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))to CO_(2)+2H_(2)O`
`C_(2)H_(6)overset(O_(2))to2CO_(2)+3H_(2)O`
1volume of `CH_(4)` of 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 = 10mL
If x mL of `C_(2)H_(6)` is taken, the volume of `CO_(2)`
obtained = 2x = `2 xx20 = 40mL`
The volume of carbondioxide obtained by the combustion of 30mL of y : x volumes of methane and ethane = 10+40 = 50mL