PHƯƠNG PHÁP QUY ĐỔI ESTE THÔNG QUA ĐỀ THI THPTQG 2019

I GI
P ESTE
Nh m
cao trong th i gian thi ch i h i h ng c i ti
truy i nh i v m t th
ng d n gi
v ng s
ng!
202-2019): t c CO2 2O. Cho
ng v i dung d ch NaOH v i .M
c t i 0,06 mol Br2 trong dung d c
A. 27,72. B. 26,58. C.27,42. D. 24,18.
ng d n gi i
L i d i d mol H 2 b
b t b ng s t pi g c)
R 1 COOCH 2
R 2 COOCH
R 3 COOCH 2
C 3 H 8 a
CH 2 b
COO 3a
- H 2 c
CH 4 a
CH 2 t
COO 3a
- H 2 0,06
a mol
C 3 H 5 a
HCOO 3a
CH 2 b
- H 2 c
HCOOCH 2
HCOOCH
a
HCOOCH 2
CH 2 b
- H 2 c
ng d n gi i
C 3 H 5 a
HCOO 3a
CH 2 b
- H 2 0,06
25,74 gam
NaOH
H 2 O 1,53 mol
HCOONa 3a
CH 2 b
- H 2 0,06
C 3 H 5 (OH) 3 3a
H:
4a + b 0,06 = 1,53 4a + b = 1,59 (I)
ng BT kh ng : 176a + 14b 0,12 = 25,74

176a + 14b = 25,86 (II)
T a = 0,03 ; b = 1,47
203-2019): n v 3,08 mol O 2 c CO 2
2 ng v i dung d ch NaOH v i
.M c t i a mol Br 2 trong dung d c
A. 0,2. B. 0,24. C.0,12. D. 0,16.
C 3 H 5 x
HCOO 3x
CH 2 y
- H 2 a
m gam
O 2
3,08 mol
NaOH
ng d n gi i
H 2 O 2 mol
CO 2 (6x + y)
HCOONa 3x
CH 2 y
- H 2 a
35,36 gam
C 3 H 5 (OH) 3 x
H:
4x + y a = 2 (I)
ng BT kh ng : 204.x + 14y 2.a = 35,36 (II)
O: 3x + 3,08 = 6x+ y + 1
(III)
T (I)(II)(III) x = 0,04 ; y = 1,96 ; a = 0,12
204-2019): n v 2,31 mol O2 c H2O
2 ng v i dung d ch NaOH v
mu i .M c t i a mol Br2 trong dung d c
A. 0,09. B. 0,12. C.0,15. D. 0,18.
C 3 H 5 x
HCOO 3x
CH 2 y
- H 2 a
m gam
O 2
2,31 mol
NaOH
ng d n gi i
H 2 O (4x + y - a) mol
CO 2 1,65 mol
HCOONa 3x
CH 2 y
- H 2 a
26,52 gam
C:
C 3 H 5 (OH) 3 x
6x + y = 1,65 (I)
ng BT kh ng : 204.x + 14y 2.a = 26,52 (II)
O: 3x + 2,31 = 2x+ 0,5.y 0,5.a + 1,65
0,5 y + 0,5.a = - 0,66 (III)
T (I)(II)(III) x = 0,03 ; y = 1,47 ; a = 0,09
207-2019): c H2O 2. Cho
ng v i dung d ch NaOH v i .M
c t i 0,04 mol Br2 trong dung d c
A. 16,12. B. 18,48. C.18,28. D. 17,72.
T
ng d n gi i
C 3 H 5 (OH) 3 3a
C:
6a + b = 1,1 (I)
ng BT kh ng : 176a + 14b 0,08 = 17,16
E
11,16 gam
O 2
0,59 mol
C 3 H 5 a
HCOO 3a
CH 2 b
- H 2 0,04
17,16 gam
H 2 O
CO 2 0,47 mol
9,36 gam
NaOH
0,52 mol
ng d n gi i 1
DLBTKL
H 2 O
CO 2 1,1 mol
HCOONa 3a
CH 2 b
- H 2 0,04
- 2014): t thu ng c
cacbon v i X; T
c t o b
h n h p E g m X, Y, Z, T c n v c. M t
ng t i dung d ch ch a 0,04 mol Br2. Kh ng mu c khi cho
ng h t v i dung d
A. 4,68 gam. B. 5,04 gam. C. 5,44 gam. D. 5,80 gam.
Nh n th y s mol H2O l mol CO2 c an col trong E ph i no
i h n h

