A `40 mL` mixture of methane and ethylene when exploded with certain volume of oxygen which is just sufficient for combustion produced `60 mL` of `CO_(2)` gas. Calculate the ratio between the volumes of `CH_(4)` and `C_(2)H_(4)` in the mixture. Wgat volume of oxygen is required if the ratio between the volumes of `C_(2)H_(4)` and `CH_(4)` is first reversed and then doubled? What volume of `CO_(2)` is produced? Assume, all the volumes being measured under identical conditions.

Explanation

Let the volumes of `CH_(4)` and `C_(2)H_(4)` in mixture be `a, b mL` respectively and `O_(2)` is just
Sufficient to burn these gases.
`{:(CH_(4)+,2O_(2),rarr,CO_(2),2H_(2)O),(a,2a,,0,0),(0,0,,a,),(C_(2)H_(4)+,3O_(2),rarr,2CO_(2)+,2H_(2)O),(b,3b,,0,0),(0,0,,2a,):}`
Given that `a+b=40` and `3a=60`
`:. a=20` and `b=20`
`:. ("Volume of"C_(2)H_(4))/("Volume of"CH_(4))=(2)/(1)`
then volume of `C_(2)H_(4)=(40xx2)/(3)= 26.67 mL`
and volume of `CH_(4)= 40-26.67 mL`
and volume of `CH_(4)= 40-26.67=13.33 mL`
Thus volume of `CO_(2)` formed in this case
`= 13.33+2xx26.67= 66.64mL`
and volume of `O_(2)` required in this case
`= 2xx13.33+3xx26.67= 106.67mL`