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

`underset("a litre")(C_(2)H_(4))+3O_(2)tounderset("2a litre")(2CO_(2))+2H_(2)O`
`underset((3.67-a)litre)(CH_(4))+2O_(2)tounderet((3.67-a)litre)(CO_(2))+2H_(2)O`
Given, `2a+3.67-a=6.11`
a=2.44 litre
Volume of ethylene in mixture =2.44 litre
Volume of methane in mixture=1.23 litre
Volume of ethylene in 1 litre mixture `=(2.44)/(3.67)=0.6649` litre
Volume of methane in 1 litre mixture `=(1.23)/(3.67)=0.3351` litre
24.45 litre of a gas at `25^(@)C` correspond to 1 mole
Thus, heat evolved by burning 0.6649 litre of ethylene thus, heat evolved by burning 0.6649 litre of ethylene
`=-(1424)/(24.5)xx0.6649=-38.69kJ`
and heat evolved by burning 0.3351 litre of methane
`=-(891)/(24.45)xx0.3351=-12.21kJ` ltBrgt so, total heat evolved by buring 1 litre of mixture.
`=-38.69-12.21`
`=-50.90kJ`