Question

Distillates of benzene containing benzene 42 wt% and Toluene mixture 3000 kg/hr shall be distilled, so that the benzene of the distillate is 95 wt % and the Tolumen of 98 %. Find the D(mole/hr) and W (mole/hr). The molecular weight of benzene and Toluene is 78 and 92 respectively.

Answer 1

Feed rate F = 3000 kg/h

Benzene enters the column = 3000 x 0.42 = 1260 kg/h

Overall mass balance

3000 = D + W

Multiply by 0.02

60 = 0.02D + 0.02W

Benzene balance

1260 = D*0.95 + W*0.02

Solve the above equations simultaneously

1200 = 0.93 D

D = 1290 kg/h

W = 3000 - 1290 = 1710 kg/h

Benzene in distillate = 1290 x 0.95 = 1225.5 kg/h

Moles of Benzene in distillate = ( 1225.5 kg/h) / (78 kg/kmol)

= 15.712 kmol/h x 1000mol/kmol

= 15712 mol/h

Toluene in distillate = 1290 x 0.05 = 64.5 kg/h

Moles of Toluene in distillate = ( 64.5 kg/h) / (92 kg/kmol)

= 0.701 kmol/h x 1000mol/kmol

= 701 mol/h

Total Moles of distillate D = 15712 + 701 = 16413 mol/h

Similarly

Benzene in bottom = 1710 x 0.02 = 34.2 kg/h

Moles of Benzene in bottom = ( 34.2 kg/h) / (78 kg/kmol)

= 0.438 kmol/h x 1000mol/kmol

= 438 mol/h

Toluene in bottom = 1710 x 0.98 = 1675.8 kg/h

Moles of Toluene in bottom = ( 1675.8 kg/h) / (92 kg/kmol)

= 18.215 kmol/h x 1000mol/kmol

= 18215 mol/h

Total Moles of bottom W = 438 + 18215 = 18653 mol/h