Question

1000 kg/h of a mixture containing equal parts by mass of benzene and toluene isdistilled to get overhead product containing 95% benzene (weight basis). The flow rate of bottom stream being 512 kg/h. Calculate :

A) flow rate of overhead product

B) The flow of benzene and Toluene in the bottom product.

C) The molar fraction of benzene in the bottom stream.

Answer 1

feed contains :1000 kg/hr of Equal parts of Benzene and toluene.

Mass flow rate of Benzene in the feed= 1000/2= 500 kg/hr= mass flow rate of toluene.

Overhead contains 95% Benzene and rest is toluene which is =100-95=5%

Flow rate of bottom stream = 512 kg/hr

Writing overall balance of the column

1000(F)= 512(B)+D

D= 1000-512=488 kg/hr

Writing Benzene balance , Benzene entering = Benzene leaving from overhead+ benzene leaving from bottom

500= 488*0.95+ Benzene in the bottoms

Benzene in the bottoms= 500-488*0.95= 36.4 kg/hr

In the bottoms, Benzene= 36.4 kg/hr and toluene= 512-36.4=475.6 kg/hr

moles = mass/molar mass, molar masses : Benzene= 78 and toluene = 92.14

Moles : Benzene = 36.4/78 kgmoles/hr= 0.47 kgmoles/hr and toluene= 475.6/92.14= 5.16 kg moles/hr

total moles of bottoms : 0.47+5.16= 5.63 kgmoles/hr

Composition ( molar): Benzene = 0.47/5.63= 0.083 and Toluene= 5.16/5.63= 0.917

Composition of bottoms ( weight %): Benzene =36.4/512=0.071 and toluene= 475.6/512=0.9289 Weight%