`{:(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`

Let volume of `C_(2)H_(4)` and `CH_(4)` be `a ` and `b` litre respectively at `25^(@)C`.
`:. a+b=3.67litre ….(1)`
Also `C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O`
`CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O`
`:. `Volume of `CO_(2)` produced by a litre `C_(2)H_(4)=2a`
and Volume of `CO_(2)` produced by a litre `CH_(4)=b`
`:. 2a+b=6.11 ……(2)`
By `eqs. (1)` and `(2)`
`a=2.44 litre, b=1.23 litre`
`:. ` Volume of `C_(2)H_(4)` in 1 litre mixture at `25^(@)C`
`=(2.44)/(3.67)xx1=0.6648litre`
and Volume of `CH_(4)` in 1 litre mixture at `25^(@)C`
`=(1.23)/(3.67)xx1=0.3352 litre`
Now, Volume of `1 mol e C_(2)H_(4) at 0^(@)C=22.4 litre`
`:. ` Volume of `1 mol e C_(2)H_(4) at 25^(@)C`
`=(22.4xx298)/(273)=24.45 litre`
Similarly,
Volume of `1 mol e C_(2)H_(4)` produces `=+1423kJ ` at `25^(@)C` ltbr. `:. 0.6648litre C_(2)H_(4) ` produces
`=(+1423xx0.6648)/(24.45)=+12.22kJ`
Similarly,
`24.45 litre CH_(4)` produces `=+891kJ at 25^(@)C`
`:. 0.3352 litre CH_(4)` produces
`=(+891xx0.3352)/(24.45)=+12.22kJ`
Total heat produced `=38.69+12.22=50.91kJ`