dich Y chiia mu6'i cua mot
axit cacboxylic don chiic va 15,4 gam hoi Z gom cac ancol. Cho toan bo Z tac
dung vdri Na du, thu duoc 5,04 lit k h i
(dktc). Co can dung dich Y , nung nong
.cha't ran thu dugc vai CaO cho den khi phan u-ng xay ra hoan toan, thu dugc 7,2
y(mol)
B. 22,60.
C. 34,30.
,
D . 34,51.
(Trich de thi tuyen sinh DH khdi B)
Huong dan gidi
_^V =
• :
'
Dap an dung la C.
,
dung dich H:S04 10%, thu dugc 2,24 lit khi H . (6 dktc). K h d i lugng dung dich
thu dugc sau phan ling la
A . 101,48 gam.
B. 101,68 gam.
C. 97,80 gam.
D . 88,20 gam.
(Trich de thi tuyen sinh DH - CD khdi A)
Hu&ng ddn gidi
2 24
"NaOH (phanurng) = 2 . n H 2 = 0 , 4 5 m o l
Theo bai ra:
=^nNoOH ('li') = 0 - 6 9 - 0 , 4 5 = 0,24mol
> RH T +Na2C0,
= 0 . 1 (mol) =>
0,1.98.100
^
nH2S04=
^ ^'"^^
3,68
mN,oH(phan fl„g) =
=^ m = 40,60(g)
= ^ ' 1 ('"O')
.
> Mu6i + Hjt . 4
Theo dinh luat bao toan khdi lugng:
mkin, loai +
Theo dinh luat bao toan khd'i lugng, ta c6: mx +
"H2
= 98 (gam)
So do phan urng: K i m loai ( A l , Zn) + H2S04(1)
0,24
=> m + 0,45.40 = 0,45.96 +15,4
=
^ d d H2SO4 =
MRH = 7 , 2 / 0 , 2 4 = 3 0 ( C 2 H 6 ) =0 mu6'i la C 2 H , ; C 0 0 N a
Dap an dung la A .
.
22,4.(x + 30y)
28 .
.
iZ. = _ . ( x + 30y)
44
55 ^
^
So d6: X + N a O H -> Y + Z
nCjHsCOONa = "NaOH
;
x + l,5y.32 = — . 4 4 + 18y z=> x + 30y = 4 4 V / 2 2 , 4
22,4
Theo bai ra: nf^^Q^ = 0 , 6 9 m o l ; n H 2 =0,225mol
^
fynii-;
Thi du 8: Cho 3,68 gam h6n hgp g6m A l va Z n tac dung vdri m6t lirgng vijfa dii
gam m6t chat k h i . Gia tri ciia m la
0,24
:lO = r''•;•:,)»!..):.'•
Theo dinh luat bao toan khoi lugng, ta c6:
Thi du 6: D u n nong m gam h6n hop X g6m cac chat c6 cung m6t loai nhom chiic
RCOONa + N a O H
'
Cn+2H2n04 + I ' 5 n 0 2 " ^ ( n + 2 ) C 0 2 + n H 2 0
nH2 (phanu-ng) = n ^ (giam)
A . 40,60.
g(x-62y)
A.V = ^(x-30y)
m„,,, + m,,,,^,
m ^ d H2SO4 =
+
=>m,,,„,i
98
=
"^ddmM
m,,„„,i
+
+
mH2
0,1.2
= 3 , 6 8 + 9 8 - 0 , 1 . 2 = 101,48 (gam).
Dap an diing la A .
^c^.; ^
!
'
^'
elm nang On luy^n thi DH 3 mign B^c - Trung - Nam mOn H(5a hpc - Cu Thanh Toan
Thi du 9: Nung nong 16,8 gam h6n hop gom Au, Ag, Cu, Fe, Zn vdi m6t luong du
khi O2, de'p khi cac phan ling xay ra hoan toan, thu duoc 23,2 gam chat ran X.
Th^ tich dung djch HCl 2M vCra dia de phan umg vdi chat rSn X la:
A. 200 ml
B. 400 ml
C. 800 ml
D. 600 ml
(Trich detuyen sink Cao dang khoiA)
Hu&ng ddn gidi
Soddxayra:
- ^ 2 0 ( o x i t ) - ^ ^ 2 H , 0
m^-^-^.
Theo djnh luSt bao toan khoi luong:
Cty TNHH MTV DWH Khang Vi$t
Hu&ng ddn gidi
88
Theo bai ra:
n^aOH =
0.2. 2 = 0,4 mol ;
ncHjCOOCaH^ = ^
PTPLT: CH3COOC2H5(l) + NaOH(r)
CHjCOONaCr) + C2H,OH t
0,1
Ta tha'y:
"
0,1 mol
nNaOH > ncH3COOC2H5
=^ NaOH du
Sau phan ung c6: CH,COONa va NaOH du.
Theo djnh luat bao toan kh6'i luong ta c6:
mo2 =23,2-16,8 = 6,4g=>no2 =0'2(mo!)
=>n_2 =0,2.2 = 0,4(mol) =:>n ^ =0,4.2 =0,8(mol) = n^ci
o
"
Dap an diing la B.
Thi du 10: Khi d6't chay hoan toan m gam h6n hop hai ancol no, don churc, mach
ha thu duoc V lit khi CO. (a dktc) va a gam H,0.
Bieu thirc lien he gii?a m, a va V la:
V
5,6
.
B. m = 2a -
V
C. m = 2a
.
D. m = 2a +
22,4
+
l,5n02
l,5n(mol)
>
nCOj
n (mol)
+
=>m+
=
22,4
V.44
+a=>m = a-
22,4
Thidu 12: Cho 13,8 gam axit A tac dung vdi 16,8 gam KOH, c6 can dung djch sau
phan ling thu duoc 26,46 gam chat ran. C6ng thiic ca'u tao thu gon ciia A la
A.C,H(,COOH
Sodo: Axit X + K O H ^ Chat ran (mudi, KOH du) + H20
5,6
Theo djnh luat bao toan khd'i luong, ta c6:
V
=^mH20
= 13,8 + 16,8-26,46 = 4,14(g)
Suy ra: M . = —
0,23
f'
" "
,, j,
+ mj^QH - "'(r)
=^nH20
=4,14/18 = 0.23 (mol)
Ne'u axit A don chtic thi n ^ = n^jO = 0,23(mol)
^.^^j ^^^^^^ ^,
= 60 ( C H X O O H )
V 3
;
^
,
' "
Dap an diing la C.
Thi du 13: Xa phong hoa hoan toan 1,99 gam h6n hop hai este bang dung dich
NaOH thu duoc 2,05 gam mu6'i ciia mot axit cacboxylic va 0,94 gam h6n hop
hai ancol la d6ng dang ke tiep nhau. Cong thiic ciia hai este do la
A. HCOOCH3 va HCOOC2H5.
B. C2H5COOCH3 va C2H5COOC2H5.
C . CH3COOC2H5 va CH3COOC3H7.
D. CH3COOCH3 va CH3COOCH5.
5,6
(Trich de thi thu:Dai hoc khoi A, B)
Hu&ng ddn gidi
Thidu 11: Xa phong hoa 8,8 gam etyl axetat bang 200 ml dung djch NaOH 2M.
Sau khi phan irng xay ra hoan toan, c6 can dung dich thu duoc chat ran khan c6
kh6'i luong la
•
.
" -S"**
V
(n+l)H20.
(n+l)(mol)
D.HCOOH
(Trich de thi du bi tuyen sinh Dai hpc khoi A)
Hu&ng ddn gidi
Dap an diing la A.
A.8,2g
C.CH3COOH
B.C2H5COOH
Gia thie't axit tac dung het v6i KOH
11,2
1 5V
Theo PTHH ta tha'y: no, = ISuro^ = -— (mol).
^
2 22,4
Theo dinh luat bao toan khoi luong: m^,,^, + TTIQ^ = m^Q^ + m^^o
1,5V.32
« 1 n-i < - il i
V
( Trich de thi thiiDai hoc khoi A, B nam 2013)
Hu&ng ddn gidi
Ancol no, don churc, mach ho: C„H2„+20.
Phuong trinh hoa hoc d6't chay:
C„H2„,20
1 (mol)
+0,1.46
Dip an dung la C.
=>VddHCi2M =0,8/2 = 0,4(1) = 400(ml)
A. m = a
8,8 + 0 , 4 . 4 0 = m
m = 20,2 gam
B. 8,56g
C. 20,2 g
D. 10,4g
(Trich de tuyen sinh Dai hoc khoi A)
Sodd phan ung: Este + NaOH
Theo djnh luat bao toan khd'i luong:
=> 1,99
+ m^aOH = 2,05
=>
= 0,025 (mol)
IN^OH
+ 0,94 ^
> mu6'i + ancol
m„,e + mN^oH " mmu«i
,^
+ ancol
m^^oH = 1 (gam)
= n^„ (don chiJc)
,
;1\H
,
Cgm nang 6n luygn thi DH 3 mign BSc - Trung - Nam mOn H6a hpc - CD Thanh To^n
Suy ra; M
Cty TNHH MTV D W H Khang Vigt
nH20(2)
= ^
- 79,6 ; M „ „ „ = ^
= 82 ( C H , C O O N a )
0,025
' •
0,025
m,„„,, = 2 5 , 6
C H 3 C O O Q H , (88)
duoc sau phan l i n g la
A.68,950g
B . 19,675g
C. 13,075g
(Trich
Huong
G i a t r i Ion nhat ciia V l a
de thi thu Dai hoc khoi A, B)
A . 22,4.
B . 11,2.
C. 5,6.
( Trich
^
Hu&ng
niHci = m d d - C % = — 100
— — = 5,4/3g
Hi A a/
5,475
,
=> n ^ c i =
= 0,15 m o l
36,5
ddn
nv
=
37,6
9,4.4
= l(mol)
—^
1(aiidehil phan lillg) ~ Idiiilro ph,iu I'nig) ~
—^
l(ancoldaiichifclaora) — 1
Ni,."
1
(mol)
(mol)
RCHO+H.
mco2 ( b a y r a ) = 0 , 1 5 . 4 4 = 6 , 6 g
R C H 2 O H + N a - > R C H 2 O N a + 0,5H2 t
vay
m
muoi clorua
"^C02
+ " 1 H 2O (>ao ra)
= 14,2 + 5 , 4 7 5 - 2 , 7 - 6 , 6
rt'i
0,5(mol)
V = 0 , 5 . 2 2 , 4 = 11,2 ( l i t )
' ^
D a p an d u n g la B .
= 10,375g
Chu
D a p an d i i n g l a D .
