PHƯƠNG PHÁP QUY ĐỔI ESTE THÔNG QUA ĐỀ THI THPTQG 2019

I GI

P ESTE

m

Nh cao trong th i gian thi ch truy

ih ih i nh

ng c i ti i v m t th ng d n gi

v ng s

ng!

202-2019): t ng v i dung d ch NaOH v ct i 0,06 mol Br2 trong dung d A. 27,72.

B. 26,58.

C.27,42.

c CO2

2O.

i .M c D. 24,18.

ng d n gi i L id b t b ng s

id t pi

mol H2 b

g c)

R1COOCH2

C3H8 a CH2 b COO 3a - H2 c

R2COOCH R3COOCH2 a mol

C3H5 HCOO CH2 - H2

CH4 a CH2 t COO 3a - H2 0,06

a 3a b c

HCOOCH2 a

HCOOCH HCOOCH2 CH2 - H2

b c

ng d n gi i

C3H5 a HCOO 3a CH2 b - H2 0,06 25,74 gam

H2O 1,53 mol

NaOH

HCOONa 3a CH2 b - H2 0,06 C3H5(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

Cho

176a + 14b = 25,86 a = 0,03 ;

T

(II) b = 1,47 A. 16,12. nv

203-2019): 2

.M A. 0,2.

B. 0,24.

ng v i dung d ch NaOH v ct i a mol Br2 trong dung d C.0,12.

D. 0,16.

3,08 mol O2

207-2019): ng v i dung d ch NaOH v ct i 0,04 mol Br2 trong dung d c B. 18,48.

C.18,28.

D. 17,72.

c CO2 i

O2 3,08 mol

C3H5 a HCOO 3a CH2 b - H2 0,04

HCOONa 3x CH2 y - H2 a

H: (I) ng : 204.x + 14y 2.a = 35,36 O: 3x + 3,08 = 6x+ y + 1 (III) x = 0,04 ; y = 1,96 ; a = 0,12

T (I)(II)(III)

204-2019): 2

mu i .M A. 0,09.

B. 0,12.

NaOH

17,16 gam

H2O 2 mol

6a + b = 1,1 ng BT kh

HCOONa 3a CH2 b - H2 0,04 C3H5(OH)3 3a

C:

35,36 gam C3H5(OH)3 x

4x + y a = 2 ng BT kh

(I) ng : 176a + 14b 0,08 = 17,16

T (II)

nv ng v i dung d ch NaOH v ct i a mol Br2 trong dung d C.0,15.

D. 0,18.

2,31 mol O2

c H2O

- 2014): t thu ng c cacbon v i X; T ct ob h n h p E g m X, Y, Z, T c n v ng t i dung d ch ch a 0,04 mol Br2. Kh ng h t v i dung d A. 4,68 gam.

B. 5,04 gam.

C. 5,44 gam.

ng mu D. 5,80 gam.

c ng d n gi i 1 DLBTKL

ng d n gi i

C3H5 x HCOO 3x CH2 y - H2 a m gam

O2 2,31 mol

H2O (4x + y - a) mol CO2 1,65 mol

NaOH

HCOONa 3x CH2 y - H2 a 26,52 gam C3H5(OH)3 x

C: (I) ng : 204.x + 14y 2.a = 26,52 O: 3x + 2,31 = 2x+ 0,5.y 0,5 y + 0,5.a = - 0,66 (III) T (I)(II)(III) x = 0,03 ; y = 1,47 ; a = 0,09 6x + y = 1,65 ng BT kh

Cho

H2O CO2 1,1 mol

CO2 (6x + y) NaOH

m gam

2.

i .M

ng d n gi i

c

ng d n gi i

C3H5 x HCOO 3x CH2 y - H2 a

c H2O

(II) 0,5.a + 1,65

E

O2 0,59 mol

CO2 0,47 mol

11,16 gam H2 O

0,52 mol

9,36 gam Nh n th y s mol H2O l ih nh

mol CO2

c an col trong E ph i no

c. M t c khi cho

CH2 = CH - COOH C3H6(OH)2

HCOOCH3 CH3- CH = CH- COOCH3

CH2=CH-COO

C3H6

CH2

CH2=CH-COO

ph n ng : CH2

a ph n ng:

CH2=CH-COO c

O2 0,59 mol

CO2 0,47 mol

Br2 0,04 mol

C3H6

CH2=CH-COO d

CH3OH 0,08 mol

b CH3- CH = CH- COOCH3

a CH2 = CH - COOH b C3H6(OH)2

H2O 0,52 mol Muoi

KOH

CH2

( THPTQG 2015): H n h p X g c, t ch c, ch a m gam X b ng dung d c h n h p mu n ,96 gam H2O. Ph

CH2

H2O 0,22 mol

B B 0,22 B ng: 60a + 100.b + 14.c = 5,88 Gi i h : a = 0,06 ; b= 0,02; c=0,02

( m gam)

