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

(Can use a basis of 100kg)

b) What is the average molecular weight of the mixture?

for (b) can use M = \sum Yi Mi where M = average molecular weight, Yi= mole fraction of individual substance, Mi = molecular weight of individual substance OR can also use 1/M = \sum Xi / Mi where 1/M = average molecular weight, Xi = mass fraction of individual substance, Mi = molecualr weight of individual substance.

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jufsanabriasa jufsanabriasa · Answer 1 · 2019-10-01T05:13:39+02:00

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 ×[tex]\frac{1mol}{16,04 g}[/tex] = 31,2 moles

Mass of ethane

350 g ×[tex]\frac{1mol}{30,07 g}[/tex] = 11,6 moles

Mass of propane

150 g ×[tex]\frac{1mol}{44,1 g}[/tex] = 3,4 moles

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

Molar fraction of n-methane:

[tex]\frac{31,2moles}{46,2moles}[/tex] =67,5%

Molar fraction of ethane:

[tex]\frac{11,6moles}{46,2moles}[/tex] = 25,1%

Molar fraction of propane:

[tex]\frac{3,4moles}{46,2moles}[/tex] = 7,4%

b) Average molecular weight:

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

I hope it helps!