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)rarr2CO_(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)`
10 mL 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 = 30 L`
But air contains only `21% O_(2)` by volume
21 L of `O_(2) = 100 L` of air
30 L of `O_(2)= ?`
The volume of air required for the combustion
`=(30)/(21)xx100=142.86L`