Thidu
>RCH,OH
l(mol)
T h e o d i n h luat bao toan kh6'i l u o n g :
muoi clorua +
, jj,,^ j j„.j|| r,, ,
Suy ra so m o l k h f g i a m 2 - 1 = 1 ( m o l )
= 0,15 m o l
" ^-'5 mol
+ "^HCl
gidi
V i khd'i l u o n g d u o c bao toan, nen: m y = m ^ = 3 7 , 6 ( g )
cua cac phan l i n g NH4HCO.,, N a H C O ^ v a K H C O , v o l H C l la:
hoc khoi A, B)
,
Suy ra: mH20 O^o ra) = 0,15. 18 = 2,7 g
"^hh
D . 13,44.
de thi thvcDqi
T a c o : m x = 2 . ( 4 , 7 . 4 ) = 37,6 (g)
> H.O + C O , t
=
, ( . ot, <,A:.
+ 0,2 . 4 0 + 0,1 .1 71 — 0,4 . 18 = 43,5 ( g )
la'y toan b o cac a n c o l t r o n g Y r o i c h o tac d u n g vdri N a ( d u ) , d u o c V l i t H , ( d k t c ) .
D . 10,375g
Theobaira:
"C02
+ "IH20
n o n g 2 m o l X ( x u c tac N i ) , d u o c h 6 n h o p Y c 6 t i k h o i h o i so v o i h e l i l a 9,4. T h u
ddn gidi
54,75.10
=
= 0 , 2 + 0,2 = 0 , 4 ( m o l )
phan tir ddu c 6 so n g u y e n t u C n h o h o n 4 ) , c 6 t i k h o i so v6\i l a 4,7. D u n
cdn vira d u 54,75 g d u n g d i c h H C l 1 0 % . Kh6'i lironig h 6 n h o p m u o i c l o r u a t h u
T h e o phan u n g : n n ^ o
HH^O
Thi du 16: X l a h 6 n h o p g o m H , v a h o i c u a hai andehit ( n o , d o n churc, m a c h h o ,
Thi du 14: D e h o a tan hoan toan 14,2 g h 6 n h o p N a H C O , , NH4HCO, v a K H C O ,
HCO," + H '
=^
D a p an d u n g l a B.
D a p an d i i n g l a D .
PTPir t h u g o n
= 2.0,1 = 0 , 2 ( m o ! )
T h e o d j n h luat bao toan k h o i l u g n g : m , , , , + m N , o H + m B a ( O H ) 2 =
V a y 2 este d o l a C H , C O O C H , ( 7 4 )
: 'J,A
=2nBn(()H)2
v: So m o l h6n h o p k h i g i a m = so m o l H , phan u n g = s6' m o l a n d e h i t d o n c h i i c ,
no, phan u n g = s6' m o l ancol d o n chiic tao ra.
'
^
15: D e t r u n g hoa 2 5 , 6 g a m h 6 n h o p 2 axit c a c b o x y l i c d a chuc cSn d u n g 1 l i t Thi du 17: D o t c h a y h o a n toan m g a m h6n h o p X g6m ba ancol ( d o n chiJc, thuoc
c u n g d a y d6ng d a n g ) , t h u d u o c 8,96 l i t k h i C O , ( d k t c ) v a 11,7 g a m H . O . M a t
d u n g d i c h h 6 n h o p N a O H 0 , 2 M v a Ba(0H)2 0 , 1 M . Sau phan l i n g c 6 c a n t h u
k h a c , neu d u n n o n g m g a m X v 6 i H2SO4 dac t h i t6ng k h o i l u o n g ete t6'i da t h u
duoc k h o i luong m u o i khan la
A . 34,4 g a m
B. 43,5 g a m
C. 4 1 , 6 g a m
( Trich
Hu&ng
Theobaira:
PTPlT:
duoc la
A . 5,60 g a m
thu Dai hoc khoi A, B)
C. 7,85 g a m
(Trich
Huong
nNaOH = 0 ' 2 n i o l ; nBa(OH)2
R(COOHX, + aNaOH
B. 6,50 g a m
ddn gidi
T h e o bai ra: n^o^ = 0 , 4 m o l ;
R(COONaX, + aH.O
2R(COOH)„ + aBa(OH), ^
Tatha'y:
dethi
D. 60,6 gam
=nNaOH = 0 ' 2 ( m o l )
ddn
dethi
D . 7,40 g a m
thu Dai hpc khoi A, B)
gidi
nH20 = ^ ' ^ 5 m o l .
(1)
l R ( C O O ) J , l B a ] , + 2aH20 ( 2 )
"
^ ' "H2O >"co2 ^
-dncol
nay no, m a c h h o ( d o n churc) d a n g
Cj^H2n+|OH
va s6 m o l h6n h o p a n c o l n^h = ny^^Q - nco2 = ^ ' ^ ^ — ^ ' ^ ~ ^ ' ^ ^ ("^"'^
dm
Cty TNHH MTV D W H Khang Vi$t
nang On luygn thi DH 3 m\in B&c - Trung - Nam mOn H6a hpc - Cii Thanh Togin
^
=> ni(bi„h brom ,a„g) = 1 , 3 2
Dap an dung la D .
0,25
'hh
Scrddtaoete: R O H + R ' O H
Theo
s0
do:
0,06.26 + 0,04.2 = 0,02. 16 + m
"^'^^
(gam).
M ' i •; -
ium)
^.^^
JBAU
v-.K'.
> R - O - R ' + H.O
—
1
0,25,
(mol):
nH20 = - l a n c o i =
H™.
^HjO
Theo djnh luat bao toan kh6'i luong: m„„„| =
=
0,25.18
^
^
= 2,25 (gam)
Phaong phap 2: PHl/ONG PHAP BAO TOAN ELECTRON
I. L I T H U Y E T
+ mH20
-
Trong phan ling oxi hoa - khu, xay ra ddng thdi qua trinh oxi hoa va qua trinh
-
Djnh luat bao toan electron: Trong phan ilng oxi hoa khir, tdng sd electron do
=> 0,25(14n + 18) = m + 2,25 => 0,25(14.1,6 + 18) = m + 2,25
=> m = 7,85
(gam)
chat k h u cho phai diing bang tdng sd electron do chat oxi hoa nhan:
Dap an dung la C.
Thi du 18:
Cho mot lucmg bot Zn vao dung dich X g6m FeCl. va CuCl.
Kh6'i
lugng chat rSn sau khi cac phan ting xay ra hoan toan nho hcfn khdi luong bdt
Zn ban dSu la 0,5 gam. C6 can pMn dung dich sau phan ling thu duoc 13,6 gam
B. 17,0 gam.
C. 19,5 gam.
D . 14,1 gam.
Huong ddn gidi
FeCl2
+ Zn-
CuCl,
-> Z n C l j +
Da'u hieu de nhan ra bai tap c6 thd' su dung phuong phap bao toan electron de
giai la cac bai tap cd phan ling oxi hoa - khu.
Thidu 1: Cho hoi nude di qua than nong do thu dugc 15,68 lit (dktc) hdn hgp khi A
( Trich de thi thu: Dai hpc khoi A, B)
Ta CO so 66 phan ling:
-
11. V A N D U N G
mu6'i khan. T6ng khoi lugng cac mu6'i trong X la
A . 13,1 gam.
2:e(cho)= Se(nhan)
gdm C O , C O o va H o . Cho toan b6 A tac dung het vdi hdn hgp M g O , CuO du, nung
nong thu dugc hdn hgp chat rdn B . Hoa tan toan bg B bang HNO3 dac, ndng, du
Fe
dugc 26,88 lit N O , (san pham khu duy nhat, dktc). Sd mol khi H , trong A la
Cu
A . 0,4.
B.0,2.
m
hh(FeCl2,CuCl2
"^Z"
~ ^ ZnCI 2
) +mz„ =
m' ,h h ( F e a 2 , c u c i 2 ) =
'3,6
-
0,5
+(mz„
Theo bai ra: n ^ = 0 , 7 ( m o l ) ; n N O 2 " ' ' ^ ( m o l )
-0,5)
=13,1
C 4- H2O ^
(gam).
Thi du 19: D u n nong hdn hgp khi gdm 0,06 mol C,H2 va 0,04 mol H . vdi xiic tac
N i , sau mot thdi gian thu dugc hdn hgp khi Y. DSn toan b6 h6n hgp Y Idi tCr ttr
qua binh dung dung dich brom (du) thi con lai 0,448 lit hdn hgp k h i Z ( d dktc)
CO ti khdi so vdi O j la 0,5. K h d i lugng binh dung dung dich brom tang la
C. 1,04 gam.
Nhan xet: K h d i lugng binh dung dung dich brom tang chinh bang khdi lugng
etilen, axetilen cd trong hdn hgp Y.
Mz = 32. 0,5 = 16; n^ = 0,448/ 22,4 = 0,02 (mol).
m c j H 2 + n i H 2 = my
= mz
+ m ,y„h brom t.i„g)
y
C
+ 2H2
_>
2y
(l)
+1
-
H2
->2x
1,2<-
rv?
2e
^
2H
(x + 2 y ) ^ 2(x + 2 y )
N
1,2
Theo djnh luat bao toan electron, ta cd: 2x + 2(x + 2y) = 1,2 => 4x + 4y = 1,2
• Hu&ng ddn gidi
Theo djnh luat bao toan khdi lugng, ta c6:
+4
C - 2e ^
N + le
D . 1,32 gam.
( Trich de thi thii Dai hoc khoi A, B)
!
-^x
+2
X
^ " •<
,^ ^ C + 2H2O ^ CO2
: ^ x + x + y + 2y = 0 , 7 ^ 2 x + 3y =0,7
Dap an dung la A .
B. 1,20 gam.
CO + H2
x
Vay tdng khdi lugng cac mudi trong X bang 13,1 gam.
A . 1,64 gam.
2013)
Hu&ng ddn gidi
'^hh(Fe, Cu)
13,6
D.'o,5.
(Trich dethi thu: Dai hpc khoi A, B nam
Theo djnh luat bao toan khdi lugng, ta c6:
r"hh(FeCl2,CuCl2)
C.0,3.
A.
= > X + y = 0,3
(2)
^
Tilf(l,2) = > x = 0 , 2 ; y = 0 , l .
'
Vay sd mol khi H2 la: x + 2y = 0 , 2 + 2.0,1 = 0 , 4 (mol)
Dap an dung la A .
,
y,,v,..
'
,
,.
