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

Calculation of total no. of moles in the gaseous mixture
Applying ideal gas equation, PV = nRT
`"1 atm"xx40L= nxx0.0821"L atm K"^(-1)"mol"^(-1)xx400K`
`"or "n=(40)/(0.0821xx400)="1.218 mole"`
Calculation of no. of moles of each component
Suppose no. of moles of `C_(2)H_(6)` in the mixture = x
Then no. of moles of `C_(2)H_(4)` in the mixture = `1.218-x`
Also, 130 g of `O_(2)=(130)/(32)" moles = 4.0625 moles"`
The reactions for complete combustion of `C_(2)H_(6)` and `C_(2)H_(4)` are
`2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O" ...(i)"`
` C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O" ...(ii)"`
From eqn. (i), no. of moles of `O_(2)` required for complete combustion of x moles of `C_(2)H_(6)=(7)/(2)xx x=3`
From eqn. (ii), no. of moles of `O_(2)` required for complete combustion of `(1.218-x)` moles of `C_(2)H_(4)`
`=3(1.218-x)`
`therefore" "3.5x+3(1.218-x)=4.0625`
`"or "0.5x=4.0625-3.654=0.4085" or "x=0.8170" mole"`
Calculation of mole fractions
`"Mole fraction of "C_(2)H_(6)=(.^(n)C_(2)H_(6))/(n_("total"))=(0.817)/(1.218)=0.67" and mole fraction of "C_(2)H_(4)=1-0.67=0.33`