30 mL of mixture of methane and ethane in `x:y` ratio by volumes on combustion gave 40 mL of carbon dioxide. If 30 mL of mixture is taken in `y: x` ratio, what is the volume of carbondioxide obtained under similar conditions? 

Explanation

The combustion of methane and ethane is given by the equations
`CH_(4)overset(O_(2))rarrCO_(2)+2H_(2)O, C_(2)H_(6)overset(O_(2))rarr 2CO_(2)+3H_(2)O`
1 volume of `CH_(4) = 1` volume of `CO_(2)`
1 volume of `C_(2)H_(6) = 2` volume of `CO_(2)`
Let, volume of `CH_(4)` in the given mixture `= x`
Let, volume of `C_(2)H_(6)` in the given mixture `= y`
Then, `x + y = 30"      "x + 2y = 40`
`y = 10"      "x = 30 - 10 = 20`
If y mL of `CH_(4)` is taken,
the volume of `CO_(2)` obtained = y = 10 mL
If x mL of `C_(2)H_(6)` is taken,
the volume of `CO_(2)` obtained `= 2x = 2 xx 20 = 40 mL`
The volume of carbondioxide obtained by the combustion of 30 mL of `y: x` volumes of methane and ethane `= 10 + 40 = 50 mL`