-.M.u
ltr,ubo'.,n
elm nang On luy^n thi DH 3 mjgn BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan
^
Thi du 2: Hoa tan hoan toan 2,44 gam h6n hop hot X gom Cu va Fe.Oy bang dung
dich H 2 S O 4 dac, nong, du. Sau phan urng thu duoc 0,504 h't khf SO, (san ph^m
khir duy nhat, dktc) va dung dich chiia 6,6 gam h6n hop muoi sunfat. Phan tram
khoi luong Cu trong X la
A. 26,23%.
B. 39,34%.
C. 65,57%.
D. 13,11%.
{Trich de thi thutDqi hoc khoi A, B nam 2013)
Huong dan gidi
Quy d6i h6n hop Cu,Fe^Oy
Cu,Fe,0
C u - 2 e ^ C u ^ * :./?•!•..<::
i
,
/
+
6
+
S+2e
c->2c
Matkhac, taco: 64a + 56b + 16c = 2,44
2Fe
-> Fe2 ( S O 4 )3
a ^
a
b
->
HCl 2M, thu duoc dung dich Z. Cho AgNOy
B. 76,70%.
du vao dung dich Z, thu duoc
C. 53,85%.
D. 56,36%.
(1)
•.-'](•
H2O
O2
0,24->0,12
"
=>y = 0,06
Thi du 3: Hoa tan hoan toan 24,8 gam h6n hop X gom Fe va Fe.Oy bang dung djch
H 2 S O 4 dac, nong, du thu duoc dung dich Y va 4,48 lit khi S O 2 (san phim khir
duy nhat, dktc). Phan tram khd'i luong nguyen to oxi trong X la
A. 20,97%.
B. 16,84%.
C. 25,73%.
D. 32,56%.
( Trich de thi thu Dai hoc khoi A, B nam 2013)
Huong ddn gidi
Theo bai ra: n^Q^ =0,2 (mol)
z
->
<-
(2)
20^"
0,12
•
Ag+ +
^ z
Cr
,
.
-> A g C U
( 2 x + 0 , 2 4 ) - > (2x +0,24)
^ 108.Z + 143,5.(2x + 0,24) = 56,69
^ 1 0 8 z + 287x = 22,25
(3)
Tir (1, 2, 3) => x = 0,07;y = 0,06;z = 0,02
U2/hh
02/hh
0,07
r
+ 0,06
Dap an diing la C.
"
Thi du 5: Cho h6n hop g6m 6,72 gam Mg va 0,8 gam MgO tac dung het v6i luong
du dung djch HNO3. Sau khi cac phan ling xay ra hoan toan, thu duoc 0,896 lit
m6t khf X (dktc) va dung djch Y. Lam bay hoi dung djch Y thu duoc 46 gam
mu6'i khan. Khf X la
Quy d6i h6n hop Fe.Fe^Oy => Fe va O .
O + 2e ^ O^"
I,
b - > 2b
A. N O ,
0,4<-0,2
+ 4e
0,06
Ag+ + l e - > A g i
Dap an dung la A.
S
g6m clo va oxi, sau phan dng chi thu duoc h6n hofp Y g6m cac oxit va muoi
2H+ + O^- ^
Vay %mcu/x =26,23%
+4
Thi du 4: Dot chay h6n hop g6m 1,92 gam Mg va 4,48 gam Fe vdi h6n hop khf X
=:>2x + 4y + z = 0,4
0,5b
T i r ( l , 2 , 3 ) =^a=0,01;b = 0,025;c-0,025
a ->3a
, ,^,.,n.
Theo nguyen tdc bao toan electron, ta c6: 0,08.2 + 0,08.3 = 2x + 4y + Iz
=^ 160a + 400.0,5b = 6,6 =^160a + 200b = 6,6 (3)
Fe^*
,
Goi x,y,z iSn luot la s6'mol ciia Cl2,02 va Ag"^ (tao ra Ag).
CUSO4
S + 2e
= 20,97%
(1)
(2)
Cu ^
+6
.
(Trich de thi tuyen sink DH khoi B)
Huong ddn gidi
^
Theo bai ra: n^, =0,08mol;np^ =0,08mol;np.p| =0,24mol
-> S
0,045 <-0,0225
Fe - 3e
Vay %mo/x =
Dap an diing la A.
A. 51,72%.
4
Theo nguydn tac bao toan electron ta c6: 2a + 3b = 2c + 0,045
,
0,325.16.100%
56,69 gam ket tua. Phan tram the tfch ciia clo trong h6n hop X la
b ->3b
0 +2e^0^-
TLr(l,2) ^ a = 0,35;b = 0,325
clorua (khong con khf du). Hoa tan Y bang mot luong vira dii 120 ml dung dich
Fe-3e^Fe^^
a -> 2a
Cty TNHH MTV DVVH Khang Vigt
J.-xc-
Theo djnh luat bao toan electron, ta c6:
3a = 2b + 0,4
(1)
Matkhac:
56a + 16b = 24,8
(2)
B. NoO
D. N O
(Trich de tuyen sinh Cao dang khoi A)
Huong ddn gidi
, ^y,,
Theo bai ra: n^^, =0,28mol; n^go =0,02mol; nx =0,04mol
j , • ,)
C.N,
15
Cty TNHH MTV DWH Khang Vi$t
Ca'm nang On luygn thi DH 3 mign BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan
•"Mg(N03),
^mo2
=nMg+nMgO=0.3mol
= 0,3.148 = 4 4 , 4 ( g ) < 4 6 g
• m'Mg(N03)2
.
^
Bi6't trong dung dich thu duoc khdng cd mudi NH4NO3, k i m loai M la
A.Zn
+5
+2
0,28
-3
N + 8e
0,56(mol)
+5
0,16
N
B. Cu
C. A g
-3
NH4NO3
0,02 ( m o l )
Hu&ng ddn gidi
*
+(S-n)
*
2N+iOe*
Dap an diing la C.
Thi du 6: Cho 24,15 gam h6n hop khi A gom clo va oxi phan ung vtra het vdi mot
x^'
x + y = 0,06
^
r > x = 0 , 0 2 ; y = 0,04
44x + 30y = 2,08
0
:
•• ^t.
Qua trinh nhudng, nhSn electron:
M
)-M'^+ne
2N
h6n hop gdm 4,8 gam magie va 8,1 gam nh6m tao ra h6n hop cac mudi clorua va
oxit ciia hai k i m loai. Thanh ph^n % theo khdi luong ciia oxi trong hdn hop A la
^'^
M
A . 26,5%
+5
+2
N + 3e ^
N
B. 32,0%
C. 20,8%
D . 16,8%
(Trich de thi thu Dai hoc khoi A, B)
Hu&ng dan gidi
^'^"-(mol)
M
(1)
Cac qua trinh xay ra trong phan ung tren:
24
Al
Thi du 8: Cho 2,16 gam M g tac dung vdi dung dich H N O , (dir). Sau khi phan umg
O. + 4e
y
W
0,4 mol
+
= 0,3 mol
xay ra hoan toan thu duoc 0,896 1ft khi N O (0 dktc) va dung djch X . K h d i luong
mudi khan thu duoc khi lam bay hoi dung dich X la
B. 13,92 gam.
C. 6,52 gam.
D . 13,32 gam.
(Trich de thi thtiDai hoc khoi A, B)
Hu&ng ddn gidi
,0 *SH,^M
Theo bai ra: n^^ =
3e
24
8J
27
4y
-> 20-
A . 8,88 gam.
> M g ^ - + 2e
= 0,2 mol
•
,
Vay n = 2, M = 65 (Zn).
4,8
<-
+1
2N
0,02.2
Theo nguyen tac bao toan electron ta cd: 9,1. n / M = 0,16 + 0,12 = 0,28
Dap an dung la A .
Mg
>
•*
T h e o b a i r a : 7 1 x + 32y = 24,15
2X
2.4e
0,16
...I
So do: hh A + hh CU, O, - > hh mudi clorua + oxit
2Cr
+
0,12 < - 0 , 0 4
Dat x, y Idn luot la sd mol CI., O, trong hdn hop A .
X
•
.,,,.n = ;
Xac djnh sd mol N2O ( x m o l ) , N O ( y m o l )
Theo nguyen tSc bao toan electron, taco: 0,56 = 0,16 + 0,08n =>n = 5
>
>* • •
XacdinhX, Y:
= > X l a N 2 0 ( 4 4 ) ; Y la N O ( 3 0 )
0,08n < - 0,04 ( m o l )
CI3 + 2e
D. A l
(Trich de thi thu Dai hpc khoi A, B)
2 N + 2ne
Vay khi X la N , :
,
0,06 mol hdn hop 2 khi X va Y , cd khdi lirong 2,08 gam vdi M^/My = 1,467.
nNH4N03 = 1-6/80 - 0 , 0 2 mol
M g + 2e
+
Thi du 7: Cho 9,1 gam k i m loai M tan hS't vao dung dich H N O , loang, d u thu duoc
Cac qua trinh oxi hod - khu xay ra:
Mg
= ^ ^ . 1 0 0 = 26,5%
Dap an diing la A .
Suy ra trong dung dich Y c6 mudi NH4NO3
mNH4N03 = 46 - 44,4 = l,6g
= 3 2 . y = 32.0.2 = 6,4g ^%mo2
Mg
Theo dinh luat bao toan electron, ta cd:
T i r ( l ) , (2) ta cd: x = 0,25 ; y = 0,2
= 0,04 (mol). ,v,
Xay ra cac qua trinh:
0,9 mol
2 x + 4y = 0,4 + 0 , 9 = 1,3
= 0,09 (mol); n^o = -ir^
22,4
(2)
-
0,09
^
2e
0,18(mol)
>
+2
Mg
0,09 (mol)
I n . , . , . , , = 0,09. 2 = 0,18 (mol)
I
r W J VicAJ ItNH i ^ l N H THUAN
,
JAiflu
JQ
WyLlSJl^L^
dm
nang On luyjn thi BH 3 niien BJc
i m n y - Nam mOn H6a hgc - Cu Thanh Tocin
+5
N
Cty TNHH MTV DWH Khang Vijt
+2
+
3e
>
0,12 (mol)
N
0,04
.
=> ms(x) = 0,025.32 = 0,8(gam) => mp^^^) = 1,92 - 0 , 8 = 1,12(gam>
,
=^np,(x) = ' ' 1 2 / 5 6 = 0,02(mol)
= > S n , ( N n h a n ) = 0,04.3 = 0,12 (mol)
Suyra: I n ,
„ho)
> 2:n,
,„hi,„
=> c6 san ph^m NH4NO3:
+5
N
' VI
'
-3
+
8e
>
(0,18-0,12)
N
'J
lO.B
.A
0,02 ->
N
mNH4NO3 = 0'09.148 +
+
-
0,025 - > 0 , 1 5 ^ _ , 5 Q
0,06
0,0075.80=13,92(gam).
Thi du 9: Hoa tan het 0,03 mol m6t oxit sat c6 c6ng thiic Fe.Oy vao dung dich
HNO3 loang du thu duoc 0,01 mol m6t oxit nito c6 c6ng thiic Nfi, (san ph^m
khir duy nha't). M6'i quan he giila x, y, z, t la
A. 2 7 x + 1 8 y = 5 z - 2 t
B. 9x - 8y = 5z - 2t
C. 3x - 2y = 5z - 2t
D. 9x - 6y = 5z - 2t.
(Trich dethi thuDai
Huong ddn gidi
Ta CO cac qua trinh nhu5ng - nhan electron nhu sau:
+2y/x
I,
>N(N02)
V
V
22,4
22,4
V
Theo nguyen tac bao toan electron, ta c6: 0,06 + 0,15 =
=> V = 4,704 (lit)
22,4
Dap an diing la A .
