Calculate the volume of air, containing 21% of oxygen by volume required for the complet combustion of 10L of ethylene under similar conditions

Explanation

Combustion of ethylene is given by the eqution `C_(2)H_(4)+3O_(2)rarr 2CO_(2)+2H_(2)O`
one volume of `C_(2)H_(4)=3` volumes of `O_(2)`
10 mL of `C_(2)H_(4)=?`
The volume of `O_(2)` required for the combustion of
10 L of `C_(2)=H_(4)=10xx3=30L`
But air contains only `21%O_(2)` by volume
`"21 L of "O_(2)="100 L of air"`
`"30 L of O"_(2)=(30)/(21)xx100=142.86L`