Sunflower oil contains 0.070 mol palmitic acid (CoH3202 /mol, 0.060 mol stearic

Question

Sunflower oil contains 0.070 mol palmitic acid (CoH3202 /mol, 0.060 mol stearic acid (C18H36O2 /mol, 0.21 sunflower oil, the liquid oleic and linoleic acids are hydrogenated in the presence of a metal catalyst to form solid stearic acid. C, H, O 18 36 2 18 H120,+ 2H 8 136 At a margarine production facility, a stream of 465.0 mol sunflower oil/hris fed into a reactor. The fresh hydrogen source available is a mixture containing 0.97 mol H2/mol and 0.030 mol N2/mol. (N2 is inert in this process.) To insure complete hydrogenation, the hydrogen gas mixture is introduced into the reactor in 55.0% excess. All reaction products leave the reactor and proceed to a separator where palmitic and stearic acids are separated from the hydrogen gas mixture. The hydrogen gas mixture is recycled to the fresh feed stream, and 115.0 mol gas/hr of this recycle stream is purged to prevent the buildup of nitrogen gas. What is the molar flow rate of the fresh hydrogen and nitrogen gas mixture to the process? Number 526.2 mol gas/hr What is the composition of the purge stream? Number mol H, mol 92.0 Hin O Previous 8 Give Up & View Solution Check Answer Next I Exit

Explanation / Answer

In this process we are hydrogenating oleic acid and linoleic acid the moles of these in the input stream are

465 x 0.21 = 97.65 mol of oleic acid this needs 97.65 mol hydrogen gas

465 x 0.66 = 306.9 mol of linoleic acid this needs 613.8 mol hydrogen gas

Total hydrogen required is 613.8 + 97.65 = 711.45 since we are adding 55% excess we are adding hydrogen

711.45 + (711.45 x 0.55) = 1102.7 mol hydrogen/hr

Since the input gas contains 0.97 mol% hydrogen the actual rate of input gas is

1102.7/0.97 = 1137 mol gas/hr (answer in first box)

Of the 1102.7 hydrogen introduced 711.45 will react and remaining are in the outlet which is 1102.7 -711.45 = 391.25 mol

Nitrogen in the outlet is 1137-1102.7 = 34.3 mol

So the purge stream will have total 391.25 mol hydrogen + 34.3 mol nitrogen

SO molH2/mol of purge gas is 391.25/(391.25+34.3) = 0.919 mol H2/mol in the purge stream (1st box of purge stream)

For nitrogen it is 1-0.919 = 0.0806 mol N2/mol in the purge stream (2nd box of purge stream)

Gas leaving the separator is (391.25 +34.3) = 425.55 mol gas /hr

In the recycle stream the gas has a flow rate of 425.55 -115 = 310.55 mol gas/hr

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