Thi du 11: Hoa tan hoan toan 8,862 gam h6n hop g6m A l va Mg vao dung djch
HNO3 loang, thu duoc dung dich X va 3,136 lit ( 0 dktc) h6n hop Y g6m hai khi
hoc khoiA,
B)
khong mau, trong do c6 m6t khi hoa nau trong khdng khi. Kh6'i luong ciia Y la
5,18 gam. Cho dung djch NaOH (du) vao X va dun nong, khong c6 khi mui khai
thoat ra. Phin tram khoi luong ciia Al trong h6n hop ban 6iu \k
+3
A. 19,53%
x Fe - ( 3 x - 2 y ) e - > x F e
B. 10,52%
C . 12,80%
D. 15,25%
(Trich de thi thvcDai hgc khoi A, B)
+2t/z
zN+(5z-2t)e->z
uTl
S - 6e—>S^(so2-)
vFe'^
+ le
Dap an diing la B.
+3
3e
; dM
0,0075 (mol)
Vay khd'i luong chat ran khan thu duoc:
m=mMg(N03)2
Qua trinh cho - nhan electron:
Fe
'
^
N
Hu&ng ddn gidi
Theo nguydn tac bao toan electron ta c6: 0,03.(3x - 2y) = 0,01 .(5z - 2t)
Theobaira: n^ =3,136/22,4 = 0,14mol
=> 9x - 6y = 5z - 2t
Trong Y c6 N O (khi bj hoa nau trong khong khi:
Dap an dung la D.
2NO + 0 2 ^ 2NO2 do nau).
Thi du 10: Nung 1,92 gam h6n hop bot X gom Fe va S trong binh km khong c6
•
Khf con lai trong Y c6 the la N , hoac N , 0 (deu kh6ng c6 mau). T a c6:
kh6ng khi, sau mot thoi gian thu duoc chat ran Y . Hoa tan het Y trong dung
djch H N O , dac, nong, du thu duoc dung djch Z va V lit khi thoat ra (dktc). Cho
Z tac dung vdri dung dich BaCK du thu duoc 5,825 gam ket tiia. Gia tri ciia V la
A. 4,704
B. 1,568
C . 3,136
D. 1,344
( Trich de thi thu Dai hoc khoi A, B)
Vi
=
5,825
Ta c6: <
= 0,025(mol)
"
So do phan ling:
•»
•
•
'-va'hup
x + y-0,14
30x + 44y = 5,18
}•
233
Giai he ta duoc: x = 0,07; y = 0,07
Cac qua trinh oxi hoa - khij xay ra:
Fe + S->.FeS,Fe,S
(X)
Tatha^y:
18
(Y)
>Fe^+,S0^-,H+,N03 ^ B a S 0 4 ^
(Z)
ng^^) f ^8*504 = 0,025(mol)
,
Xac dinh s6 mol N O (x mol), NoO (y mol) trong Y .
Hu&ng ddn gidi
Theo bai ra: n^^^^
= 30 < 37 =^ khi con lai c6 M > 37, vay khi do la N.O ( M = 4 4 ) . ,
0
Al
a(mol) - >
. c<•
•''
0
>
AP+ +3e
3a
Mg
b(mol) ->
> Mg^+ + 2e
2b /* f
19
Cty TNHH MTV D W H Khang Vigt
Ca'm nang Qn luyQn thi DH 3 mign B j c - Trung - Nam mOn H6a hgc - Cu Thanh To^n
+3
N
+2
+ 3e - »
+S
N(NO)
0,21
2N
+ 8e
0,07
0,56
+1
N2 (NjO)
0,07
.i«
Ta c6: 27.a + 24.b - 8,862
3a + 2b = 0,21+0,56 = 0,77
=^
24,3n= 2,7M =^ 9n = M .
^s^p
_
DapandiinglaC.
y
y
^ 1 i
Thi du 14: Chia m gam A l thanh hai phSn bang nhau:
- PhSn m6t tac dung vori luong du dung dich NaOH, sinh ra x mol khi H2;
- PhSn hai tac dung v6\g du dung djch HNO3 loang, sinh ra y mol khi N^O
(san phaim khir duy nha't).
Quan he giCra x va y la
f ,n
A . x = y.
B.y = 2x.
C. x = 2y.
D . x = 4y.
(Trich dethi thi Dai hoc khoi A, B)
Huofng ddn gidi
So d6 cac qua trinh:
+5
2H" + 2e
Goi X la s6'mol N , => n N o = 2 x ; n N 2 0 = x ( m o I )
A l - 3e ^
2N+10e^N2
Al^^
0,4-). 1,2 (mol)
+2
)•
+1
2 N + 8e
> IN
(RO)
,
8y <• y (mol)
X (mol)
2x x = 4y.
Cac qua trinh nhucmg - nhan electron.
+5
3
27(A1)
thoa
a = 0,042; b = 0,322
,
0,042.27.100%
d0.0 . - M
Vay: %m,,=—
= 12,80%
.
, ' )»*
3+
8,862
r r n « > ' / . ,: .,! .
V
Dap an dung la C.
Thidu 12: Cho 10,8 gam b6t Al tan hoan toan trong dung djch H N O 3 tha'y thoat ra
3 khi Nn, NO va N2O c6 ti le mol tuong l i n g la 1: 2: 1. Trong dung dich thu duoc
khdng CO NH4NO3. The tich 3 khi tren (dktc) la
A. 2,24 lit
B. 3,36 lit
C. 4,48 lit
D. 6,72 lit
( Trich de thi tkuDqi hoc khoi A, B)
f
Hu&ngddngidi
; Theo bai ra n ^ , =0,4(mol)
'
2
18
loai
1
9(Be)
loai
n
M
KL
Dap an diing la D.
lOx <-x
+5
'
+1
Phaong phap 3: P H a O N G P H A P B A O T O A N DIEN T I C H
2 N + 8 e ^ N 2 (NjO)
N + 3e-).N(N0)
6x <-2x
8x <-x
Theo nguyen tSc bao toan electron: 1,2 = lOx + 6x + 8x = 24x
=> \ 0,05 (mol)
L
LITHUYET
Trong he c6 lap, t6ng so didn ti'ch am va t6ng s6' dien tich duong c6 tri s6' bang
nhau: |Eq(-)| = | l q ( + )
vay V = ( X + 2x + x). 22,4 = 4. 0,05 . 22,4 = 4,48 (lit)
Dap a n diing la C.
,
n,.q,
Thidu 13: Cho 24,3 gam m6t kim loai M (c6 hoa trj n duy nha't) tac dung v6i 5,04
lit khi O. (6 dktc) thu duoc chat rSn A. Cho A tac dung het v6i dung dich HCl
tha'y CO 1,8 gam khi
thoat ra. Kim loai M la
A. Mg.
B.Zn.
C. A l .
D. Ca.
( Trich dethi thi Dai hoc khoi A, B)
Hit&ng dan gidi
5^^^^^^,,
Theo bai ra: no^ = - ^ = 0,225(mol);
22,4
1,8
n^^ = — = 0,9(mol).
Theo dinh luat bao toan electron, ta c6:
= 0,9 + 1,8 = 2,7
+
+n2q2
+
+n3q3
_
_
+... = nf.q, +n2'.q2
+n3'.q3
+...
=> Sqj^.nj = SqT-"!
II. VAN DUNG
Thi du 1: Dung dich X c6 chiia: 0,07 mol Na"; 0,02 mol S O 4 " va x mol OH".
Dung djch Y c6 chiia C104,N03 va
y mol
t6ng s6' mol C I O 4 va NO3 la
0,04. Tr6n X va Y duoc 100 ml dung djch Z. Dung djch Z c6 pH ( bo qua sir
dien li cua H^O) la
A. 1
B.2
C. 12
D. 13 '
'
(Trich de thi thi Dai hoc khoi A, B nam 2013)
M
21
d m nang On luy$n thi DH 3 mign Bjc - Trung - Nam mOn H6a hpc - Cii Thanh Toan
Cty TNHIi MIV DWH Khang Vi^t
Huong ddn gidi
Ap dung dinh luat bao loan dien tich, ta c6:
;
Dung dich X: 0,07.1 = 0,02.2+ x . l =>x=0,03(mol) .
i / V j . .
Dung dich Y: 0,04.1 = y. 1 => y = 0,04 (mol).
PTPlT:
Ml
+ OH"
Ban dSu:
,;J
y
Phan ung:
""
0,03 -> 0,03
C6n:
=0,01/0,1 = 0 , 1 M
.,3
X
0,01
m=>
"i
U
mi'f^
Huong ddn gidi
Theo djnh luat bao toan diSn tich, ta c6: 0,02.2 + 0,03.1 = x. 1 + y. 2
,
=^ x + 2y = 0,07
(1)
Mat khac: T(5ng khd'i luong muoi bang t6ng khO'i luong cac ion:
,
0,02. 64 + 0,03. 39 + 35,5. x + 96. y = 5,435
'
^ 35,5 X + 96y = 2,985
(2) ^'^ S«o •'
T i r ( l , 2 ) : x = 0,03; y = 0,02.
,
Dap an dung la C.
,
,(''"!•
Thi du 4: Dung djch E g6m x mol Ca'*, y mol Ba'*, z mol HCO3 . Qio tif tiT dung
0
djch Ca(OH)2 nong d6 a mol/1 vao dung djch E den khi thu duoc luong k6't tiia
^ p H = -lgO,l = l
\6n nha't thi vira het V lit dung dich Ca(0H)2. Bieu thiic lien he giira cac gia tri
Dap an diing la A.
Thi du 2: Hoa tan h6n hop mu6'i vao nu6c duoc dung dich X chiia 3 ion trong s6'
cac ion sau:
Ion
Na"
Ba^^
cr
NO3-
Ag^
SO4--
S6'mol
0,2
0,1
0,3
0,4
0,1
0,2
V, a,
X,
y la
A.V = ^ ^ .
a
B. V = a(2x + y)
C. V = 2a(x + y ) .
(Trich de thi tuyen sinh Dai hoc khoi A)
Hu&ng ddn gidi
Dung djch X chiia 3 ion la
Theo djnh luat bao toan didn tich ta c6: z = 2x + 2y
A. Na^ A g ^ N O , -
B . Na\^ C f
C. Ba-^ Ag\-
D. V = ^ ^ ^ ^ .
a
D. Na^ Ba-^ NO,"
(Trich de thi thii Dai hoc khoi A, B nam 2013)
HCO3 + OH" -> CO^" + H2O
I
z -> z
=>z = 2.V.a
Huong ddn gidi
(l)
(2)
i:.V;<,
•
Phuong phap di giai nhanh bai nay la thir cac phuong an:
+ Cac ion c6 ciing t6n tai trong mot dung djch duoc khong (dieu kiSn 1).
