36 ml of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion to produce 16 ml of${\mathrm{CO}}_2,24 ml$ of water vapor and 8 ml of $N_2$. (The volumes are measured at STP conditions), volume of $O_2$required for complete combustion is ____?

$$\begin{gathered} \text{ Volume of }C{O}_2=16ml \\ 22400 ml\text{ of }{\mathrm{CO}}_2\text{ contain }32\text{ gram of Oxygen } \\ \therefore 16 ml\text{ of }{\mathrm{CO}}_2\text{ contain }\frac{16\times 32}{22400}\text{ gram of }{O}_2=0.0228\text{ gram } \\ \text{ Volume of }{\mathrm{H}}_2\mathrm{O}=24 ml \\ 22400 ml\text{ of }{\mathrm{H}}_2\mathrm{O}\text{ contain }16\text{ gram of Oxygen } \\ 24 ml\text{ of }{\mathrm{H}}_2\mathrm{O}\text{ contain }\frac{24\times 16}{22400}\text{ gram of oxygen }=0.0171\text{ gram } \\ \mathrm{Total} \mathrm{mass} \mathrm{of} \mathrm{oxygen} = 0.0228+0.0171 = 0.0399 \mathrm{gram} \\ 32 \mathrm{gram} \mathrm{of} \mathrm{oxygen} \mathrm{has} \mathrm{volume} 22400 ml \mathrm{at} \mathrm{STP} \\ \text{ Then }0.0399\text{ gram of oxygen has volume }\frac{22400\times 0.0399}{32}=27.9\simeq 28ml \end{gathered}$$