Question

A mixture of $2.3g$ formic acid and $4.5g$ oxalic acid is treated with $conc.H_{2}S{O}_4$. The evolved gaseous mixture is passed through $KOH$ pellets. Weight (in g) of the remaining product at STP will be _____.

• A. $2.8$
• B. $1.4$
• C. $4.4$
• D. $3.0$

Answer 1

Answer: (A)

$2.8$

$HCOOH\stackrel{H_{2}S{O}_4}{\to }CO+H_{2}O$ .......(i)

$(COOH{)}_2\stackrel{H_{2}S{O}_4}{\to }CO+C{O}_2+H_{2}O$ .......(ii)

Conc. $H_{2}S{O}_4$ is a strong dehydrating agent.

Moles of $HCOOH=\frac{2.3}{46}=0.05$ mole

Moles of $(COOH{)}_2=\frac{2.3}{46}=0.05$ mole

From reaction (i),

Number of $CO$ formed $=0.05$ mole

From reaction (ii),

Number of $CO$ formed $=0.05$ mole

Number of $C{O}_2$ formed $=0.05$ mole

Hence, total $CO$ formed $=$ $0.05+0.05=0.1$ mole

total of $C{O}_2$ formed $=0.05$ mole

$KOH$ pellets absorbs all $C{O}_2$, $H_{2}O$ is absorbed by $H_{2}S{O}_4$ thus $CO$ is remaining product.

Thus, the weight of the remaining product $=0.1\times 28=2.8$ g.

Hence, the correct option is $A$