+ Dung dich c6 trung hoa dien kh6ng (dieu kien 2).
Phuong an A, B, D thoa man di^u kien 1.
T i r ( l , 2 ) =>2x + 2y = 2.Va=>x + y = V.a ^ V = ^ ^
a
Dap an diing la A.
Phuong an C khdng thoa man diSu kitn 1 (loai).
Chu v: Luong ket tiia lom nha't khi toan bo luong HCO3 chuye'n het thanh CO3" .
Ap dung dinh luat bao toan didn tich:
Thidu 5: Dung dich X g6m 0,1 mol H", z mol A l ' " , t mol NO3 va 0,02 mol SO4".
-
dphuomgan A: |0,2.1(+) + 0,l.l(+)|7i|0,4.1(-)| (loai)
-
6 phuong an B: |0,2.I(+) + 0,1.2(+)|^|0,3.1(-)| (loai)
-
d p h u o n g a n D |0,2.1(+) + 0,1.2(+)| = |0,4.1(-)| (thoaman)
504-^. Tdng kh6'i luong cac mu6'i tan c6 trong dung dich la 5,435 gam. Gia trj
Cho 120 ml dung dich Y g6m KOH 1,2M va Ba(0H)2 0,1M vao X, sau khi cac
phan ling ket thiic, thu duoc 3,732 gam ke't tua. Gia trj ciia z, t Mn luot la
A. 0,120 va 0,020
B. 0,012 va 0,096
C. 0,020 va 0,120
D. 0,020 va 0,012
(Trich de thi tuyen sinh Dai hoc khoi A)
Hu&ng ddn gidi
Theo bai ra: n^oH =0,144mol; ng^lQPij^ =0,012mol
ciia X va y lin luot la
=>n
Vay dap an dung la D.
Thi du 3: M6t dung dich chiJa 0,02 mol Cu"*, 0,03 mol K*, x mol CP va y mol
A. 0,01 \k 0,03.
B. 0,05 va 0,01.
^
C. 0,03 v i 0,02.
D. 0,02 va 0,05.
(Trich de thi thu! Dai hoc khoi A, B)
=0,144 + 0,012.2 = 0,168mol;n
OH~
,+=0,012mol
-
Ba^+
Theo djnh luat bao toan dien tich, ta c6: 0,1.1 + z.3 = t.l +0,02.2
23
ca'm nang On luy^n thi DH 3 m'lin BSc - Trung - Nam mOn H6a hpc - CO Thanh Toan
= > t - 3 z = 0,06
^
(1)
.
"ea^- < " S 0 2 - ^ " B a S 0 4 i = " B a 2 - =0,012mol
=^"^8.80,1
Cty TNHII MIV UVVII Kliaiig Vi$t
_ ^
vay trong X c6: 0,1
=0,012.233 = 2,796(g)<3,732(g)
=> Trong ket tua c6 A 1 ( 0 H ) , :
m^^^Q^^^ =
Theo (2): n^^^_ = nc,co3 = 0,03(mol)
mol CP; 0,04 mol Ca-^ 0,06mol HCO3
,,,.ui>; ,
Theo djnh luat bao toan dien tich, ta tinh dugc s6' mol Na* :
t.. • i v
n
,0
^ =0,l.l+0,06.1-0,04.2 = 0,08(mol).
•
3,732 - 2,796 = 0,936(g)
* Dun soi dung djch X:
=>"AI(OH), =0,936/78 = 0,012(mol)
H+
+ O H " ^ H2O
0,1 ^ 0 , l ( m o l )
2HCO3
>C02 T +co^-
+ H2O
0,03 (mol)
Al'+ + 3 0 H - ^> Al(OH)3 i
0,06
0,012 <-0,036 <-0,012(mol)
vay dun can dung djch X thu dugc CaCO,, CaCU, NaCl (chat ran),
vay m = m^^2+ + r"^n2cot + "^r,cr +
Al''+ + 4 0 H " ^ r A l ( O H ) ^ " "
0,008 <-(0,168-0,1-0,036)
Na"'
=> m = 0,04.40 + 0,03.60 + 0,1.35,5 + 0,08.23
-
m = 8,79 (gam).
Dap an diing la D.
=>z = 0,012+ 0,008 = 0,020
(2)
T i r ( l , 2 ) = > t = 0 , 0 6 + 3.0,02 = 0,120
;
Thi du 7: Dung djch X chiia cac ion: Fe'", SO4-", N H / , CP. Chia dung djch X
thanh hai phdn bang nhau:
-
Vay z, t Idn lugt la 0,020 va 0,120
Dap an diing la C.
Chu y; M6t each gdn diing, coi H* kh6ng do nu6c phSn li ra (bo qua sir didn l i cua
nude).
Thi du 6: Dung dich X chiia cac ion: Ca'"^, Na*, HCO3 va CI", trong do so nnol cua
ion Cr la 0,1. Cho 1/2 dung djch X phan ihig vdi dung dich NaOH (du), thu
ducKc 2 gam ket tiia. Cho 1/2 dung dich X con lai phan ling v6i dung dich
Ca(OH)2 (du), thu dugc 3 gam ket tua. Mat khac, neu dun s6i de'n can dung djch
X thi thu dugc m gam chat rSn khan. Gia tri ciia m la
A. 9,21
B. 7,47
C. 9,26
D. 8,79
(Trich de thi tuyen sink Dai hoc khoi B)
Hu&ng dan gidi
-
Phan mot tac dung vori lugng du dung dich NaOH, dun nong thu dugc 0,672 lit
khi (0 dktc) va 1,07 gam ket tua;
Phdn hai tac dung vai lugng du dung djch BaCU, thu dugc 4,66 gam ket tua.
T6ng khoi lugng cac mu6'i khan thu dugc khi c6 can dung djch X la (qua trinh
c6 can chi c6 nu6c bay hai)
jom 6« r Fe(OH),
NHi t + H , 0
. i"
H003H
> BaS04 i
*) X€x 1/2 dung dich X:
Ca2+ + HCO3 + NaOH
CaCO^ i +Na+ + H j O
(l)
Theo bai ra: n 3^ = 2.n'Fe(OH)3 = 2 . — = 0,02(mol)
Fe
107
0,02mol
Ca^* + 2HCO3 + Ca(0H)2 ^ 2CaC03 i +2H2O
= 2 . ^ ^ = 0,06(mol)
22,4
(2)
0,03mol
so
Chiing to trong phan ling (1) c6 Ca"^ het (con du H C O 3 )
^
94
,_ =2.n BaS04 - 2 . — - 0 , 0 4 ( m o l )
233
Theo djnh luat bao toan dien tich: 0,02.3 + 0,06.1 = 0,04.2 + n^^_ . 1
"Ca2+ " "CaC03 = 0,02(mol) .
25
Cty TNHH MTV DWH Khang Vigt
cam rang 6n luygn thi DH 3 mjgn Bac - Trung - Nam mfln H6a hgc - Cu Thanh Toan
Khoi lircmg muoi bang long kh6'i luong cac ion:
m„„«i = 0,02. 56 + 0,06. 18 + 0,04. 96 + 0,04. 35,5= 7,46 (gam)
Dap an diing la D.
no(x) + " 0 ( 0 2 ) = n 0 ( C 0 2 )
+"0(H20)
=^0,1.2 + 0,24.2 = nco2-2 + 0,2.1
nco2 =0'24{mol)
Thi du 8: Hoa tan h6n hop mud'i vao nuofc duoc dung djch chiia 0,1 mol N H / ; 0,2
mol k?*\3 mol NO", 0,1 mol CF va 0,15 mol ion X. Di6n tich ciia ion X la
A. 1 -
B. 1+
C.2D.2+
(Trich de thi thi Dai hgc khoi A, B)
_ "CO2 _ 0,24
0,1 " " ^ ' ^ . ^ ' ^ ' ' V '
S6 nguyfin tir cacbon trung blnh trong phan tir: x =
'hh
• X khdng phai la 2 axit no, mach ho, don chiic
Vi nco2 >"H20
=> Loai A va C.
Huong dan gidi
Ta tha'y n€u kh6ng c6 ion X thi dung dich du cac hat mang diSn tich duong
Mat khac, vl nc = x = 2,4 => loai D.
(jO,l. 1(+) + 0,2.3(+)| > |0,3.1(-) + 0,1.1(-)|)
Vay 2 axit trong X co the la CH3COOH va CHj = CH - COOH
,
Do do X la ion am (anion).
i
Dap an diing la B.
Thi du 2: D6't chay hoan toan 20 cm' hoi hop chat hOu co X (chi g6m C, H, O) cSn
Dat dien tich cua ion X la a ( - ) .
vira du 110 cm^ khi O 2 , thu duoc 160 cm' h6n hop Y g6m khi va hoi. Dfin Y
Ap dung dinh luat bao toan dien tich ta c6:
0,1.1(+) + 0,2.3(+)| = |0,3. l ( - ) + 0,1. l ( - ) + 0,15.a(-)| =^ a = 2
qua dung djch H2SO4 dac (du), con lai 80 cm' khi Z. Bife't cac th^ tich khi va
hoi do 6 C l i n g di^u kien. Cong thiic phan tir ciia X la
Dap an diing la C.
A. CjHgO
B. C4Hg02
I. L I T H U Y E T
-
C. C4H,oO
Trong phan ting hoa hoc, cac nguyen to duoc bao toan. Dieu nay c6 nghla la
V^' V c o 2 = V ^ ^ ^ o ( , ) = > X d a n g C „ H 2 „ 0 ,
t6ng s6' mol nguyen tir ciia cac nguy6n t6' trudc va sau phan ling lu6n bang nhau.
=> Loai A v a C .
C&n chu y, han che viet PTHH ma thay vao do nen viet so do phan ihig (c6 he
So dd: C „ H 2 „ 0 , + 0 2 ^ CO2 + H2O
s6') d^ bieu didn cac bien doi co ban ciia cac nguyen t6' quan tam.
II. VAN DUNG
Thi du 1: H6n hop X g6m hai axit cacboxylic don chiic. Dot chay hoan toan 0,1
mol X cSn 0,24 mol O, thu duoc COj va 0,2 mol H^O. Cong thiic hai axit la
A. HCOOH va C2H,COOH
D. C4H8O ^ ' ^ ' "
(Trich dethi thu Dai hgc khoi A, B nam 2013)
Huong ddn gidi
'I •
PHirONG PHAP 4: PHlTONG PHAP BAO TOAN N G U Y E N T O
-
i^(ii(>)fA
>
' ' f
20 -> 110
80
80
Vl nguyen t6' oxi duoc bao toan nfen ta co: 20.a +110.2 = 80.2 + 80.1
a= 1
CTPT ciia X la C „ H 2 „ 0 =^ loai B.