CH 2 = CH - COOH
C 3 H 6 (OH) 2
CH 2 =CH-COO
CH 2 =CH-COO
CH 2
ph n ng:
a
b
c
d
C 3 H 6
CH 2 = CH - COOH
C 3 H 6 (OH) 2
CH 2
Muoi ( m gam)
KOH
H 2 O a mol
(a + 2c )
CH 2 =CH-COO
C 3 H 6
CH 2 =CH-COO
O 2
0,59 mol
Br 2
0,04 mol
CO2
H 2 O
0,47 mol
0,52 mol
11,16 gam
B (I)
D ch s mol CO2 2O : -a + b 3c = 0,05 (II)
B t pi g
B
ng: 72.a + 76.b + 184.c + 14.d =11,16 (IV)
Gi i h : a= 0,02 ; b=0,1 ; c=0,01 ; d = 0,02
c ch 3H 6(OH) 2.
: m = 4,68 gam
ROH
0,08
Na
H 2
0,04 mol
ng d n gi i
m ancol = 32 ( CH 3OH )
i X:
C 3 H 6 (OH) 2
( b + c )
( THPTQG 2015): H n h p X g c, t t ancol Y v i 3 axit
ch - ng k ti t axit
c, ch a m ). Th y ph
gam X b ng dung d c h n h p mu Y. Cho m gam Y ng Na
n
,96 gam H2O. Ph ng c a este
A. 34,01%. B. 38,76%. C. 40,82%. D. 29,25%.
A.
B.
C.
D.
HCOOCH 3
CH 3 - CH = CH- COOCH 3
CH 2
ph n ng :
a
b
c
HCOOCH 3
CH 3 - CH = CH- COOCH 3
CH 2
5,88 gam
NaOH
0,08
B
B
0,22 (II)
B
ng: 60a + 100.b + 14.c = 5,88 (III)
Gi i h : a = 0,06 ; b= 0,02; c=0,02
O 2
2
V c : 34,01% ch n A
M
5,07 gam
DLBTKL
O 2
0,1975 mol
H 2 O
CO 2 0,175 mol
7,7 gam
3,69 gam
0,205 mol
Muoi
CH 3 OH 0,08 mol
H 2 O 0,22 mol
ch 2 2
: X, Y X < M Y
2 2 2
ng d n gi i
Nh n th y s mol H2O l mol CO2 c an col trong M ph i no
i h n h

HCOOH
C 2 H 4 (OH) 2
HCOO
HCOO
C 2 H 4
x
y
CH 3 COOH
C 2 H 5 COOH
G
CH 2
HCOOH
a
b
c
d
TH1 :
C 2 H 4 (OH) 2
HCOO
HCOO
CH 2
5,07 gam
C 2 H 4
O 2
0,1975 mol
KOH
0,1 mol
(I)
(II)
2 2 c = 0,03 (III)
5 (IV)
0,02 ; b=0,04 ; c=0,01 ; d=0,035
2H 4(OH) 2
CO 2
H 2 O
0,175 mol
0,205 mol
0,04 C 2 H 4 (OH) 2
0,01
CH 3 COO
C 2 H 5 COO
x + y =0,02
C 2 H 4
x +2.y + 0,03 = 0,035 (Bao toan CH 2 )
( Loai )
-2017). M t h n h p E g ch u no ,m ch
h n v 560 ml dung d c hai mu
kh n h p T g c
2 2 c a a g n nh t v
A.43,0. B. 37,0. C.40,5. 13,5 .
i E:
HCOOCH 3
HCOO -CH 2
ng d n gi i 1
x
y
HCOOH
CH 3 COOH
HCOO- CH 2
CH 2
ph n ng:
0,04 C 2 H 4 (OH) 2
HCOO
0,01
C 2 H 4
CH 3 COO
Ta d
x + y =0,02
y + 0,01 = 0,035 (Bao toan CH 2 )
TH2:
x= 0,005
y = 0,015
x
y
z
HCOOCH 3
HCOO -CH 2
HCOO- CH 2
CH 2
40,48 gam
B
B
n = 0,98 mol
O2
NaOH
0,56 mol
HCOONa
CH 2
CH 3 OH
C 2 H 4 (OH) 2
CH 2
2y = 0,56 (I)
v
t
x
y
0,56 mol
a gam
Bao toan O
O 2
CO 2
H 2 O
0,72 mol
1,08 mol
51,12 gam

