X = 7 .5000 = 35000 mg = 35 gam 1.64. Dap so dung la C.
Cac phuong trinh phan umg :
(R-COOgCjHs +3NaOH 3R-C00Na + C3H5(OH)3 (i)
NaOH + HCl -> NaCl + HjO (2)
S6'"mol NaOH ban diu = - 34,25 mol.
40
S6'mol NaOH dir = 0,5 .1 = 0,5 mol.
So mol NaOH tham gia phan irng (1) = 34,25 - 0,5 = 33,75 mol. Theo(l) : mg,jx„ol = -92 = 1035 gam = 1,035 kg.
nixa phong = 10-000 + (33,75 .40) - 1035 gam = 10315 gam.
1.65. Dap an dung la D.
1.66. NhOng cau phat bleu diing la C. 1.67. Dap an dung la B. 1.67. Dap an dung la B.
Kl hieu hai g6'c axit tircfng iJng la Ă R-COO" ) va B ( R'-COO"). Viet CTCT cua cac triglixerit thu dugc :
2 triglixerit chi c6 1 loai g6'c axit. 4 triglixerit chi c6 2 loai gfíc axit.
1.68. Cong thiJc dung la B.
Phuong trinh phan irng xa phong hoa :
(R-C00)3R' +3NaOH -» 3R-C00Na + R'(0H)3 Imol 28,2 gam 9,2 gam
(0,3 mol) (0,1 mol) Ki hieu R = C^Hy ; R' = C^Hb.
C^Hy -COONa = ^ = 9 4 - > x = 2 ; y = 3. 156
C,Hh(OH)3 = ^ = 92 a = 3 ; b = 5.
Vay CTCT cua este la (C2H3COO)3C3H5.
1.69. Dap an diing la D.
Cho Na2C03 vao 4 chát. Noi nao c6 khi bay ra la axit axetic, nai nao tao thanh hai 16p chát long (néu Ik kl, de lau c6 tan m6t it) la đu an. Con lai rugu va nu6c phan bidt bang each dot chaỵ
1.70. Dap so dung la B.
Phirong iiinh phan irng xa phong hoa :
(C|7H33C00)3C3H5 +3NaOH ^ 3C|7H35C00Na + C3H<s(0H)3 ^ ^ ^ ^ = 11,24 mol 3.11,24 mol 11,24 mol
890
phong = 3 .11,24 .306 = 10.318 gam = 10,318 kg.
mgiixerol = 11'24 .92 = 1034 gam = 1,034 kg. 1.71. Cong thurc diing la D.
S6 mol h6n hop hai este đng phan = ^ ^ = 0,02 mol.
88
C4H8O2 la c6ng thiic ciia cac este no, don churc, mach họ Do do s6
mol muoi tao thanh bang s6' mol este va bang 0,02 mol. Goi Mm la khC'i lugng mol trung binh ciia hai mu6'i tao thanh
Mm = — = 75 gam
Trong do McH^^coONa = gam -> Muoi thur hai c6 M < 75. Do la muoi natri ciia axit fomic (HCOONa).
Til CTPT C4H8O2 c6ng thiJc cáu tao cua 2 este : ^, „. ,
C H 3 - C O O - C H 2 - C H 3 va H - C O O - C H 2 - C H 2 - C H 3 . 1.72. C6ng thurc diing la Ạ
Dat CTCT cua X Ik : C^Hy-COO- C 2 H 5 .
Phuong trinh phan ling :
C^Hy - C O O- C 2 H 5 + NaOH CxHyCOONa + C2H5OH (
S6 mol NaOH tham gia (1) : nN^oH = 0-125 .1 .75% = 0,09375 mol.
Theo (1) : nx = nN^oH = 0,09375 mol. 6,9375
-> Mv = = 74 gam. ^ 0,09375
12x + y + 44 + 29 = 74 ^ x = 0 ; y = 1. Du do CTCT cua X la : H - COO - CjH^ .