C3H6(OH)2 ( b + c )

B D ch s mol CO2 3c = 0,05 2O : -a + b B t pi g ng: 72.a + 76.b + 184.c + 14.d =11,16 B Gi i h : a= 0,02 ; b=0,1 ; c=0,01 ; d = 0,02 c ch 3H6(OH)2. : m = 4,68 gam

O2

5,88 gam

H2O a mol

(a + 2c ) 11,16 gam

c

Muoi

NaOH 0,08

HCOOCH3

(II) (III) ch

(I) (II)

2

2

2

V

c : 34,01% ch n A

(IV)

: X, Y

X

2

t ancol Y v i 3 axit ng k ti t axit ). Th y ph Y. Cho m gam Y ng Na

2

< MY

2

A. B. C. D.

ng c a este

ng d n gi i

DLBTKL A. 34,01%.

B. 38,76%.

C. 40,82%. ng d n gi i

Na ROH 0,08

D. 29,25%.

M

O2 0,1975 mol

H2O ancol

i X:

7,7 gam

5,07 gam

H2 0,04 mol

m

CO2 0,175 mol

0,205 mol

3,69 gam

= 32 ( CH3OH ) Nh n th y s mol H2O l ih nh

mol CO2

c an col trong M ph i no

HCOOH C2H4(OH)2

x

HCOO C2H4

CH3COOH

HCOO

y C2H5COOH

CH2

0,04 C2H4(OH)2

a HCOOH b C2H4(OH)2

CO2

O2 0,1975 mol

0,01

0,175 mol

C2H4

C2H5COO x + y =0,02

HCOO c

CH3COO

C2H4

H2O

HCOO

( Loai )

0,205 mol

x +2.y + 0,03 = 0,035 (Bao toan CH2)

KOH 0,1 mol

d CH2

-2017). M t h n h p E g h kh

5,07 gam (I) (II) c = 0,03 (III) 2 2 5 (IV) 0,02 ; b=0,04 ; c=0,01 ; d=0,035 2H4(OH)2

G TH1 :

ch 560 ml dung d

nv

u no ,m ch c hai mu

nh pTg 2

A.43,0.

c 2

B. 37,0.

c a a g n nh t v C.40,5.

13,5 .

ng d n gi i 1 i E: HCOOCH3

HCOO -CH2 HCOO- CH2

x

HCOOH

y

CH3COOH

CH2 ph n ng: x HCOOCH3

0,04 C2H4(OH)2 HCOO 0,01

y

C2H4

z

Ta d

y + 0,01 = 0,035 (Bao toan CH2)

TH2:

CH2

0,56 mol a gam

v

Bao toan O

HCOO- CH2

CH3COO

x + y =0,02

HCOO -CH2

HCOONa NaOH 0,56 mol

CH3OH

CH2

x

C2H4(OH)2 y

x= 0,005 40,48 gam

CH2

y = 0,015

O2

CO2

0,72 mol

H2 O

1,08 mol

t 51,12 gam

B B n = 0,98 mol O2

2y = 0,56

(I)

= 19,76 gam ch n A

ancol

HCOO C 3H6 L i Gi i 2

16,128 0,72mol 2 22, 4 19, 44 n H 2O 18 1, 08mol

nCO

CH3COO CH2=CH-COO

n CO

2

n

H 2O

CH2=CH-COO CH2=CH-COO ch h

n

T

C 3H5

CH2

1,08 0,72 0,36

bi

CnH2n+2Ox

O2

0,36

n

0, 72 0,36

Bao toan H

HCOO

n CO2 + ( n+1) H2O 0,72

C3H6

x

CH3COO

H2O (5x + 7y +z) O2 0,5 mol

2

V y ancol T g m C2H5

2H4(OH)2

CH2=CH-COO CH2=CH-COO

ph n

CO2 0,45 mol

y

C3H5

CH2=CH-COO x mol C2H5OH E 40,48

Muoi +

+ NaOH 0,56 mol

gam

22,4

gam

a gam

y mol C2H4(OH)2

x+ y = 0,36

x= 0,16

x + 2y = 0,56

y = 0,2

(Ta coi ph n TN1 g p t l n ph n TN2) Ta d x = 0,06t

x + y = 0,16t 2x + 3y =0,42 t

ch n A 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 nh pEg nv 0,5 mol O2 c 0,45 mol CO2 .