VayCTPTXla C4H8O.
Dap an diing la D.
B. CH3COOH va CH2 = CHCOOH
-
Thi du 3: Hoa tan hoan toan 0,3 mol h6n hop gdm A l va AI4C3 vao dung djch
C. CH3COOH va CjH^COOH
KOH (du), thu duoc a mol h6n hop khf va dung djch X. Sue k h i CO2 (du) vao
D. CH2 - CHCOOH va CHj = C ( C H 3 ) C 0 0 H .
dung djch X, lugng ket tiia thu duofc la 46,8 gam. Gia trj ciia a la:
( Trich de thi thu Dai hgc khoi A, B nam 2013)
A. 0,55
B. 0,60
C.0,40
D. 0,45
(Trich de thi thucDgi hgc khoi A, B nam 2013)
Huong dan gidi
Dat cong thiic 2 axit don chiic trong X la C^HyOj
Sor d6 dot chay: C , H y 0 2 + O2 -> CO2 + HjO
Huong ddn gidi
£
So d6 cac phan ling xay ra:
Al
0,1
0,24
0,2(mol)
[A14C3'
KOH
CH4
T;H2 t
KAl(OH)Jdd)
CO2
Al(OH), i
KHC03(dd)
Vi nguyen to oxi duoc bao toan ntn ta co:
27
dm
Cty TNHH MIV iiVVH Khang Vi$t
nana On luyQn thi BH 3 m\6n Bic - Trunp - Nam mOn H6a hqc - Cii Thanh Toan
Ke't tiia thu duoc la A l ( O H ) ,
= 46,8/78 = 0,6(mol)
;n^|,OH)
Al->.A!(OH)_,i
AI4C3 ^ 4 A l ( O H ) 3 i
X
y
X
2 H 2 0 + 2 e ^ H , +20H"^
Theo djnh luat bao toan electron ta c6: 1,5 = 8x +0,3 =>8x = 1,2 => x =0,15
4y
= > m p e 3 O 4 = 0 ' ' 5 - 2 3 2 = 34,8(g)
V I nguydn to nhom duoc bao loan, nfen ta c6 he:
'x + y = 0 , 3
fx-0,2
v a y m = 13,5 +34,8 = 48,3(g)
x + 4 y = 0 , 6 ^ [ y =0,1
Dap an diing la A .
2 A I + 2 K O H + 6 H 2 O ^ 2 K A 1 ( O H ) ^ + 3H2 t
^ 0,2
^
Chuji: PTHH cac phan iJng xay ra:
9Fe + 4 A I 2 O 3
8A1 + 3Fe,04
0,3
. AI4C3 + 4 K O H + I2H2O
0,1
0,3<-0,15
-iL.lS^^
2A1 + 2 N a O H + 6 H 2 O
4 K A l ( O H ) ^ + 3CH4 T
->
2 N a A l ( O H ) ^ + 3H2 t
N a A l ( O H ) ^ + C O 2 - > NaHCO^ + A 1 ( 0 H ) 3 i
0,3
Vay a = 0,3 + 0,3 = 0 , 6 0 ( m o l ) .
Thi du 5: D6't chay hoan toan h6n hop X gom hai ancol no, hai chuc, mach ho can
vua du V | li't khi O,, thu duoc V , Ii't khi C O . va a mol H j O . Cac khi diu do a
Dap an diing la B.
Thi du 4: Nung nong m gam h6n hop gom A l va Fe304 trong di^u kifen khong c6
dieu kien tieu chuan. Bieu thiic lien he giua cac gia trj V|,V2, a la
khong k h i . Sau khi phan ung xay ra hoan toan, thu duoc h6n hop rSn X . Cho X
A . V , = V , + 22,4a.
B. V , = 2 V , - 1 1 , 2 a .
tac dung v6i dung djch N a O H (du) thu duoc dung dich Y , chat ran Z va 3,36 lit
C.V, = 2V,+11,2a.
D. V, = V , -22,4a.
khi H 2 (a dktc). Sue khi C O j (du) vao dung djch Y , thu duoc 39 gam kd't tua.
(Trich de tuyen sinh Cao dang
Gia tri ciia m la:
A . 48,3
C. 45,6
B. 57,9
D . 36,7
(Trich de thi thvcDai hoc khdi A, B nam 2013)
Hu&ng dan gidi
l a CO. nhh(ancol)X - " H 2 0 ~"co2
22,4
V i nguyen t6' oxi duoc bao toan ntn ta c6: n^^^^j + n ^ Q ^ ^ = no(co2) + " O ( H 2 0 )
T h e o b a i r a : n^^ = 0 , 1 5 ( m o l )
a--
So 66 cac phan ling xay ra:
ddY
Al
Fe
Fe,04
->X
(IdNaOH
AI2O3
H2
>Al(OH)3;
t
2V,
4V,
22,4
22,4
-a
Dap an dung la B.
gam X phan ling he't voti dung dich N a H C 0 3 t h u duoc 1,344 lit C 0 2 ( d k t c ) . Dot
chay hoan toan m gam X cin 2,016 lit 0 2 ( d k t c ) , thu duoc 4,84 gam C O 2 va a
V i nguyen t6' nh6m duoc bao toan nen: n ^ , = n^,^Qj^j = 0 , 5 ( m o l )
gam H 2 O . Gia trj ciia a la
A . 1,44.
=^ m A, =0,5.27 = 13,5(gam)
B. 1,62.
C.3,60.
D . 1,80.
(Trich de thi tuyen sinh DH khdi A nam 2012)
Cac qua trinh cho - nhan electron:
0 , 1 5 - > 1,5
22A,
^ ' . 2 . ^ . 2 + a.l
22,4
22,4
Thidu 6: H6n hop X gom axit fomic, axit acrylic, axit oxalic va axit axetic. Cho m
nH2 = 3 , 3 6 / 2 2 , 4 = 0,15(mol)
- 3e - > A l ^ ^
.2+
=>V, = 2 V 2 - 1 1 , 2 a
"^"2
Z
Ket tua la A l ( O H ) 3 : n ^ , ( O H ) ^ = 3 9 / 7 8 = 0 , 5 ( m o l )
Al
khoiA)
Hu&ng ddn gidi
Huong ddn gidi
+8/3
Fe3 + 8e
X
->8x
0
3Fe
•
Theo bai ra: nco2 = 1,344/22,4 = 0 , 0 6 m o l ;
=0,09mol;nco2 = 4 , 8 4 / 4 4 = 0,11 mol
29
Cty 1MHH MIV L.'VVH Khang Vigt
ca'm nang 6n luyjn thi DH 3 mign B&c - Trung - Nam mOn H6a hpc - CCi Thanh ToSn
V i cac nguyen t6' (Fe, Cu va S) dugc bao toan ntn ta c6 so do:
Sa 66 p h a n i r n g :
- C O O H + N a H C 0 3 - > - C O O N a + CO2 t + H 2 O
0,06
*-
;0
0,06
2FeS2
2x
H.ii!;OJ>rff
=> "-cooH(x) = " C O 2 = 0 , 0 6 ( m o l ) ^n^^^^
-0,06.2 = 0,12(mol)
CU2S
<-
X
Fe2 ( 5 0 4 ) 3
->
=^ n H 2 0
A . 18,0.
B.22,4.
Sod6:
D. C 3 H 4 .
(Trich de thi tuyen sinh DH khoi A)
Huong dan gidi
D . 24,2.
^
Fe^^Fe203,Fe304,FeO,Fe-^^^^^!^Fe(N03)3
0,1
T a c o : n g ^ ^ o , ! =0>2mol
C. 15,6.
,j
Hu&ng ddn gidi
A. C H 4 .
>
Bao toan nguydn t6' s i t :
n^^^p^^ =
0,1
f
Of
^.^^.^^^^^^^^ ^""H^o,)^ = "^^ =
v a y m =0,1.242 = 2 4 , 2 ( g )
= 39,4 - 1 9 , 9 1 2 = 19,488(g)
"'"^
„
Dap an diing la D .
(l)
Thi du 10: D6't chay hoan toan 0,2 mol h6n hgp g6m mdt ancol va m6t axit don
V I nguyen t6' cacbon, hidro dugc bao toan nen: m^^ = nij^^^Q^j +
chiic C O cung so nguydn t6' cacbon cin diing 0,45 mol O , , thu dugc 0,4 mol CO,
^^^^^^Q^
va 0,5 m o l H^O. Phdn tram khoi lugng ciia axit trong h6n hgp trdn la:
(2)
A . 46,5%
B. 32%
C. 50%
D.49,18%
(Trich dethi thvcDai hoc khoi A, B)
Tir ( 1 , 2) x = 0,348; y = 0,232
Hu&ng ddn gidi
Dat X l a Q H b ta c6 t i le: a: b = x : 2y = 0,348:0,464 = 1,5:2 = 3 : 4
Taco: n c = n c o 2 / n h h =0,4/0,2 = 2
Vay X la C 3 H 4 .
Dap an dung la D .
Thi du 8: Hoa tan het m gam h6n hop FeS^ va C U 2 S trong dung dich H N O 3 , sau cac
phan ling hoan toan thu dugc dung dich X chi c6 2 chat tan, \6i tong khoi lugng
cac chat tan la 72 gam. Gia tri ciia m la
B.40
=> axit don chuc phai la C H 3 C O O H ; ancol c6 dang C^Hfi,
C. 20
(Trich de thi thitDai hoc khoi A,
f7-a^
+^
D . 80
B)
V i chl thu dugc hai cha't tan trong dung dich sau phan ling => Hai ch^t tan do la
(a = 1 hoac a = 2)
C H 3 C O O H + 2O2 - > 2 C O 2 + 2 H 2 O
C2H,0, +
Hu&ng ddn gidi
C U S O 4 va Fe2(S04)3:
0:1
(Trich dethi thUDai hoc khoi A, B)
Theo bai ra: np^ =0,1 ( m o l )
I
.jg m vnios n i o . i / '
duy nha't) va dung djch chiia m gam mu6'i. Gia trj cua m la
dung dich giam bot 19,912 gam. C6ng thiic phan tijf cua X la
A . 60
=> x = 0,1
^rB.V> iM
d j c h B a ( O H ) 2 . Sau cac phan umg thu dugc 39,4 gam ket tua va khoi lirgng ph&n
C. C 2 H 4 .
;'.
b6 X tac dung vdi dung djch H N O 3 loang (du), thu dugc khi NO (san ph^m khir
thuofng) roi dem toan b6 san phdm chay ha'p thu h6't vao binh dung dung
=>4,64 = 12x + 2y
i.
Thidu 9: D6't 5,6 gam Fe trong khong k h i , thu dugc h6n hgp cha't ran X . Cho toan
=0-08 ^ a =0,08.18 = l , 4 4 ( g ) ^
Thi du 7: Dot chay hoan toan 4,64 g a m mot hidrocacbon X (cha't khi of di6u kifin
^ 4 4 x + 18y = 19,488
2x ( m o l )
DapandiinglaB.
