9.21. Chát nao la nhien lieu sach cho tuong lai ? A . K h i than kh6 ; A . K h i than kh6 ;
B. K h i hidro ; C. K h i thien nhien ; C. K h i thien nhien ; D. Khf than ir6t.
9.22. Chat long nao la nhien lieu tuong lai cho 6t6, xe miy vi it gay 6 nhiem
mOi truemg ? A . Benzen ; B. Xang diu ; C. Diu diezel; D. Ancol etylic. 140 \ Phanll
HL/dNG DAN TRA LOi
I Chifafng 1 Chifafng 1 € S T € - U P I T 1.1. Dap an dung la C. 1.2. Dap an diing la Ạ 1.3. Dap an diing la B.
Thi du : CH3 - COO - C2H5 + NaOH ^ CH3 - COONa + C2H5OH.
1.4. Dap an dung la D.
Axit panmitic C15H3, - C O O H , axit stearic C17H35 - C O O H .
1.5. T6n gpi diing la A . 1.6. Dap an dung la C. j 1.6. Dap an dung la C. j 1.7. Dap dn diing la C. 1.8. Dap an dung la Ạ 1.9. C6ng thilc dung la B. 1-10. Dap an dung la D. 1-11. Dap an dung la B. 1-12. Dap an dung la C. 1-13. Dap dn diing la C. ' ' ' •••14. Dap an diing la Ạ - < Phuong trinh phan ling este hoa :
CH.COOH + CsH^OH :f=± C H 3 - C O O C 2 H 5 + HjO 60 , ,
S6 mol axil, ancol, este : ncH,COOH = 77; = ^ ' nc,H,= ^ =2,174 mol ; n,,,e= f = 0'625 mol.
Hieu suat phan iJng : H = ^^^.100% = 62,5 %.
1.15. Dap s6'dung la B.
Phirang tiinh phan ling thuy phan :
(C,7H35COO)3C,H,+3NaOH ^ 3CnHgsCOONa + C3H5(OH)3 (1)
S6 mol NaOH : n^.^oH = ^ =
Theo (1) : nc,H,(OH), = \H =0-01
-> mc,H,(OH), =0.01-92 = 0,92 kg.
Hieu suat phan urng 80% ntn kh6'i lucfng glixerol thu ducfc thirc td' la :
0,92 . — = 0,736 kg. 100
1.16. Dap an dung la C.
Dat cong thiirc phan tir ciia este la CnHn^Ot.
C n H ^ i O i + (n+m/4 + t/2) 0 2 ^ nCOj + (m/2) H 2 0
P2O5 + 3 H 2 O -> 2 H 3 P O 4
C O 2 + CăOH)2 -> CaC03i+ H 2 O nH,o = ^ = 0,345 mol ; nco, = ^ = 0,345 mol
So mol H2O = s6' mol CO2 = 0,345 mol
_^ n = — m = 2n. Do do 2 este đu no, don chilc, cung g6'c rugfụ
2
.| 17. C6ng thirc diing la D.
- Kh6'i luong NaOH dem dung : mN^oH = ^^T^^T^ = 3,751 gam. (adsbygoogle = window.adsbygoogle || []).push({});
- Kh6-i lugng NaOH phan dng : mNaOH.pu = ^iZ^'^^l = ^ 8^^- (100 + 25)
-> Mg = 88 gam. Vay : R + 4 4 + R' = 88 -> R + R' = 44.
Khị R = 1 -> R' = 4 3 (C 3 H 7 ) ^ E la H C O O C 3 H 7 : propyl fomiat. Khi R = 15 -> R' = 29 (C 2 H 5 ) -> E la C H 3 C O O C 2 H 5: etyl axetat,
1.18. Cong ihirc diing la C.
Cong thirc phan tir ciia X la C7H,204 :
nx = 0,1 , n^aOH = 0,2 -> X la este hai chu-c. R -COO-^
Rl -COO'^2+2NaOH -> R-COONa + R, -COONa + R2(OH)2
0,1 mol 0,2 mol 0,1 mol 0,1 mol ^ (RCOONa).0,l + (RCOONa).0,l = 17,8 gam. R + R,=:44 ^ R = 15(-CH3),R, = 2 9 ( - C 2 H 5 )
X CO c6ng thti-c cáu tao : C H 3 - C O O- C H 2 ' ^^^^ Y : C H 2 - OH.
C 2 H 5- C O O- C H 2 C H 2- O H
l-IS. C6ng thu:c diing la Ạ
S6 mol cac chát : ncô = ^ = 0,3 mol ; nn^o = ^ = 0.3 mol.
CTPT cua este la C n H 2 n 0 2 . Este don chii-c nen z = 2. Do do c6ng thiirc phan tir ciia este la C n H 2 n 0 2 v6i n > 2.
•••20. C6ng thu-c dung la Ạ
C6ng thurc tdng quat cua A : Rn(COO)n,nR'n,