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

`DeltaH_(C)(CH_(4))=-890"kJmol"^(-1)`
`DeltaH_(C)(C_(2)H_(4))=-1423"kJmol"^(-1)`
Let the mixture contain x lit of `CH_(4)` and (3.67-x)` lit of ethylene.
`CH_(4) +2O_(2)rarrCO_(2)+2H_(2)O`
`x" lit "` x` lit
`C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O`
`(3.67-x)" lit "2(3.67-x)` lit
Volume of Carbondioxide formed
`=x+2(3.67-x)=6.11` lit
`x+7.34-2x=6.11`
`7.34 -m x = 6.11 =1.23` lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence
`DeltaH_(C)=[(DeltaH_(C)(CH_(4)))/(22.4lit)xx(x)lit]+`
`[(DeltaH_(C)(C_(2)H_(4)))/(22.4lit)xx(3.67-x)lit]`
`DeltaH_(C)=[(-890"kJ mol"^(-1))/(22.4lit)xx1.23lit]+`
`[(-1423)/(22.4lit)xx(3.67-1.23)lit]`
`DeltaH_(C)=[-48.87"kJ mol"^(-1)]+[-155 "kj.mol"^(-1)]`
`DeltaH_(C)=-203.87 "kJ.mol"^(-1)`