A mixture of ethane `(C_(2)H_(6))` and ethene `(C_(2)H_(4))` occupies `40 L` at `1.00 atm` and at `400 K`. The mixture reacts completely with `130 g` of `O_(2)` to produce `CO_(2)` and `H_(2)O`. Assuming ideal gas behaviour, calculate the mole fractions of `C_(2)H_(4)` and `C_(2)H_(6)` in the mixture.

Answer

`C_(2)H_(6)=0.66,C_(2)H_(4)=0.34`

Explanation

Apply PV = nRT and calculate n of the mixture. We get n = 1.22 moles.
Suppose `C_(2)H_(6)="x mole. Then "C_(2)H_(4)=(1.22-x)" mole"`
`2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O`
`C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O`
`O_(2)" used for x mole of "C_(2)H_(6)=(7)/(2)xx x=3.5x" mole"`
`O_(2)` used for `(1.22-x)` mole of `C_(2)H_(4)=3(1.22-x)` mole
Total `O_(2)` used `=3.5x+3(1.22-x)=(130)/(32)" (given)"`
Calculate x. We get x = 0.805
`"Mole fraction of "C_(2)H_(6)=(0.805)/(1.22)=0.66`