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Distillates of benzene containing benzene 42 wt% and Toluene mixture 3000 kg/hr shall be distilled, so...
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.
Solutions
Expert Solution
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
Amanda K answered 2 years agoRelated Solutions
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