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),(a,3a,2a),(CH_(4)+,2O_(2)rarr,CO_(2)+2H_(2)O),(b,2b,b):}`
Let volume of `C_(2)H_(4) = a`
Let volume of `CH_(4) = b`
`a +b = 3.67 L`
Volume of `CO_(2) = 2a +b = 6.11 L`
Solving, we get `a = 2.44L, b = 1.23 L`
Volume of `C_(2)H_(4)` in `1L` of mix at `25^(@)C`
`= (2.44)/(3.67) xx1 = 0.6648 L`
Volume of `CH_(4)" in" 1L of mix = (1.23)/(3.67) xx 1 = 0.3352 L`
Volume of `1 mol C(2)H_(4) of at 0^(@)C = 22.4 L`
Volume of `1mol of C_(2)H_(4) at 25^(@)C`
`= (22.4 xx 298)/(273) = 24.45 L`
Similarly,
Volume of `1 mol of CH_(4)at 25^(@)C = (22.4 xx 298)/(273) = 24.45 L` ltbr. `24.45 L produces C_(2)H_(4) = 1423 kJ at 25^(@)C`
`0.6648 L of C_(2)H_(4) = (1423 xx 0.6648)/(24.45) = 38.69 kJ`
Similarly,
`24.45 L of CH_(4) "produces" = 89.1 kJ`
`0.3352 L of CH_(4) "produced" = (89.1 xx 0.3352)/(24.45) = 12.22 kJ`
Total heat produced `= 38.69 +12.22 = 50.91 kJ`