Question

A mixture of $HCOOH$ and $H_2{C}_2{O}_4$ is heated with concentrated $H_{2}S{O}_4$. The gas produced is collected and on treating with $KOH$ solution, the volume of the gas decreases by $\frac{1}{6^{th}}$. The molar ratio of the two acids ($HCOOH$ : $H_2{C}_2{O}_4$) in the original mixture is:

The corresponding reactions are:
$HCOOH\stackrel{H_{2}S{O}_4}{\to }{H}_{2}O+CO$
$H_2{C}_2{O}_4\stackrel{H_{2}S{O}_4}{\to }{H}_{2}O+C{O}_2+CO$

Assume that the water produced is in liquid state.

• A. $2:1$
• B. $4:1$
• C. $6:1$
• D. $8:1$

Answer 1

The correct option is B $4:1$


$HCOOH\stackrel{H_{2}S{O}_4}{\to }{H}_{2}O+CO\uparrow$
$amol$ $amol$
From stoichiometry, moles of $CO$ produced = a mol
$\underset{bmol}{H_2{C}_2{O}_4}\stackrel{H_{2}S{O}_4}{\to }{H}_{2}O+\underset{bmol}{C{O}_2}\uparrow +\underset{bmoles}{CO}\uparrow$
From stoichiometry, moles of $C{O}_2$ and $CO$ formed is b mol.
Total number of moles of gas formed = $a+2b$
Again, the gas evolved is treated with $KOH$ and it reacts only with $C{O}_2$ in ordinary conditions.
Thus, moles of $C{O}_2$ absorbed by $KOH$ = $b$
Hence, $b=\frac{1}{6}\times \left(a+2b\right)$
$\therefore \frac{a}{b}=4:1$