ancol = 19,76 gam
ch n A
L i Gi i 2
HCOO
CH 3 COO
C 3 H 6
n
n
n T
-
n
B
CO
2
H 2O
16,128
22,4
0,72mol
19,44
18 1,08 mol
1,08 0,72 0,36
C n H 2n+2 O x
O 2
0, 36 0 , 72
0,72
2
0,36
ph n
x+ y = 0,36
x + 2y = 0,56
i E:
n
C O
2
n
H
n CO 2 + ( n+1) H 2 O
E + NaOH Muoi +
0, 56 mol
40, 48 gam 22, 4 gam
x= 0,16
y = 0,2
ch n A
2
O
ch h
V y ancol T g m C 2H 5 2H 4(OH) 2
a gam
ng d n gi i 1
x mol C 2 H 5 OH
y mol C 2 H 4 (OH) 2
x + y = 0,36
201-2018): Este X hai ch c, m ch h , t o b i m t ancol no v
ch c. Este Y ba ch c, m ch h , t o b i glixerol v i m
n h p E g n v 0,5 mol O 2 c 0,45 mol CO 2 .
M t c n v 210 ml dung d
n h p ba mu ng kh ng mu i c a hai axit no
c
A. 13,20. B 20,60. C 12,36. D 10,68.
CH 2 =CH-COO
CH 2 =CH-COO
CH 2 =CH-COO
CH 2
bi
HCOO
CH 3 COO
CH 2 =CH-COO
CH 2 =CH-COO
CH 2 =CH-COO
CH 2
0,16.t
C 3 H 6
C 3 H 5
x
C 3 H 5
(Ta coi ph n TN1 g p t l n ph n TN2)
Ta d
x + y = 0,16t
2x + 3y =0,42 t
z
y
x = 0,06t
y = 0,1 t
O 2
Bao toan H
0,5 mol CO 2
0,45 mol
NaOH
0,42.t
H 2 O (5x + 7y +z)
x
y = 3
5
(I)
B :6x + 12y +z = 0,45 (II)
B : 4x + 6y + 0,5.2 =0,45.2 + 5x + 7y + z
x + y + z = 0,1 (III)
Gi i h : x = 0,015 ; y = 0,025
ch c ch n trong g 2
trong g c axit no. V y mu i axit
0,015
HCOONa
CH 2 0,06
CH 3 COONa 0,015

D c a a = ch n C
ng d n gi i 2
( C n H 2n-2 O 4 )
( C m H 2m-10 O 6 )
R 1 COO
R 2 COO
x mol
R 3 COO
R 3 COO C 3 H 5
R 3 COO
y mol
C 3 H 6
( Gi s ng ch t TN1 g p k l m 2)
Ta d
x + y = 0,16k
2x + 3y =0,42 k
x = 0,06k
y = 0,1 k
O 2
0,5 mol
NaOH
0,42 k mol
t = 4x + 6y + 0,1
Theo s ch s mol CO2 2 s :
x 5y
CO 2 H 2O
T (1)(2) gi i h c
x
y
=
3
5
(1)
CO 2
0,45 mol
H 2 O
t mol
n n 0,45 - 4x - 6y - 0,1 = x + 5y 5x + 11y = 0,35 (2)
x = 0,015
y = 0,025
Ch c c p nghi m :
n = 10
m = 12
0,015.n + 0,025.m = 0,045
D c a = 12,36
202-2018): H n h p X g
gam X thu c 1,56 mol CO 2 2O. M ng v v i 0,09 mol NaOH trong
d ch ch ch a a gam h n h p mu i natri panmitat, natri stearat. c
A. 25,86. B 26,40. C 27,70. D 27,30.
ng d n gi i 1
i X:
C 15 H 31 COOH
(C 15 H 31 COO) 3 C 3 H 5
CH 2
bi
x
y
z
C 15 H 31 COOH
(C 15 H 31 COO) 3 C 3 H 5
CH 2
B
D ch s mol CO2 2
2y =1,56-
B
V y mu i :
C 15 H 31 COONa 0,09
CH 2 0,06
C 15 H 31 COOH
x mol
C 17 H 35 COOH
y mol
(RCOO) 3 C 3 H 5
z mol
m=24,64 gam
(C n H 2n - 4 O 6 )
Ta de co : x + y + 3z = 0,09
O 2
CO 2
NaOH
0,09 mol
H 2 O
1,56 mol
1,52 mol
Muoi
H 2 O
c a =25,86 gam ch n A
ng d n gi i 2
O
4,46 mol
( 1)
(2)
c : nO = 4,46
x
C 3 H 5 (OH) 3
CO 2
1,56 mol
H 2 O
1,52 mol
NaOH
0,09 mol
y
C 3 H 8 O 3
z
H 2 O ( x + y)
Muoi

