10mL of a gaseous hydrocarbon is exploded with 200 mL of oxygen. The gaseous product was then allowed to cool and attain room temperature and pressure, the volume was then found to be 180 mL. This mixture of gases was then passed through KOH solution followed by anhydrous CaCl_2,the resulting gas measured 100 mL. The hydrocarbon is

Let the hydrocarbon be C_xH_y. The reaction that occurs can be represented as 

$$\begin{gathered} {\mathrm{C}}_x{\mathrm{H}}_y+\left(x+\frac{y}{4}\right){\mathrm{O}}_2 ⟶ x{\mathrm{CO}}_2+\frac{y}{2}{\mathrm{H}}_2\mathrm{O} \\ 10\mathrm{mL} 10(x+ y/4) \mathrm{mL} 10\mathrm{xmL} \end{gathered}$$

1 mole of hydrocarbon reacts with (x + y/4) moles of O₂(g) to produce x mole of CO₂ and y/2 moles of H₂O.
Volume of CO₂ produced from 10 mL of hydrocarbon = 10xmL
Volume of O₂ consumed by 10 mL of hydrocarbon = 10x(x+y/4)mL
At a pressure of 1 atm and room temperature water vapour is condensed to liquid state. The residual gases are CO₂ and unreacted O₂.
Hence, volume of CO₂ and left out O₂ = 180 mL On passing the mixture of gases through aqueous KOH, CO₂ is absorbed leaving behind O₂.
Hence volume of unreacted O₂ = 100 mL (given) Volume of O₂ reacted= 200— 100 = 100 mL Volume of CO₂ produced = 180—100= 80 mL
Then, we have 10x=80 or x=8
And 10x(x+y/4)=100
or 8+y/4=10 $\Rightarrow$ y=8

Thus the hydrocarbon is C₈H₈