A gas mixture of `3.67L` of ethylene and methane on complete combustion at `25^(@)C` produces `6.11 L` of `CO_(2)`. Find out the heat evolved on buring `1L` of the gas mixture. The heats of combustion of ethylene and methane are `-1423` and `-891kJ mol^(-1)`, respectively, at `25^(@)C`.

Explanation

`{:(C_(2)H_(4)+,3O_(2)rarr,2CO_(2)+,2H_(2)O,),(aL,,2aL,,):}`
`{:(CH_(4)+,2O_(2)rarr,CO_(2)+,2H_(2)O,),((3.67-a)L,,(3.67-a)L,,):}`
Given, `2a + 3.67 -a = 6.11`
`a = 2.44 L`
Volume of ethylene in mixture `= 2.44 L`
Volume of methane in mixture `= 1.23 L`
Volume of ethylene in `1L` mixture `=(2.44)/(3.67) = 0.6649L`
Volume of methane in `1L` mixture `= (1.23)/(3.67) = 0.3351L`
`24.45L` of a gas at `25^(@)C` correspond to `1mol`.
Thus, heat evolved by buring `0.6649L` of ethylene
`=- (1423)/(24.5) xx 0.6649 =- 38.69 kJ`
and heat evolved by buring `0.3351L` of methane
`=- (891)/(24.45) xx 0.3351 =- 12.21 kJ`
So, total heat evolved by buring `1L` of mixture
`=- 38.69 - 12.21`
`=- 50.90 kJ`