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.

Explanation

Volume of the mixture at NTP `=(40xx1)/(400)xx(273)/(1)=27.3"litre"`
Let the volume of ethane=x litre
Volume of ehtene=`(273-x)"litre"`
Balanced equations:
`underset(1"vol")(C_(2)H_(6))+underset(7//2"vol")(7//2O_(2)) to 2CO_(2)+3H_(2)O`
`underset(1"vol")(C_(2)H_(4))+underset(7//2"vol")(3O_(2)) to 2CO_(2)+2H_(2)O`
Total volume of oxygen required for complete combustion of the mixture is: `[(7)/(2)x+(27.3-x)xx3]"litre"`
`or [((7x+(27.3-x)xx6))/(2)]"litre"`
`"Mass of oxygen "[(7x+(27.3-x)xx6)/(2)]xx(32)/(22.4)`
`130=(x+16.8)xx(16)/(22.4)`
`therefore x=18.2`
Hence, mole friction of ethane =`(18.2)/(27.3)xx100=66.66`
Mole fraction of ethane=33.34