1.73. Cong thuc dung la D.
nx =0,1 mol (tính theo cong thirc pV=nRT). 8 6
Mx = —0,1 = 86 ->• X phai la este dan chiic R - C O O - R ' . R - C O O - R ' +NaOH R-COONa + R'OH
0,1 mol 0,1 mol
MR-COONa= 82 ^ R = 82 -67 = 15 - ^ R l a CH3. ^ R ' - 8 6 - 15 - 4 4 = 27 R ' l a C H 2= C H - .
Vay CTCT cua X la : CH3 - COO - CH = CH2.
1.74. C6ng thiic diing la B.
Trong 4 dap an A, B, C, D, chi c6 A va B tac dung vdfi NaOH tao r;! andehit, trong do B tao mu6'i H-COONa c6 M < 70 :
H - COO - CH = C H - C H 3 + N a O H H - C O O N a + CjH^-CHO
1.75. C6ng thiJc dung la C.
C H 3- O O C - C O O- C 2 H 5 + 2NaOH->
^ NaOOC - COONa + CH3OH + C2H5OH
-176. C6ng thu-c dung la C.
C H 3- C O O- C H 2 - C H 2- O O C - H + 2 N a O H ^
^ CH3 - COONa + H - COONa + HO - CH2 - CH2 - OH
-1.77. C6ng thiic diing la D.
H - COO - CH = CH - CH3 + NaOH ^ H - COONa + C2H<; - CHO Hai san pham cia phan ling nay d6u c6 kha nang tham gia phan iJng trang guang.
1.78. Dap an diing la B.
Phi/ang trinh phan lirng :
(R-COO)3C3H5 +3NaOH ^ 3 R-COONa + C3H3(OH)3 (1) Theo (1) s6' mol glixerol = ^ s6' mol NaOH = ^ 0,1 mol.
-> mo = 0,1 .92 = 9,2 gam.
Theo dinh luat bao toan khd'i lugng :
^ m ^ p - 89 + 0,3.40-9,2 = 91,8 gam.
1.79. Dap so duiig la Ạ ' '
Co nhicu each tra Idi, nhixng phuang phap bao toan kh6'i luong la dan gian nhát :
S6' mol ancol trong - h6n hap = 2. nu, = 2. = 0,1 mol. 2 ^ 22,4 30
S6 mol axit du = — = 0,5 mol ntn phai tinh theo ancol. 60
CH3COOH + C2H5OH ^ CH3COOC2H5 + H2O ( I )
CH3COOH + ROH -» CH3COOR + H2O (2)
0,1 mol 0,1 mol 0,1 mol
Hieu suát 100% nen ta CO : 0,1 .60 + 3,9 = meste + 0,1 .18
mesie = 6 + 3 , 9 - 1,8 = 8,1 gam. 80
Thirc te m'este ^ ^'^"[55 ^ ^''^^
1.80. C6ng thirc diing la D.
Tdng s6' mol este = n^aOH = 0,1.1 = 0,1 mol Meste = ^ = Este duy nhát c6 M < 67 la H - COOCH3.
Este thu- 2 phai la C3H6O2 •
Vi: R-COOR^ + NaOil -> R-COONa + R^OH 0,1 mol 0,1 mol
' R , = l
R2 =15
vay este thii hai phai la C H 3 - C O O C H 3 .
Meste = — = 7 5 -» R = 75 - 67 = 8 ^ 0,1 1.81. Dap s6'diing la B.
Phuong trinh phan ling :
R-COOH + R'OH < > R-COO-R' + HjO Trirdc p,ir: 1 mol Imol
p,u: X mol X mol
Sau p,ir: 1-x 1-x x x x_ Ji_ Ta c6 : Kp = . ^ Y = 4 (V la th^ tich h6n hap). ~V~' V ^ = 2 - > X = - tiic la 66,7% 1-x 3 1.82. C6ng thiJc diing la C.
R-COO-R' + NaOH ^ R-COONa + R'OH 0,1 mol 0,1 mol 0,1 mol 0,1 mol
10,4 + 0,1.40 = 9,8 + m„ m = 4,6 M„ = — = 46.
d d d QJ
^ CTPTancol la C2H5OH.
vay CTCT cua este la : HO-CHj -COO-CjHg .
1.83. Dap an diing Ik C.