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.