Question

A mixture containing 10%ethanol and 90% water by weight is fed into a distillation column at the rate of 114.1kg/h.The distillate contains 76.4%ethanol and the distillate is produced at the rate of one tenth of that of the feed.perfom material balance and calculate the percentage of ethanol in the bottom.

Answer 1

mass of mixture = 106.26kg/hr

10% of the mixture is ethanol = 10.626 kg

number of moles of ethanol = $\dfrac{10626\text{g} }{46\text{g/mol}}$

= 231 moles

90% of the mixture is water = 95.634 kg

number of moles of water = $\dfrac{95634\text{g}}{18\text{g/mol}}$

= 5313 moles

Total number of moles = 231 + 5313 = 5544 moles

mole fraction of ethanol = 231/5544 = 0.0417

mole fraction of water = 5313/5544 = 0.9583