M t c nv 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.

y = 0,1 t

0,015

CH3COONa 0,015

ng d n gi i 1

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 trong g c axit no. V y mu i axit HCOONa

i E:

NaOH 0,42.t

0,16.t

x + y = 0,36

B

z

CH2

CH2

0,06

2

D

c aa=

ch n C

i X:

ng d n gi i 2

C15H31COOH

R1COO C3H6

( CnH2n-2O4 )

(C15H31COO)3C3H5

R2COO x mol R3COO

( CmH2m-10O6 )

R3COO

CO2 0,45 mol

O2 0,5 mol C3H5

H2 O

y mol

1,56 mol

y (C15H31COO)3C3H5 H2O 1,52 mol

z CH2

0,42 k mol

ng ch t TN1 g p k l

NaOH 0,09 mol

m 2)

Muoi H2O x

x = 0,06k

x y

y = 0,1 k

=

3

C3H5(OH)3 y

5

(1) t = 4x + 6y + 0,1 Theo s ch s mol CO2 nCO 2 n H 2O x 5 y c T (1)(2) gi i h

CO2

O2

t mol NaOH

x + y = 0,16k 2x + 3y =0,42 k

bi x C15H31COOH

R3COO

( Gi s Ta d

CH2

s : 0,45 - 4x - 6y - 0,1 = x + 5y

2

5x + 11y = 0,35 (2)

B D 2y =1,56B V y mu i :

ch s mol CO2

2

C15H31COONa 0,09 CH2

x = 0,015

0,06

c a =25,86 gam ch n A

y = 0,025

ng d n gi i 2 0,015.n + 0,025.m = 0,045 Ch

c c p nghi m :

n = 10 m = 12

D

c a = 12,36

C15H31COOH x mol C17H35COOH y mol (RCOO)3C3H5 (CnH2n - 4 O6) z mol

CO2 O 4,46 mol ( 1) (2)

m=24,64 gam

gam X thu d A. 25,86.

202-2018): H n h p X g c 1,56 mol CO2 ng v v i 0,09 mol NaOH trong 2O. M ch ch ch a a gam h n h p mu i natri panmitat, natri stearat. c B 26,40.

C 27,70.

D 27,30.

H2O 1,52 mol H2O ( x + y) NaOH Muoi 0,09 mol

Ta de co : x + y + 3z = 0,09

c : nO = 4,46 ng d n gi i 1

1,56 mol

C3H8O3 z

S

ch s mol CO2

mol H2O : z = 0,02 ; x + y = 0,03 x

CH2= C -COO - CH2-C = CH O2

CH3

H n h p E g m ba este m ch h gam E b ng O2 NaOH 2,5M, t

t este

c 0,37 mol H2O. M ch nh pXg

n ng v

CO2

COOCH3 y CH2 = C

v i 234 ml dung d ch

H2O 0,37 mol

COO CH2 -CH =CH2

ic ng m1

kh

ng m2 gam. T l m1 : m2 A. 2,7. . B 1,1.

g n nh t v . C 4,7.

z

. D 2,9.

ng d n gi i 1 Theo gi thi

i E:

t

CH2= C -COO - CH2-C = CH

CH

COOCH3

CH

COO CH2 -CH =CH2

NaOH 0,585.a

CH2 12,22 gam

CH3 x + y + z = 0,36.a

COOCH3 CH2 = C COO CH2 -CH =CH2 CH

COOCH3

CH

COO CH2 -CH =CH2

Ta d : n - n CO2 H2O

n CO2

= 3 (x + y + z)

= 1,08.a + 0,37

= n + 0,5.n H O x + 2y + 2z + n 2 O2 CO2 n O = 0,495.a + 0,555 2

CH2

c a = 2/9

bi x = 0,03 y + z =0,05

V

ct=0 p ch

ng:

m2 = 1,6 gam

CH3OH 0,05 mol

m1 m2

CH2=CH - CH2-OH 0,05 mol m1= 4,58 gam CH = C - CH2 - OH 0,03 mol Ch n D

ng d n gi i 2

= 2,8625

222-2019):H n h p E g m ba este m ch h d ng v v i dung d gam h n h p T g m ba mu i c c Na2CO3 ,CO2 ?

A. 9.

B. 12.

CO2 O2

CnH2n -6 O2a t mol

( 0,37 + 3 t ) mol

( 1) (2)

12,22 gam

H2 O 0,37 mol

Tc nv 2O

.Ph

ng c D. 6

C.5. ng d n gi i

CH2 = CH- COOCH3 COOCH3 COOCH3 CH2

: n= 7,625.

