Calculate the volume of air, contained `21%` of oxygen by volume required for the complete combustion of 10 L of ethylene under similar conditions 

Explanation

Combustion of ethylene is given by the equation
`C_(2)H_(4)+3O_(2) to 2CO_(2)+2H_(2)O`
one mole of `C_(2)H_(4)="3 moles of "O_2`
one volume of `C_(2)H_(4)="3 volumes of "O_(2)`
10mL of `C_(2)H_(4)=?`
The volume of `O_(2)` required for the combustion of 10 L of `C_(2)H_(4)=10 xx 3=30L`
But air contains only 21% `O_(2)` by volume
21 L of `O_(2)` =100 of air 30 L of `O_(2)=?`
The volume of air required for the combustion
`=(30)/(21) xx 100=142.86L`