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

Combustion of ethylene is given by the equation
$C_2+H_4+3O_2\to 2C{O}_2+2H_{2}O$ 
One volume of $C_2{H}_4=3$ volumes of $O_2$  
10L of $C_2{H}_4=?$ the volume of $O_2$ required for the combustion of 10L of $C_2{H}_4=10\times 3=30L$ but air contains only 21% $O_2$ by volume 21L of $O_2=100Lofair$  
30L of $O_2=\frac{30}{21}\times 100=142.86c$