"o(H20)
Dap an dung la A .
m c o 2 + niH20
->
X
v a y m = 2.0,1.120 + 0,1.160 = 4 0 ( g )
V i nguydn t6' o x i dugc bao toan n&n ta c6: nQ^^) + " 0 ( 0 2 ) = " 0 ( 0 0 2 )
B. C 4 H , o .
2CUSO4
Theo bai ra: 400.x + 160.2x = 72 => 720x = 72
C O 2 + H 2 O
=>0,12 + 0,09.2 = 0,11.2 + n H 2 o l
+
2
;
0
H20 ~ n c o 2
V i nguyen tO'oxi dugc bao toan nen ta c6: 0,1.2+ 0,1.a+ 0,45.2 = 0 , 4 . 2 + 0,5.1
=^a = 2 ( C 2 H , 0 2 )
31
Cty \N\iH MIV UVVH Khang Vi^t
dm nang On luy^n thi DH 3 mign B&c - Trung - Nam mOn H6a hqc - CCi Thanh Toan
0,1.60 + 0,1.62
Dap an diing la D. .
,
,.
,
„
y'Thidu 11: H6n horp X g6m axit axetic, axit fomic va axit oxalic. Khi cho m gam X
tac dung vdri N a H C O , (du) thi thu dugc 15,68 lit khi C O , (dktc). Mat khac, d6't
chay hoan toan m gam X cdn 8,96 lit khf O, (dktc), thu duoc 35,2 gam C O , va y
mol H i O . Gia tri ciia y la
A. 0,3.
B. 0,8.
C.0,2.
D. 0,6.
ffli du 13: Dot chay hoan toan mot luong h6n hop khi X g6m m6t ankan va mot
anken cin diing vira dii 0,7 mol O,, thu duoc 0,4 mol CO,. C6ng thiJc ciia ankan la
AQH,o
B.QH,
C.CH,
D. CH4
(Trich de thi thi Dai hoc khoi A, B)
Huong ddn gidi
Sod6: C,Hy + O 2 ^ C 0 2 + H 2 O
Vi nguyen to oxi dugc bao toan ndn:
no(02) = " o ( c o 2 ) + " o ( H 2 0 ) ^ " H 2 O = 0-6 (mol)
1,
MO
(Trich de thi thUDqi hoc khoi A, B)
Suyra: n,,,,,^^ = n H 2 0 - " 0 0 2 = 0 , 6 - 0 , 4 = 0,2 (mol)
Huong ddn gidi
'
Av*.
Theobai ra: nco2 ( l ) = 15,68/22,4 = 0,7(mol) ;
=>nx>nankan^nx>0,2(mol)
no2 = 0,4(mol); nco2 (2) = 35,2/44 = 0,8(mol)
Do do: nc(x) < 0,4/0,2 = 2 => Trong X c6 CH4 (ankan).
CH3COOH
a
+ HCOOH + HOOC - COOH ^ ^ ! ^ y ^ ^ C 0 2
b
c
—•
(a + b + 2c)(mol)
Suy ra, s6' mol nguyfen tir nguyen to oxi trong X :
"o(x) = 2a + 2b + 4c = 2(a + b + 2c) = 2.nco2
= 2.0,7 = l,4(mol)
X + O2 - > C O 2 + H 2 O
VI nguyen to oxi duoc bao toan nen:
"o(X) + "0(02) ^ "0(C02)
Dap an dung la D.
Thidu 14: H 6 n hop X g6m CiH, va H , c6 cung s6' mol. La'y m6t lugng h6n hop X
cho qua chat xuc tac, nung nong dugc h6n hgp Y g6m C,H4, CiH^ va C,H2, H ,
du. DSn Y qua nuorc brom tha'y binh nu6c brom tang 10,8 gam va thoat ra 4,48
lit h6n hgp khi (do b dktc), c6 ti khd'i so vdi H , la 8. The ti'ch khi O, (khi do a
dktc) vira dii de dot chay hoan toan h6n hgp Y la
A. 33,61ft.
B. 22,4 lit.
C. 26,88 lit.
D. 44,8 lit.
(Trich dethi thi Dai hoc khoi A, B)
Huong ddn gidi
Saddphanung:
-p>bv,
"0(H20)
X (C,H,, H,)
=> 1,4+ 0,4.2 = 0,8.2+ y.l : ^ y = 0,6
-
Dap an diing la D.
2FeS,
CU3S
+
0,12 (mol)
=^a=
^
a (mol)
^
= 0,06(mol).
> C,H„ H,"'^*
Goi X la so mol C,H, (cung nhu H,) trong h6n hgp X .
>
Fe,(S04)3 + CuS04
Theo dinh luat bao toan kh6'i lugng, ta c6: mx = my
-
Mat khac, theo bai ra: my = m ,bi„h b^om .a„g) + " i (khf
4 48
my = 1 0 , 8 + ^ — . 8 . 2 = 14 (gam)
22,4
-
• X. i
(1)
-
.hoi.
(2)
n)
(3)
/ ,oMi iish'
,
Tir (1, 2, 3) suy ra: x = 0,5 (mol).
'
Theo dinh luat bao toan nguyen t6': V i nguy6n t6' C, H dugc bao to^n ntn thi
tich khf O, d^ dot chay hoan toan h6n hgp Y cung bang the' tfch O, de d6't chay
hoan toan h6n hgp X :
C2H,
+
5/20,
>
1,25 (mol)
0,5
H.
Dap an diing la D.
) Y (C,H4, C,H„ C,H,, H,)
mx = (26 + 2). x = 28x (gam)
Thi du 12: Hoa tan hoan toan h6n hop g6m 0,12 mol FeS, va a mol CujS vao axit
HNO, (viJra dii), thu duoc dung djch X (chi chiia hai mu6'i sunfat) va khi duy
nha't NO. Gia tri ciia a la
A. 0,04.
B. 0,075.
C.0,12.
D. 0,06.
(Trich de thi tuyen sinh Dai hoc khoi A)
Huong ddn gidi
Vi cac nguyen to (Cu, Fe va S) duoc bao toan ntn ta c6 so 66:
'
0,5
+
1/20,
> 2C0, + H,0
,?;/HtX)3H /
:D-,1I-
y
> H,0
> 0,25 (mol)
33
Ctv iMMii M i v iJVVH Khang V i § t
Ca'm nang On luygn thi B H 3 mfo.f ^ a c - Trung - Nam mOn H6a hi?c - Cu Thanh To^n
=> X"o2
=0,25+1,25 = 1,5 (mol) =i> Vo^
,dk.c) =
Huong ddn giai
1,5. 22,4 = 33,6 (lit).
Vi axit don chiic => phan tir X c6 2 nguySn til oxi.
Dap an diing la A.
Nguyen to oxi dupe bao toan nen: 0,1.2 + 0,24.2 = 0,2.1 + n^^^^ .2
Chu y: Sir dung dinh luat bao toan nguyen to de giai nhanh cac bai toan hoa hoc
phiic tap:
C2H2
^
^„^^
nco2 = 0,24 (mol)
,ii;);-sr
Nhan xet:
C
+
C2H4, C2H,
^
> CO,
^
.0.50,
^
Q
.
nco > " H , o c6 axit khOng no (loai A, C).
.
nc =0,24/0,1 = 2,4 (=> loai B).
>»A H
61 nSv,,,
•
Dap an diing la C
Thi du 15: H6n hop M g6m andehit X, xeton Y (X, Y c6 ciing s6' nguy6n tir
cacbon) va anken Z. Dot chay hoan toan m gam M cdn diing 8,848 lit O2 (dktc)
sinh ra 6,496 lit CO, (dktc) va 5,22 gam HjO. Cong thuc ciJa andehit X la
A. CH3CHO
B.C3H7CHO
C. C4H,CHO D. C2H5CHO
(Trich de thi thi Dai hoc khoi A, B)
Huong dan giai
Theo bai ra: no2 = 0,395 (mol); ncoj = 0,29mol; ni^^Q = 0,29mol
Vi T\QQ^ = n^^o => X, Y deu no, her, dcrti chiic.
:
Thi du 17: Cho h6n hpp khi X gom HCHO va H , di qua 6'ng sii dung b6t Ni nung
nong. Sau khi phan ling xay ra hoan toan, thu dupe h6n hpp khi Y g6m hai chat
h&u CO. Dot chay het Y thi thu dupe 11,7 gam H2O va 7,84 lit khi CO, (o dktc).
PhSn tram theo the tich ciia H , trong X la
„:.,
A. 65,00%.
B. 46,15%.
C. 35,00%.
D. 53,85%.
(Trich dethi thu: Dai hoc khoi A, B)
Hu&ng ddn giai
7 84
117
Theo bai ra: nco2 = ^
= 0,35 (mol); nH20 " " f ^ =
("'"'^
' ''
VI X, Y CO Cling s6' nguyen tir cacbon n6n diu c6 ciing CTPT la C„H2nO(x mol).
Ta CO so do cac phan ling xay ra:
HCHO
> HCHO
D a t Z l a C ^ H 2 ^ (ymol)
H,
So dd phan ling: X, Y,Z + O 2 - > C O 2 + H 2 O
(X)
V i nguyen t6' oxi dupe bao toan nen: n ^ ^ . y ) + "0(02) = "0(002) ^ "O(H20)
HCHO + H2
=> 1.x+ 2.0,395 = 2.0,29+ 1.0,29
x = 0,08(mol)
Vi nguyen t6' cacbon diroc bao toan nen ta c6: n.0,08 + m.y = 0,29
> 1,. > •
> CO, (0,35 mol)
CH3OH
H , 0 (0,65 mol)
(V)
J I.
> CH3OH
u .
> CO, + H , 0
HCHO + O2
> 2CO, + 4 H , 0
2CH3OH + 3 0 ,
Ap dung djnh luat bao toan nguyen t6', cho nguyen to cacbon va hidro, ta c6:
Vi m.y > 0 = > n < 3,625
"QHCHO)
Mat khac, Y la xeton ntn n > 3.
Vay n = 3
Do do andehit X la C3H6O
rr;i
"H(H2)
= nc ( C O 2) = 0,35 (mol) =
" H (HCHO) -
(C2H5CHO)
=> n H ( H . ^ +
Dap an diing la D.
Thi du 16: H6n hop X g6m hai axit cacboxylic dcm chiic. D6't chay hoan toan 0,1
vay
B. CH,=CHCOOH va CH2=C(CH3)COOH
C. CH3COOH va C2H5COOH
D. CH3COOH va CH2=CHC00H.
H (H
2
O)
2. 0,35 = 2.0,65
nH(H,) = 0,6 (mol)
mol X cin 0,24 mol O,, thu dirac CO2 va 0,2 mol H2O. C6ng thiic hai axit la
A. HCOOH va C 2 H 5 C O O H .
1
nHCHO
% V H , , =
"2/x
M 1 ^
nH2 = ^
=46,15%.
