Question

A 60% mole Benzene and 40% mole Toluene mixture is sent to a distillation column, where it is separated into two equal streams. Calculate the composition of each stream, if the amount of benzene out of the top is 3 times greater than out the bottom.

Answer 1

let the feed to the distillation column= 100 moles, it contains 60% benzene and 40% toluene, hence benzene in the feed= 60 moles and toluene in the feed= 40 moles

100 moles are seperated into Distillate (D) and bottoms (W) and it is given that D= W

hence 100= D+W, 2D= 100, D= W= 50 moles

it is given that the benzene at the top is 3 times greater than at the bottom

let xf= mole fraction of benzene in the feed, xD= mole fraction of benzene in the distillate and Xw= mole fraction of benzene in the bottoms

writing overall benzene balance, 40 = D*xd+W*xw

but it is given that D*xd= 3*WXw hence 60 = 3Wxw+Wxw, 4Wxw= 60, but W= 50, xW= 60/200=0.3

since D*xD= 3*W*xw, 50*Xd= 3*50*0.3, Xd= 0.9

since, there are only two components entering the column and sum of mole fraction of components either in the distilalte or in the residue = 1

mole fraction of toluene in the distillate= 1-0.9=0.1, mole fraction toluene in the bottom= 1-0.3= 0.7

Distillate (D): flow = 50 moles. Benzene =0.9 and toluene= 0.1, bottoms: 50, benzene =0.3 and toluene =0.7