A liquid mixture containing OO mo% benzene 250 mol% tol ere, and the balance rye

Question

A liquid mixture containing OO mo% benzene 250 mol% tol ere, and the balance ryere is fed to a distillation column. The bottom s product contains Romo% rrere and no benzene, and 960% of the xylene in the feed is recovered in this stream. The overhead product is fed to a second column The overhead pract from the second colu n contains 970 % of the benzene in the feed to this col mn. The composition of this stream is 940 mol% benzene and the balance toluene Determine the percentage of toluene fed to the first column that emerges in the bottom of the second column. o 9868% 96.98% O 89.82% ? 88.92%

Explanation / Answer

Let 100 mol of feed enters into the first column

Toluene in feed = 100 x 0.250 = 25.0 mol

Moles of xylene in feed = 0.450 x 100 = 45.0 mol

Moles of xylene recovered = 45.0x 0.960 = 43.20 mol/hr

First column

Xylene balance

Xylene in feed = xylene in bottom of first column

43.20 = moles of bottom in first column x 0.980

moles of bottom in first column = 44.082 mol

Overall balance

100 = overhead mol + bottom mol

overhead mol = 100 - 44.082 = 55.918 mol

Benzene balance

Benzene in feed = Benzene in overhead of first column

0.30 x 100 = 55.918 x mol fraction of benzene in first overhead

mol fraction of benzene in first overhead  = 30/55.918

= 0.5365

Column 2

Benzene balance

Benzene in overhead of first column = Benzene in overhead of first column

0.5365 x 55.918 x 0.970 = overhead mol x 0.940

overhead mol  = 30.957 mol

Overall balance of both columns

Bottom mol of column 2 = 100 - 44.082 - 30.957

= 24.961 mol

Toluene balance in column 2

Toluene inlet = toluene out in overhead + toluene in bottom

0.4310 x 55.918 = 30.957 x 0.060 + 24.961 x mol fraction of toluene in bottom

mol fraction of toluene in bottom 2 = 0.89111

% toluene

= (100*moles of toluene in bottom 2) / (moles of toluene in feed)

= (100*0.89111*24.961)/(100 x 0.250)

= 88.97 %

Option D is the correct answer

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