Question

A mixture of $2.3g$ formic acid and $4.5g$ oxalic acid (anhyd.) is treated with conc. $H_{2}SO_{4}$ . The evolved gaseous mixture is passed through $KOH$ pellets. Weight (in g) of the remaining gaseous product at $STP$ will be

• A. $2\cdot 8$
• B. $3\cdot 0$
• C. $1\cdot 4$
• D. $4\cdot 4$

Answer 1

Answer: (A)

$HCOOH \rightarrow H_{2}O\left(\right.l\left.\right)+CO\left(\right.g\left.\right)$
$2.3gm\left(\frac{1}{20} mole\right)$
$H_{2}C_{2}O_{4} \rightarrow H_{2}O\left(\right.l\left.\right)+CO\left(\right.g\left.\right)+\left(CO\right)_{2}\left(\right.g\left.\right)$
$\frac{1}{20}\text{ mole }$
$\left(\frac{1}{20} + \frac{1}{20}\right)moleCO,\frac{1}{20}mole\left(CO\right)_{2}$
$CO_{2}$ is absorbed by $KOH$
So, $\frac{1}{10}mole$ of $CO$ is left $=2.8gm$ of $CO$