S ch s mol CO 2 mol H 2O : z = 0,02 ; x + y = 0,03
x
CH 2 = C -COO - CH 2 -C = CH
H n h p E g m ba este m ch h t este
CH 3
COOCH 3
O 2
CO 2
gam E b ng O2 c 0,37 mol H2O. M n ng v v i 234 ml dung d ch
NaOH 2,5M, t c h n h p X g i c
ng m1
kh ng m 2 gam. T l m 1 : m 2 g n nh t v
A. 2,7. . B 1,1. . C 4,7. . D 2,9.
ng d n gi i 1
Theo gi thi i E:
y
z
t
CH 2 = C
COO CH 2 -CH =CH 2
CH COOCH 3
CH COO CH 2 -CH =CH 2
CH 2
NaOH
0,585.a
H 2 O 0,37 mol
CH 2 = C -COO - CH 2 -C = CH
12,22 gam
CH 3
COOCH 3
x + y + z = 0,36.a
CH 2 = C
COO CH 2 -CH =CH 2
Ta d :
n
CO2
- n
H2 O = 3 (x + y + z) n
CO2
= 1,08.a + 0,37
CH
CH
CH 2
bi
COOCH 3
COO CH 2 -CH =CH 2
x + 2y + 2z + n = n + 0,5. n
O2 CO2
H2 O
n O2
= 0,495.a + 0,555
c a = 2/9
x = 0,03
y + z =0,05
c t = 0
V p ch
ng:
Ch n D
CH 3 OH 0,05 mol
CH 2 =CH - CH 2 -OH 0,05 mol
CH = C - CH 2 - OH 0,03 mol
m 2 = 1,6 gam
m 1 = 4,58 gam
m 1
m 2
= 2,8625
ng d n gi i 2

222-2019):H n h p E g m ba este m ch h u t o b i axit cacboxylic v i ancol :
C n H 2n -6 O 2a
t mol
12,22 gam
CO 2
O 2 ( 0,37 + 3 t ) mol
(2)
( 1)
H 2 O
0,37 mol
NaOH
0,585 mol
Theo gt ta d = 1,625
ng DDLBTKL : 12,22 = mC + mH + mO = (0,37 + 3t ).12 + 0,37.2 + t.3,25.16
d ng v v i dung d c 12,88 gam h n h 24,28
gam h n h p T g m ba mu i c T c n v 0,175 mol O 2 , thu
c Na2CO3 ,CO2 2O .Ph ng c g n nh t v
?
A. 9. B. 12. C.5. D. 6
ng d n gi i
D a theo d ki
HCOOCH 3
CH 2 = CH- COOCH 3
COOCH 3
COOCH 3
V t trong E :
x
y
z
: n= 7,625.
CH 2 = C -COO - CH 2 -C = CH
CH 2 = C
CH
CH
CH 3
COOCH 3
COO CH 2 -CH =CH 2
COOCH 3
COO CH 2 -CH =CH 2
a nhau ; t o ra mu ; ancol t o
C ph ;
x
y
CH 2
E
bi c:
HCOOCH 3
CH 2 = CH- COOCH 3
COOCH 3
z
COOCH 3
t CH 2
E
+ NaOH
ROH (12,88 gam )
x + y + z = 0,2 (I)
24,28 gam
Bao toàn khôi luong : 24,28 + 0,175.32 = 0,055.18 + 106 u + 44 v
Bao toàn Na: x + y + 2z = 2u
Bao toàn Oxi: 2x + 2y + 4z + 0,175.2 = 3u +2v + 0,055
x + y + 2z = 0,35 (II)
Bao toàn C : x + 3y + 2z + t' = u + v =0,41
Bao toàn H: x + 3y + 2t' = 0,055.2
x HCOONa
y CH 2 = CH- COONa + O 2
COONa
0,175 mol
z
COONa
t' CH 2
x + 3y + 4z = 0,71 (III)
28,89 = 106u + 44v
0,295 = - u + 2v
Na 2 CO 3
u
CO 2 v
H 2 O 0,055 mol
u= 0,175
v= 0,235
B
Gi i h : x= 0,03 ; y + z =0,05
ng:
Ch n D
CH 3 OH 0,05 mol
CH 2 =CH - CH 2 -OH 0,05 mol
CH = C - CH 2 - OH 0,03 mol
m 2 = 1,6 gam
m 1 = 4,58 gam
m 1
m 2
= 2,8625
T
x= 0,02
y = 0,03
z = 0,15
t' = 0
c t = 0,12
2 ch thu c c
0,12 = 0,02n + 0,03 .m + 0,15.l ( n,m,l l 2 Trong X,Y,Z )
duy nh
c
2n + 3m = 12
m 0 1 2 3 4
n 6 4,5 3 1,5 0

