Question

A mixture of 50 wt% methane, 35 wt% ethane, and 15 wt% propane. Determeine the mole fraction of methane.

Answer 1

Answer:

a) Molar fraction:

Methane: 67,5%

Ethane: 25,1%

Propane: 7,4%

b) Molecular weight of the mixture: 25,16 g/mol

Explanation:

with a basis of 100 kg:

Moles of methane:

500 g ×$\frac{1mol}{16,04 g}$ = 31,2 moles

Mass of ethane

350 g ×$\frac{1mol}{30,07 g}$ = 11,6 moles

Mass of propane

150 g ×$\frac{1mol}{44,1 g}$ = 3,4 moles

Total moles: 31,2 moles + 11,6 moles + 3,4 moles = 46,2 moles

Molar fraction of n-methane:

$\frac{31,2moles}{46,2moles}$ =67,5%

Molar fraction of ethane:

$\frac{11,6moles}{46,2moles}$ = 25,1%

Molar fraction of propane:

$\frac{3,4moles}{46,2moles}$ = 7,4%

b) Average molecular weight:

$0,5*16,04g/mol + 0,35*30,07g/mol+0,15*44,1g/mol}$ = 25,16g/mol

I hope it helps!