40 g of a sample of carbon on combustion left 10% of it unreacted.  The volume of oxygen required at STP for this combustion reaction is   

$\text{C}+{\text{O}}_2\underset{}{\overset{}{\to }}{\text{CO}}_{\text{2}}$

12g   22.4L

            The amount of carbon unreacted =

$40\times \frac{10}{100}\text{g}=4\text{g}$

So, the amount of carbon reacted =

$\left(40-4\right)\text{g}=36\text{g}$

$\because$  At STP, for the combustion of 12g of C, oxygen required = 22.4 L

$\therefore$  For the combustion of 36g of C, oxygen required will be

$\frac{22.4}{12}\times 36\text{L}=67.2\text{L}$