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_(2) = 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 10mL of `C_(2)H_(4) = 10xx3 = 30mL`
but air contains only 21% `O_(2)` by volume
21mL of `O_(2)` = 100mL of air
30mL of `O_(2)` = ?
The volume of air required for the combustion = `(30)/(21)xx100 = 142.86 mL`