The molar conversion of C_6 H_2 n + 2 in reaction (a) is 96% and of CO in reacti

Question

The molar conversion of C_6 H_2 n + 2 in reaction (a) is 96% and of CO in reaction (b) is 92%. Calculate: the average molecular mass of the off-gas; the mass of gas fed to the reformer, kgh^-1; the mass of hydrogen produced, kg h^-3. Allyl alcohol can be produced by the hydrolysis of allyl chloride. Together with the main product, allyl alcohol, dialyl ether is produced as a byproduct. The conversion of allyl chloride is typically 97% and the selectivity to alcohol is 90%. both on a molar basis. Assuming that there are no other significant side reactions, calculated masses of alcohol and ether produced, per 1000 kg of allyl chloride fed to the reactor. Aniline is produced by the hydrogenation of nitrobenzene. A small amount of cyclo-hexylamine is produced as a byproduct. The reactions are: Nitrobenzene is fed to the reactor as a vapor, with certain quantity of hydrogen. The conversion of the nitrobenzene, to all products, is 96%. and the selectivity for aniline is 95%. The unreacted hydrogen is separated from the reactor products and recycled to the reactor (as shown in the figure below). A purge is taken from the recycle stream to maintain the inerts in the recycle stream below. The fresh hydrogen feed to the reactor is 99.5% pure, the remainder being inerts. All percentages are molar, For a feed rate of 100 kmol h of nitrobenzene, calculate: the fresh hydrogen feed; the purge rate required; the composition of the reactor outlet stream. Determine the pinch temperature and the minimum utility requirements for the process set out below. Take the minimum approach temperatures as 15 degree C.

Explanation / Answer

7 )

Given:

Conversion of Allyl chloride: 0.97

Selectivity to alcohol: 0.90

Total amount of allyl chloride fed: 1000 kg.

Molar mass of allyl chloride (NA0): 0.0765 kg/mol

Total moles of allyl chloride fed = 1000/0.0765 = 13071.9 mol

Molar mass of allyl alcohol: 0.05808 kg/mol

Molar mass of diallyl ether: 0.09814 kg/mol

Reactions are as follows:

C3H5Cl + H2O = C3H5OH + HCl .................(i) Desired reaction

C3H5Cl + 1/2 H2O = 1/2 C6H10O + HCl .................(ii) Undesired reaction.

So,

Convrsion of Allyl chloride is = 0.97 = (NA0-NA) / NA0

NA is unreacted allyl chloride after both the reactions

Then

0.97 = (13071.9 - NA)/ 13071.9

NA= 392.16 mol

So, total amount of allyl chloride reacted in both reactions is= 13071.9 - 392.16 = 12679.74 mol

Also given that selectivity to alcohol is 90%, which means 90% of total reacted in both reaction has been converted to desired product i.e allyl alcohol

Amount of allyl chloride reacted in first reaction is = 0.90 * 12679.74 = 11411.77 mol

And Amount of allyl chloride reacted in second reaction is = 0.10 * 12679.74 = 1267.97 mol

Stoichimetry of first reaction tells us that,

1 mol allyl chloride forms 1 mol of allyl alcohol

hence, 11411.77 mol of allyl chloride will form 11411.77 mol of allyl alcohol

Hence amount of allyl alcohol produced = 11411.77 mol or 11411.77 mol * 0.05808 kg/mol = 662.80 kg

Now from 2 nd reaction stoichiometry,

1 mol of allyl chloride will form 1/2 mol of diallylether

and Amount of allyl chloride reacted in second reaction is = 0.10 * 12679.74 = 1267.97 mol

Then, amount of diallylether produced= 1267.97 / 2 = 633.99 = 634 mol or 634 mol *0.09814 kg/mol= 62.22 kg

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