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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. $H_2SO_4$. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
Explanation
Total CO = 1/10 mol; $CO_2$ is absorbed by KOH.
Detailed Solution
$HCOOH \xrightarrow{\text{Conc. } H_2SO_4} CO(g) + H_2O(l)$
2.3 g or $\frac{1}{20}$ mol HCOOH gives $\frac{1}{20}$ mol CO
$(COOH)_2 \xrightarrow{\text{Conc. } H_2SO_4} CO(g) + CO_2(g) + H_2O(l)$
4.5 g or $\frac{1}{20}$ mol oxalic acid gives $\frac{1}{20}$ mol CO and $\frac{1}{20}$ mol $CO_2$
The gaseous mixture formed is CO and $CO_2$; when it is passed through KOH, only $CO_2$ is absorbed. So the remaining gas is CO.
Weight of remaining gaseous product CO $= \frac{2}{20}\times28 = 2.8$ g
2.3 g or $\frac{1}{20}$ mol HCOOH gives $\frac{1}{20}$ mol CO
$(COOH)_2 \xrightarrow{\text{Conc. } H_2SO_4} CO(g) + CO_2(g) + H_2O(l)$
4.5 g or $\frac{1}{20}$ mol oxalic acid gives $\frac{1}{20}$ mol CO and $\frac{1}{20}$ mol $CO_2$
The gaseous mixture formed is CO and $CO_2$; when it is passed through KOH, only $CO_2$ is absorbed. So the remaining gas is CO.
Weight of remaining gaseous product CO $= \frac{2}{20}\times28 = 2.8$ g