Question

Two hundred kg of liquid contains 30% butane, 40% pentane, and the rest hexane (mass %) Determine: The mole fraction composition of the liquid The mass fraction composition on hexane free basis 1. 2.

Answer 1

Answer:

The mole fraction composition of the liquid is :

Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.

Explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane = $\frac{30}{100}\times 200 g=60 g$

Moles of butane = $n_1=\frac{60 g}{58 g/mol}=1.0345 mol$

Percentage of pentane= 40%

Mass of pentane= $\frac{40}{100}\times 200 g=80 g$

Moles of pentane= $n_2=\frac{80 g}{58 g/mol}=1.1111 mol$

Percentage of hexane = 100% - 30% - 40% = 30%

Mass of hexane = $\frac{30}{100}\times 200 g=60 g$

Moles of hexane = $n_2=\frac{60 g}{86 g/mol}=0.6977 mol$

Mole fraction of butane, pentane and hexane : $\chi_1, \chi_2 \& \chi_3$

$\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638$

$\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908$

$\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454$