= 0,3 (mol)
*
,
0,3 + 0,35
Dap an diing la B.
(Trich de thi thicDqi hoc khoi A, B)
Thi du 18: Dot chay hoan toan m gam h6n hpp g6m n hidrocacbon khac nhau, thu
dupe 11 gam CO, va 9 gam H2O. Gia trj ciia m la
,
CtyTN f III MTV UVVH Khang Vi^l
d m nang On luygn thi DH 3 mi^n BJlc - Trung - Nam mOn H6a hqc - Cu Thanh ToSn
A . 4,0.
B. 6,2.
C. 8,0.
Theo nguyfin tSc bao toan electron, ta c6:
D. 13,6.
(Trich de thi thuDai hoc khoi A, B)
Huong ddn gidi
PTPl/:C,H,
) xCO,+
^H.O
*
r(X>^ '^^
Tir(l,2)tac6:
1,5 = 3a + 8b
AKNO,)., + N O t + N p t + H j O
Theo djnh luat bao toan kh6'i luong: m^i +
+ m^dHNOj =mddsau +0.1-30 +0,15.44
^.12
'
^
,
,
,
Dap an dung la B.
^ m =
^
mddHN03 = " ^ d d s a u + ' " N O + ' " N 2 0
=^in„ds.„- niddHNOj =13,5 - 3 - 6 , 6 =3,9 (gam)
n i c t C x H y ) + mH(CxHy) = n i c x H y = ITl0(C02) + " 1 H ( H 2 0 )
(2)
0,15.
So 66 phan ling: A l + HNO3 ^
=> 13,5
V i nguydn t6' C, H duoc bao toan, nen:
a = 0,1, b =
•
'
Thi du 2: Hoa tan he't 52 gam k i m loai M trong 739 gam dung djch H N O 3 , ke't thiic
+ ^ . 2 . 1 = ^ 4 , 0 (gam).
phan iJng thu duoc 0,2 mol N O ; 0,1 mol N2O va 0,02 mol N , . Bie't kh6ng c6
Dap an diing la A .
phan ling tao mu6'i NH4NO3 va HNO3 da la'y d u 15% so vdri luong c^n thie't. K i m
loai M va n6ng d6 phSn tram ciia HNO3 ban dSu Mn luot la
A.Crva21,96
B.Znva20
D.Znv^
17,39
Sa 66 cho - nhan electron
LITHUYET
M-ne^M""
toan (bao toan khoi lirong, bao toan electron, bao toan dien tich va bao toan
X^
nguydn i6).
N+3e
nx
+5
(mol)
0,6 < - 0,2
+1
2N+8e
II. V A N D U N G
+2
>N
+5
Rat nhi^u bai toan phiic tap c6 the' giai nhanh bang each t6 hofp cac djnh luat bao
(mol)
+5
0
2 N + lOe
>N2(N20)
0,8 <- 0,1 (mol)
Thi du 1: Hoa tan hoan toan 13,5 gam b6t A l vao dung dich H N O , loang du, sau
00
i.f;
>N2
0,2 < -
0,02 (mol)
cac phan u-ng hoan toan thu duoc 5,6 lit h6n hop 2 khf N O va N . O (dktc). Trong
Theo nguyen tac bao toan electron, t h i : nx = 0,6 + 0,8 + 0,2 = 1,6 (mol)
dung djch kh6ng c6 N H 4 N O , . Kh6'i luong dung djch sau phan ling thay d6i so
=>x = 0 , 1 6 / n (mol)
vdi kh6'i lircfng dung djch H N O , ban ddu la
52
52n
^
Suyra:M^,= — =
= 32,5n
X
^1 ^
A. khong thay doi
B. tang 3,9 gam
C. tang 13,5 gam
D . giam 9,6 gam
(Trich de thi thvcDai hoc khoi A, B nam
Huong ddn gidi
Theo bai ra:
n^, = 13,5 / 27 = 0,5 (mol)
nNO.N20 = 5 , 6 / 2 2 , 4 =0,25 (mol)
2013)
Gia trj thoa man la n = 2,
MM
= 65 (Zn)
K V
A p dung djnh luat bao toan nguyfen t6' cho nguydn t6 nito, ta c6:
j ''^ ^
^N(HN03) = " N { N O ) + "N(N20) + "N(N2)
Taco:
=f
a + b =0,25
(1)
Cac ban phan ting oxi hoa - khu:
+3
Al
-
3e
0,5
->1,5
-» A l
'
N
+2
+
3e ->
3a
+5
2N
+3
N(NO)
a
Sifi'' - V
52
"zn = ^ = 0,8(mol)=>n2„(NO3)2 = 0 , 8 ( m o l )
65
Goi a, b l l n luot la s6' mol N O , N , 0 .
0
2013)
Huong ddn gidi
PHL/ONG PHAP TO HOP CAC 0!NH LUAT BAO TOAN
I.
C.Crva20
(Trich de thi thic Dai hoc khoi A, B nam
P h a o n g p h a p 5:
HHNOJ
"N(zn(N03)2)
= 0,2.1 + 0,1.2 + 0,02.2 + 0,8.2 = 2,04(mol)
V I HNO3 la-y d u 15% nen:
mH,03<.H-«
O H N O J '"-VC
= 2.346. 6 3 =
=
2,04.(100+15)
, , {p,, i
>•
C'^t • ,
^^^^^
= 2,346(mol)
147,798 (g)
+1
+
8e
Sb
^
N2 (N2O)
b
= > C % ( d d H N O 3 ) = ^ ^ ^ ^ . 1 0 0 % = 20%
>u' (liA s , v x i o
31
Cty iNHH r\yii ; L V V H MicHig
Ca'm nang On luy$n thi DH 3 mign Bjc - Trung - Nam mOn H6a hpc - CCi Thanh Toan
Vay M la Zn va C % ( H N O , ) = 20%
T a c 6 : 24x + 65y = 8,9
Dap an diing la B.
V I nguyfen t6' nito duoc bao toan nen ta c6: 2x + 2y + 2z + 0,045.2 = 0,5.1
Thi du 3: Cho 18,4 gam h6n horp X g6m Cu2S,CuS,FeS2 va FeS tac dung hd't vdi
^x
(1)
+ y + z = 0'205
(2)
^
HNO3 (dac nong, du) thu duoc V lit khi chi c6 NO2 (a dktc, san ph^m khu duy
Theo nguyen tSc bao toan electron, ta c6: 2x + 2y = 0,045.8 + 8z
nha't) va dung dich Y . Cho toan bo Y vao mot luomg d u dung djch BaCi2 , thu
2x + 2 y - 8 z = 0,36
duoc 46,6 gam ket tua; con khi cho toan bo Y tac dung vori dung djch NH3 du
thu duoc 10,7 gam ket tua. Gia trj cua V la
A . 11,2.
B. 38,08.
f S ' = y;: x 'il ahx nh::i'.
m = ni^g(N03)2 + '"zn(N03)2 + "INH4NO3
."q ^_
D . 24,64.
(Trich de thi tuyen sinh DH
. ,.,
* x
=0,1.148 + 0,1.189 + 0,005.80 = 34,10(g)
khoiA)
Hu&ng ddn gidi
,
xn •
Dap an dung la A .
Theobai ra: n^^^^^ - 4 6 , 6 / 2 3 3 = 0,2mol
Thi du 5: De phan ling h6't a mol kirn loai M cSn 1,25a m o l
H 2 S O 4
va sinh ra khi X
(san ph^m khir duy nha't). Hoa tan hd't 19,2 gam k i m loai M vao dung djch
np^(OH)^^= 10,7/107 = 0 , l m o l
H.S04tao ra 4,48 lit khi X (san ph^m khir duy nha't, dktc). K i m loai M la
A.'cu
Quy d6\ hop X thanh x mol Fe; y mol Cu va z mol S
=>56x + 64y + 32z = 18,4
;
( 1 , 2, 3) => X = 0, l ; y = 0,l;z = 0,005
Vay
C. 16,8.
(3)
B. M g
C.Al
D.Fe.
(Trich de tuyen sinh dqi hoc khoi A)
(l)
Hu&ng ddn gidi
Mat khac: z =
= ng^^Q^ = 0 , 2 (bao toan nguydn t6' liru huynh)
Ta CO cac qua trinh:
0
+n
M-ne->M
Ta lai c6: x = np^ = npg^Qp^^^ = 0,1 (bao toan nguydn to sSt)
64
a-> na-> a
+3
+2
T6ng s6' m o l electron do Fe - > Fe; Cu - > Cu
\ 2 - , o»«n+
n S O f + 2M"^
an/2
S - > S la ne (cho) = 0 , 1 . 3+0,1. 2+0,2. 6 = 1,7 mol
+.'>
^^.^^f,
Theo bai ra: nx = 4,48/ 22,4 = 0,2(mol)
(^.^ff
+6
+(6-m)
S(H2S04) + m e ^
—
m
M2 (SO4 )„
<-na
"
(Bao toan electron)
(Bao toan didn tich)
(JJ
«- a
V I nguyfen t6' luu huynh dircrc bao toan n6n ta c6:
+4
/
'
an .
n n ,
^
^ _
— + — = l,25a =>—+—=1,25 => 2n + nm = 2,5m
m
2
m 2
=> T6ng s6' m o l electron do N + le - > N la 1,7 mol
=> nN02 = l , 7 m o l ^ VNO2 = 1,7.22,4 = 38,08(l)
vr.ri j
Suy ra cap gia trj phii hop: n = 2, m = 8 (HjS)
PTPU: •
•
•
Dap an diing la B.
Thi du 4: Hoa tan hoan toan 8,9 gam h6n hop g6m M g va Z n bang luong vifa du
4M + 5H2SO4 ^ 4 M S 0 4 +H2S + 4H2O
I K
500 m l dung djch HNO3 I M . Sau khi cac phan iJng ket thuc, thu duoc 1,008 lit
khf N2O (dktc) duy nha't va dung dich X chiia m gam mud'i. Gia trj ciia m la
A . 34,10.
B. 31,22.
C. 33,70.
D . 34,32.
(Trich de tuyen sinh Cao dang khoi A)
Hu&ng ddn gidi
19 2
,9 2
= > — = 0 , 2 . 4 = 0,8 = > M = — = 2 4 ( M g )
M
0,8
••••^
-rx
.
Vay M la k i m loai M g .
Theobai ra: n^NOj = 0 , 5 m o l ; n N 2 o =0.045mol
Dap an diing la B.
Goi X, y , z Mn luot la so mol M g ( N 0 3 ) 2 ; Z n ( N 0 3 ) 2 , N H 4 N 0 3 .
Chu
v: San phdm k h u X (khi) c6 th^ la SO,, HjS.
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