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

Let the mixture contain x litre of `CH_4 and 3.67 - x` litre of ethylene.
`CH_(4) + O_(2) rarr CO_(2)`
`x " "x`
`C_(2)H_(4) + O_(2) rarr 2CO_(2)`
`3.67 - x " "2(3.67 - x)`
Given : `x + 2(3.67 - x) = 6.11 L implies x = 1.23 L`
Volume of ethylene = 2.44 L
Total moles of gasses in 1 litre = `(pV)/(RT) = (1 xx 1)/(0.082 xx 298) = 0.04`
Also, `CH_4` and ethylene are in 1 : 2 volume (or mole) ratio, moles of `CH_4 = (0.04)/3` and moles of ethylene = `(2 xx 0.04)/(3)`
`implies "Heat evolved due to methane" = (0.04)/3 xx 891 = 11.88 kJ`
Heat evolved due to ethylene `= (2 xx 0.04)/3 xx 1423 = 37.94 kJ`
`implies` Total heat evolved on combustion of `1.0 L` gaseous mixture at `25^@C `is `11.88 + 37.94 = 49.82 kJ`.