A mixture of ${\mathrm{C}}_3{\mathrm{H}}_6,{\mathrm{C}}_3{\mathrm{H}}_8$ and ${\mathrm{C}}_4{\mathrm{H}}_{10}$ having total volume 90ml is subjected to complete combustion liberating 320ml of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ at same temperature and pressure. Calculate volume % of ${\mathrm{C}}_4{\mathrm{H}}_{10}$.

$\begin{array}{r}{\mathrm{C}}_3{\mathrm{H}}_6+\frac{9}{2}{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}\\ (90-\mathrm{a}-\mathrm{b}) 3(90-\mathrm{a}-\mathrm{b})\end{array}$

$\underset{\mathrm{b}}{{\mathrm{C}}_3{\mathrm{H}}_8}+5{\mathrm{O}}_2\to \underset{3\mathrm{b}}{3{\mathrm{CO}}_2}+4{\mathrm{H}}_2\mathrm{O}$

$\underset{\mathrm{a}}{{\mathrm{C}}_4{\mathrm{H}}_{10}}+\frac{13}{2}{\mathrm{O}}_2\to \underset{4\mathrm{a}}{4{\mathrm{CO}}_2}+5{\mathrm{H}}_2\mathrm{O}$

Total ${\mathrm{CO}}_2$

$\begin{array}{l}\Rightarrow 3(90-a-b)+3b+4a=320\\ \Rightarrow a=50\end{array}$

or $= 55.55\mathrm{\%}$

Volume of ${\mathrm{C}}_4{\mathrm{H}}_{10}$ in total = $\frac{50}{90}\times 100$

                                              $=\frac{500}{9}=55.5$