$\underset{\downarrow }{{\mathrm{C}}_x{\mathrm{H}}_y+(x+y/4){\mathrm{O}}_2\to }\to \underset{\downarrow }{\downarrow }{\mathrm{CO}}_2+y/2+{\mathrm{H}}_2\mathrm{O}$
$20{\mathrm{cm}}^{3}100{\mathrm{cm}}^{3}60{\mathrm{cm}}^3$ After explosion, the gases that will be found are ${\mathrm{CO}}_2,{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{g})}$ and excess oxygen. After cooling. ${\mathrm{H}}_2\mathrm{O}$ will no longer exist as a gas. instead it transforms to water in the liquid form. Since $110{\mathrm{cm}}^3$ of gases remains after cooling, the volume of ${\mathrm{CO}}_2+{\mathrm{O}}_2=110{\mathrm{cm}}^3$. Only ${\mathrm{CO}}_2$ can be absorbed by $\mathrm{NaOH}$ therefore, the decrease in volume on absorption with $\mathrm{NaOH}$ indicates volume of ${\mathrm{CO}}_2$ while the volume remaining is that of excess oxygen
Volume of excess $0xygen=50{\mathrm{cm}}^3$
Volume of ${\mathrm{CO}}_2=110{\mathrm{cm}}^3-50{\mathrm{cm}}^3=60{\mathrm{cm}}^3$
Total volume of oxygen
= volume of oxygen used + volume of excess oxygen
$\therefore 150{\mathrm{cm}}^3=$ volume used $+50{\mathrm{cm}}^3$
Volume of oxygen used
$=(150-50){\mathrm{cm}}^3=100{\mathrm{cm}}^3$vogadro's law states that equal volume of gases under the same conditions of temperature and pressure contains the same number of molecules. The implication of this is that if you are given two gases A and B, mole of $\mathrm{A}/$ mole of $\mathrm{B}⟹$ volume of A/volume of B
You can apply this to any pair of gases From this and the equation of reaction, we have:
mole of ${\mathrm{C}}_x{\mathrm{H}}_y/$ mole of ${\mathrm{O}}_2=\frac{20}{100}=\frac{1}{5}$ $\frac{1}{x+\frac{y}{4}}=\frac{1}{5},x+\frac{y}{4}=5$ $\therefore \frac{4x+y}{4}=5,4x+y=20\dots$ (i) mole of ${\mathrm{C}}_x{\mathrm{H}}_y/$ mole of ${\mathrm{CO}}_2=\frac{\text{ volume of }{\mathrm{C}}_2{\mathrm{H}}_y}{\text{ volume of }{\mathrm{CO}}_2}$ $\therefore \frac{1}{x}=\frac{20}{60},20x=60,x=3$
Substitute $x=3$ in equation (i) $4(3)+y=20,y=20-12⟹8$ substitute for $x$ and $y$ in the hydrocarbon ${\mathrm{C}}_x{\mathrm{H}}_4\Rightarrow {\mathrm{C}}_3{\mathrm{H}}_8$