E a nhau ; t o ra mu C ph ;

; ancol t o

bi

c:

x HCOOCH3 y CH = CH- COOCH 2 3 COOCH3 z COOCH3

t trong E :

CH2= C -COO - CH2-C = CH CH3

ROH (12,88 gam ) + NaOH

t CH2

y CH2 = C

x HCOONa y CH2 = CH- COONa z

COONa

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

COO CH2 -CH =CH2

+ O2 0,175 mol

COOCH3 COO CH2 -CH =CH2

0,295 = - u + 2v

v= 0,235

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 + 3y + 4z = 0,71 (III)

T

x=0,02

B Gi i h : x= 0,03 ; y + z =0,05 ng:

CH3OH 0,05 mol

y = 0,03

m1= 4,58 gam Ch n D

m1 m2

CH2=CH - CH2-OH 0,05 mol CH = C - CH2 - OH 0,03 mol

t' = 0

z = 0,15 m2 = 1,6 gam

= 2,8625

c t = 0,12 ch thu cc 0,12 = 0,02n + 0,03 .m + 0,15.l ( n,m,l l duy nh 2n + 3m = 12 m 0 1 2 3 4 n 6 4,5 3 1,5 0 2

2

c

v

u=0,175

Bao toàn Oxi: 2x + 2y + 4z + 0,175.2 = 3u +2v + 0,055

CH

CO2

28,89 = 106u + 44v

Bao toàn Na: x + y + 2z = 2u

CH

Na2CO3 u

H2O 0,055 mol

COONa t' CH2

E

COOCH3

z

24,28 0,175 mol O2 , thu g n nh t v

HCOOCH3 NaOH 0,585 mol

x

c 12,88 gam h n h

D a theo d ki

Theo gt ta d = 1,625 ng DDLBTKL : 12,22 = mC + mH + mO = (0,37 + 3t ).12 + 0,37.2 + t.3,25.16

V

u t o b i axit cacboxylic v i ancol :

Trong X,Y,Z )

T

Lo i Lo i Ch n Lo i Lo i : s l ho mb ot : HCOOC4H9 2,04 . 100% = 8,81% %X = 23,16 Ch n A

Lo V

x=0,05 y = 0,03

t' = 0

z = 0,5

v E A. 8.

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 Tc n 0,365 mol O2 c Na2CO3 , H2 ng c a Y trong 2 .Ph g n nh t v ?

B. 5.

C.7.

D. 6

c mE = 68,36 gam ; t = 0,27

213-2019):H n h p E g m ba este m ch h

2

Trong X,Y,Z )

c

5n + 3m = 27 0 1 9 7,3 Lo i Lo i : s l ho

Lo

2 5,6 Lo i mb

3 4 Ch n ot

4 2,3 Lo i

5 0,67 Lo i

: CH2 = CH - COOC5H11

V

CH2 = CH- COOCH3 COOCH3

2

duy nh

ng d n gi i

HCOOCH3

cc

0,27 = 0,05n + 0,03 .m + 0,5.l ( n,m,l l

n m

D a theo d ki

ch thu

%X = 4,26 . 100% = 6,23 % 68,36 Ch n D

COOCH3 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 nh pQg ch h .

2 2O. Ph i ng c H

CH2 E bi

c:

x HCOOCH3 y CH = CH- COOCH 2 3 COOCH3 z COOCH3

A. 9,38%.

ROH (38,34 gam ) + NaOH

x HCOONa y CH2 = CH- COONa

t CH2

z

+ O2 0,365 mol

COONa

ng d n gi i

CO2 0,6 mol

RCOONa 0,47 mol 58,5 = 106u + 18v

u= 0,54

- 0,47 = - u + v

v= 0,07

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)

NaOH 0,47 mol T

H2O 0,8 mol O2

48,28 gam CnH2n + 2Ox 0,2

n= 3

CO2 0,6 mol

Bao toàn C : x + 3y + 2z + t' = u + 0,6 =1,14 Bao toàn H: x + 3y + 2t' = v.2 =0,14

x + 3y + 4z = 2,14 (III) x + 3y + 4z = 2,14 (III)

D. 7,55%.

Na2CO3 u

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

C. 6,52%.

H2O v

COONa t' CH2

E

B. 8,93%.

Bao toan OH: x =2,35 ancol

Muoi

Muoi

= 108

( CH2=CH-CH2

COONa)

CH2=CH- CH2- COOC3H7 CH2=CH- CH2- COO C 3H 6 CH2=CH- CH2- COO CH2=CH- CH2- COO CH2=CH- CH2- COO

C 3H 5

CH2=CH- CH2- COO

% KL H trong Y : 7,55% ch n D

( Thi th Vinh 2018): este no, hai ch u m ch h mol O2. M gam E c nh n h p ch a mu i kali c a hai axit cacboxilic. T ng s A. 16.

B. 14.

C. 18.

( Thi th c 2018): X, Y u hai ch m t C=C); Z este thu n ch c t o b i X, Y T Y, Z (s mol c a Y g p 2 l n s mol c a Z) c 2. M c m t ancol T duy nh 440 ml dung d ch NaOH 1M (v b gam mu i B (MA< MB). D T y kh th H2 l a : b g n nh t v i A. 3,6.

B. 3,9.

C. 3,8.

am n h p E ch a X, Y c c m t ancol duy D. 12

X no, Y

a n h p E ch a X, Ev i n h p F g m a gam mu i A ng

D. 3,7.