Some Basic Concepts of Chemistry - Result Question 58
####57. $5.00 mL$ of a gas containing only carbon and hydrogen were mixed with an excess of oxygen $(30 mL)$ and the mixture exploded by means of electric spark. After explosion, the volume of the mixed gases remaining was $25 mL$.
On adding a concentrated solution of $KOH$, the volume further diminished to $15 mL$, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas. $\quad(1979,3 M)$
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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 :
$C _x H _y(g)+\left(x+\frac{y}{4}\right) O _2(g) \longrightarrow x CO _2(g)+\frac{y}{2} H _2 O(l)$
volume of $CO _2(g)+O _2(g)$ (remaining unreacted) $=25$
$\Rightarrow$ Volume of $CO _2(g)$ produced
$=10 mL\left(15 mL O O _2\right.$ remaining $)$
$\because \quad 1 mL C _x H _y$ produces $x mL$ of $CO _2$
$\therefore 5 mL C _x H _y$ will produce $5 x mL$ of $CO _2=10 mL$
$\Rightarrow \quad x=2$
Also, $1 mL C _x H _y$ combines with $\left(x+\frac{y}{4}\right) mL$ of $O _2$
$5 mL C _x H _y$ will combine with $5\left(x+\frac{y}{4}\right) mL^{\text {of } O _2}$
$\Rightarrow 5\left(x+\frac{y}{4}\right)=15\left(15 mL\right.$ of $O _2$ out of $\left.30 mL\right)$
$\Rightarrow y=4$, hence hydrocarbon is $C _2 H _4$
(remaining unreacted)
