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第3章 习题及答案

2021-07-03 来源:爱问旅游网
[Thinking questions]

1. After sulfonation of 100mol of aniline with concentrated sulphuric acid, the products are composed of 89 mol of p-aminobenzenesulfonic acid, 2 mol of aniline and some other byproducts.

NH2Concentrated H2SO4NH2SO3H

(1) Conversion of aniline, X.

(2) Selectivity of sulfonation of aniline to p-aminobenzenesulfonic acid, S. (3) Theoretical yield of sulfonation of aniline to p-aminobenzenesulfonic acid, Y.

2. Sulfonation of 100kg of aniline (purity 99%, molecular weight 93) to obtain 217 kg of sodium p-aminobenzenesulfonate (purity>97%, molecular weight 231.2). Calculation the theoretical yield and mass yield of sodium p-aminobenzenesulfonate based on aniline.

NH2SulfonationSO3Na NH2(1) Theoretical yield of sulfonation of aniline to sodium p-aminobenzenesulfonate Y (2) Mass yield of sulfonation of aniline to sodium p-aminobenzenesulfonate Yw (3) Consumption rating of aniline

3. During chlorination of benzene to chlorobenzene, 100mol of benzene react with 40

mol of chloride in order to reduce the generation of byproduct dichlorobenzene. Products are composed of 38 mol of chlorobenzene, 1mol of dichlorobenzene and 61mol of unreacted benzene. After separation, 60mol of benzene can be recycled and 1mol of benzene is lost.

Cl+ Cl2+ HCl

C6H6 60C6H6 1 C6H5Cl 38 C6H4Cl2 1C6H6 40100 ChlorinatorC6H6 61Benzene-extracting tower C6H5Cl 38C6H4Cl 2 1(1) Single pass conversion of benzene XS and total conversion of benzene XT

(2) Selectivity of chlorination of benzene to chlorobenzene S

(3) Single pass yield (Ys) and total yield (YT) for chlorination of benzene to chlorobenzene.

Answers:

[Thinking question1]

After sulfonation of 100mol of aniline with concentrated sulphuric acid, the products are composed of 89 mol of p-aminobenzenesulfonic acid, 2 mol of aniline and some other byproducts.

NH2NH2Concentrated H2SO4SO3H

(3) Conversion of aniline, X.

(4) Selectivity of sulfonation of aniline to p-aminobenzenesulfonic acid, S.

(3) Theoretical yield of sulfonation of aniline to p-aminobenzenesulfonic acid, Y.

1002100%98.00% 100189(2) S1100%90.82%

1002(1) X

11100%89.00%(3) Y 100YSX90.82%98.00%89.00%89

[Thinking question 2]

Sulfonation of 100kg of aniline (purity 99%, molecular weight 93) to obtain 217 kg of sodium p-aminobenzenesulfonate (purity>97%, molecular weight 231.2). Calculation the theoretical yield and mass yield of sodium p-aminobenzenesulfonate based on aniline.

NH2SulfonationSO3Na NH2

(1) Theoretical yield of sulfonation of aniline to sodium p-aminobenzenesulfonate Y (2) Mass yield of sulfonation of aniline to sodium p-aminobenzenesulfonate Yw (3) Consumption rating of aniline

Theoretical yield:

21797%231.2 y100%85.6% 10099%93

Mass yield:

217100%217% yw100

The mass yield of p-aminobenzenesulfonate is greater than 100%. It is mainly due to the higher molecular weight of desired products than that of the reactant.

Consumption rating of aniline: 1/Yw =100/217 = 0.461t = 461kg

[Thinking question 3] During chlorination of benzene to chlorobenzene, 100mol of

benzene react with 40 mol of chloride in order to reduce the generation of byproduct dichlorobenzene. Products are composed of 38 mol of chlorobenzene, 1mol of dichlorobenzene and 61mol of unreacted benzene. After separation, 60mol of benzene can be recycled and 1mol of benzene is lost.

Cl+ Cl2+ HCl

C6H6 40

C6H6 60 100 ChlorinatorC6H6 61 C6H5Cl 38C6H4Cl 2 1C6H6 1C6H5Cl 38C6H4Cl2 1Benzene-extracting tower

(1) Single pass conversion of benzene XS and total conversion of benzene XT

100%, r: reacted nA,innA,inRRnA,innA,out X 单   100 R: reactor %R nA,in n s A, in s A ,out S: system  nX100%s 总nA,in

XAnA,rnA,innA,outXs10061100%39.00% 100401XT100%97.50%

40

(2) Selectivity of chlorination of benzene to chlorobenzene S anp pS100% nA,innA,out

138

1100% 10061

97.44%

(3) Single pass yield (Ys) and total yield (YT) for chlorination of benzene to chlorobenzene.

np y100%p nA,ina

SX

Ys381100138.00%

YsSXS97.44%39.00%38.00%

11100%95.00%YT 10060or YT97.50%97.44%95.00%38

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