chemical process

2012-05-03 11:49 pm
1.Chlorobenzene (C6H5Cl) is an important solvent and intermediate in the production of many other chemicals. It is produced by bubbling chlorine (Cl2) gas through liquid benzene (C6H6) in the presence of a catalyst to produce C6H5Cl and HCl. In an undesired side reaction, the product is further chlorinated to produce dichlorobenzene (C6H4Cl2), which can be further chlorinated to produce trichlorobenzene (C6H3Cl3).

The reactions are as follow:
C6H6 (l) + Cl2(g) → C6H5Cl (l) + HCl (g)
C6H5Cl (l) + Cl2(g) → C6H4Cl2(l)+ HCl (g)
C6H4Cl2 (l)+ Cl2 (g) → C6H3Cl3(l) + HCl (g)

The feed to a chlorination reactor consists of pure liquid benzene and a technical grade of chlorine gas (98 wt% of Cl2, the rest is a gaseous impurity with an average molecular weight of 25.0). The liquid output from the reactor contains 65.0 wt% C6H6, 32.0 wt% of C6H5Cl, 2.5 wt% of C6H4Cl2, and 0.5 wt% of C6H3Cl3. The gaseous output contains only HCl and the impurity that enters with chlorine gas.

(a) Determine the percentage by which benzene is fed in excess, and the conversion of benzene. (Hints: use liquid output as a basis).
(b) Determine the yield of chlorobenzene. (Yield is defined as the ratio of the amount produced to the theoretical amount that could have been produced).
(c) Determine the mass ratio of the gas feed to the liquid feed.
(d) In this process, why would benzene be fed in excess?
更新1:

想問下有無大大識做?感激不盡!!

更新2:

原來有個方法可以敢做架,唔駛block diagram, 重要快敢多,但係米依種題目都可以敢做?

回答 (3)

2012-05-04 7:36 pm
✔ 最佳答案
a) Taking 100 g of the liquid output:

65.0 g is C6H6, with no. of moles = 0.833
32.0 g is C6H5Cl, with no. of moles = 0.284
2.5 g is C6H4Cl2, with no. of moles = 0.017
0.5 g is C6H3Cl3, with no. of moles = 0.00275

Hence total no. of moles of C6H6 used to react to produce this 100 g of liquid output is 0.833 + 0.284 + 0.017 + 0.00275 = 1.138

Thus % by which benzene is fed in excess = 0.833/1.138 x 100% = 73.26%

b) Referring to 100 g of liquid output in (a), no. of moles of Cl2 used to produce it was 0.284 + 2 x 0.017 + 3 x 0.00275 = 0.327

Hence yield of chlorobenzene = 0.284/0.327 x 100% = 87.06 %

c) Mass ratio of Cl2 to benzene = 0.327 x 2 x 35.5/100 = 0.232

Hence mass ratio of gas to liquid feed = 0.232/0.98 = 0.237

d) To make the chlorobenzene being the major product since chlorination is uncontrollable that we cannot stop it immediately at the chlorobenzene stage.

2012-05-06 09:55:19 補充:
chem 計數以我黎講係唔用 flow chart 快過用, 我以前最易搞錯係 Contact Process 成日以為每 mole 的 sulphur 可以 produce 2 moles of H2SO4, 但 actually 只係 1 mole.
參考: 原創答案
2012-05-04 3:40 am
mass balance 吖嘛, 當年做唔少.
你畫咗block diagram 未?

2012-05-05 17:51:30 補充:
IIRC, at the time when I was doing this kind of questions, I ALWAYS use this method, to avoid confusion and for easier checking.

2012-05-05 17:51:37 補充:
IIRC, at the time when I was doing this kind of questions, I ALWAYS use this method, to avoid confusion and for easier checking.
2012-05-04 1:53 am
唔知點解我計完之後個total kmole/h(in) 唔等於 total kmole/h(out)

2012-05-03 20:38:11 補充:
有,Block diagram:

In Dis Gen Out Out(kg/h)
C6H6 1.137 0.304 0 0.833 65
C6H5Cl 0.304 0.02 0 0.284 32
C6H4Cl2 0.02 0.003 0 0.017 2.5

2012-05-03 20:42:28 補充:
Cl2 33.56 33.56 0 0 0
C6H3Cl3 0 0 0.003 0.003 0.5
HCL 0 0 33.56 33.56 1225
total In(kmole/h)=35.021
out(kmole/h)=34.697
唔 balance

2012-05-03 20:52:27 補充:
實際上我唔係好清楚 Cl2 同 HCl 點計個weight,
如果balance,b ,c 都應該計到

2012-05-08 19:09:30 補充:
sorry ar,我唔記得發問到期前處理,麻煩各位見到請投翻雷滾天-風卷殘雲個位博士一票
真係唔該各位既回答同意見,感激不盡!!


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