Lo i Lo i Ch n Lo i Lo i
Lo : s l ho m b o t
V : HCOOC 4H 9
2,04
%X =
. 100% = 8,81%
23,16
Ch n A
T
x= 0,05
y = 0,03
z = 0,5
t' = 0
213-2019):H n h p E g m ba este m ch h u t o b i axit cacboxylic v i ancol :
c).Cho 0,58
ng v v i dung d c 38,34 gam h n h
n h p T g m ba mu i c T c n
v 0,365 mol O 2 c Na 2CO 3 , H 2 2 .Ph ng c a Y trong
E g n nh t v ?
A. 8. B. 5. C.7. D. 6
D a theo d ki
HCOOCH 3
ng d n gi i
2 ch thu c c
c m E = 68,36 gam ; t = 0,27
0,27 = 0,05n + 0,03 .m + 0,5.l ( n,m,l l 2 Trong X,Y,Z )
5n + 3m = 27
duy nh
n 0 1 2 3 4 5
m 9 7,3 5,6 4 2,3 0,67
Lo i Lo i Lo i Ch n Lo i Lo i
Lo : s l ho m b o t
V : CH 2 = CH - COOC 5H 11
c
x
y
CH 2 = CH- COOCH 3
COOCH 3
COOCH 3
CH 2
E
bi c:
HCOOCH 3
CH 2 = CH- COOCH 3
COOCH 3
z
COOCH 3
t CH 2
E
+ NaOH
ROH (38,34 gam )
x + y + z = 0,58 (I)
73,22 gam
Bao toàn khôi luong : 73,22 + 0,365.32 = 0,6.44 + 106 u + 18 v
Bao toàn Na: x + y + 2z = 2u
Bao toàn Oxi: 2x + 2y + 4z + 0,365.2 = 3u +v + 0,6.2
x + y + 2z = 1,08 (II)
Bao toàn C : x + 3y + 2z + t' = u + 0,6 =1,14
Bao toàn H: x + 3y + 2t' = v.2 =0,14
x HCOONa
y CH 2 = CH- COONa + O 2
COONa
0,365 mol
z
COONa
t' CH 2
x + 3y + 4z = 2,14 (III)
x + 3y + 4z = 2,14 (III)
58,5 = 106u + 18v
- 0,47 = - u + v
Na 2 CO 3
u
CO 2 0,6 mol
H 2 O v
u= 0,54
v= 0,07
%X =
T
4,26
68,36
48,28 gam
. 100% = 6,23 %
NaOH
0,47 mol
Ch n D
o BGD 2019): H n h p T g m ba este X, Y, Z m ch h (MX < MY < MZ). Cho 48,28
ng v a v i dung d ch ch c m t mu i duy nh t c a axit
n h p Q g ch h .
2 2O. Ph i ng c H
A. 9,38%. B. 8,93%. C. 6,52%. D. 7,55%.
n= 3
Bao toan OH: x =2,35
ng d n gi i
RCOONa 0,47 mol
H 2 O 0,8 mol
O 2
C n H 2n + 2 O x
0,2
CO 2
0,6 mol
ancol Muoi Muoi = 108

( CH 2=CH-CH 2 COONa)
CH 2 =CH- CH 2 - COOC 3 H 7
CH 2 =CH- CH 2 - COO
CH 2 =CH- CH 2 - COO
C 3 H 6
CH 2 =CH- CH 2 - COO
CH 2 =CH- CH 2 - COO C 3 H 5
CH 2 =CH- CH 2 - COO
% KL H trong Y : 7,55% ch n D
( Thi th Vinh 2018): a m
este no, hai ch u m ch h n h p E ch a X, Y c
mol O2. M gam E c c m t ancol duy
nh n h p ch a mu i kali c a hai axit cacboxilic. T ng s
A. 16. B. 14. C. 18. D. 12
( Thi th c 2018): X, Y u hai ch X no, Y a
m t C=C); Z este thu n ch c t o b i X, Y T n h p E ch a X,
Y, Z (s mol c a Y g p 2 l n s mol c a Z) c 2. M E v i
440 ml dung d ch NaOH 1M (v c m t ancol T duy nh n h p F g m a gam mu i A
b gam mu i B (MA< MB). D T y kh ng
th
H2 l a : b g n nh t v i
A. 3,6. B. 3,9. C. 3,8. D. 3,7.