Some Basic Concepts of Chemistry 1 Question 37
57. $5.00 \mathrm{~mL}$ of a gas containing only carbon and hydrogen were mixed with an excess of oxygen $(30 \mathrm{~mL})$ and the mixture exploded by means of electric spark. After explosion, the volume of the mixed gases remaining was $25 \mathrm{~mL}$.
On adding a concentrated solution of $\mathrm{KOH}$, the volume further diminished to $15 \mathrm{~mL}$, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas. $\quad(1979,3 \mathrm{M})$
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Solution:
- In the present case, $V \propto n$ ( $\because$ all the volumes are measured under identical conditions of temperature and pressure) Hence, the reaction stoichiometry can be solved using volumes as :
$\mathrm{C}{x} \mathrm{H}{y}(g)+\left(x+\frac{y}{4}\right) \mathrm{O}{2}(g) \longrightarrow x \mathrm{CO}{2}(g)+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}(l)$
volume of $\mathrm{CO}{2}(g)+\mathrm{O}{2}(g)$ (remaining unreacted) $=25$
$\Rightarrow$ Volume of $\mathrm{CO}_{2}(g)$ produced
$=10 \mathrm{~mL}\left(15 \mathrm{~mL} \mathrm{O} \mathrm{O}_{2}\right.$ remaining $)$
$\because \quad 1 \mathrm{~mL} \mathrm{C}{x} \mathrm{H}{y}$ produces $x \mathrm{~mL}$ of $\mathrm{CO}_{2}$
$\therefore 5 \mathrm{~mL} \mathrm{C}{x} \mathrm{H}{y}$ will produce $5 x \mathrm{~mL}$ of $\mathrm{CO}_{2}=10 \mathrm{~mL}$
$\Rightarrow \quad x=2$
Also, $1 \mathrm{~mL} \mathrm{C}{x} \mathrm{H}{y}$ combines with $\left(x+\frac{y}{4}\right) \mathrm{mL}$ of $\mathrm{O}_{2}$
$5 \mathrm{~mL} \mathrm{C}{x} \mathrm{H}{y}$ will combine with $5\left(x+\frac{y}{4}\right) \mathrm{mL}^{\text {of } \mathrm{O}_{2}}$
$\Rightarrow 5\left(x+\frac{y}{4}\right)=15\left(15 \mathrm{~mL}\right.$ of $\mathrm{O}_{2}$ out of $\left.30 \mathrm{~mL}\right)$
$\Rightarrow y=4$, hence hydrocarbon is $\mathrm{C}{2} \mathrm{H}{4}$
(remaining unreacted)
