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

The mole fraction composition of the liquid is :

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

Step-by-step explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane =
`(30)/(100)* 200 g=60 g`

Moles of butane =
`n_1=(60 g)/(58 g/mol)=1.0345 mol`

Percentage of pentane= 40%

Mass of pentane=
`(40)/(100)* 200 g=80 g`

Moles of pentane=
`n_2=(80 g)/(58 g/mol)=1.1111 mol`

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

Mass of hexane =
`(30)/(100)* 200 g=60 g`

Moles of hexane =
`n_2=(60 g)/(86 g/mol)=0.6977 mol`

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

`\chi_1=(n_1)/(n_1+n_2+n_3)=(1.0345 mol)/(1.0345 mol+1.1111 mol+0.6977 mol)=0.3638`

`\chi_2=(n_2)/(n_1+n_2+n_3)=(1.1111 mol)/(1.0345 mol+1.1111 mol+0.6977 mol)=0.3908`

`\chi_3=(n_1)/(n_1+n_2+n_3)=(0.6977 mol)/(1.0345 mol+1.1111 mol+0.6977 mol)=0.2454`