Some Basic Concepts of Chemistry - Result Question 40
####40. A mixture of $2.3 g$ formic acid and $4.5 g$ oxalic acid is treated with conc. $H_2 SO_4$. The evolved gaseous mixture is passed through $KOH$ pellets. Weight (in g) of the remaining product at STP will be
(a) 1.4
(b) 3.0
(c) 4.4
(d) 2.8
[2018]
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Solution:
- (d) $HCOOH \xrightarrow[\substack{\text{ Deydrating } \\ \text{ agent }}]{H_2 SO_4} CO+H_2 O$
$ \begin{matrix} \text{ At start } \\ \text{ (moles) } \end{matrix} =\frac{2.3}{46}=\frac{1}{20} \quad \begin{matrix} {[H_2 O \text{ absorbed by } H_2 SO.} \\ 0 & 0 \end{matrix} $
$ \begin{aligned} & \begin{matrix} \text{ Final moles } & 0 & \frac{1}{20} & \frac{1}{20} \end{matrix} \\ & H_2 C_2 O_4 \xrightarrow{H_2 SO_4} CO+CO_2+H_2 O \\ & \text{ At start }=\frac{4.5}{90}=\frac{1}{20} \\ & \text{ (moles) } \\ & {[H_2 O \text{ absorbed by } H_2 SO_4]} \\ & \frac{1}{20} \quad \frac{1}{20} \quad \frac{1}{20} \end{aligned} $
$CO_2$ is absorbed by $KOH$.
So, the remaning product is only $CO$.
Moles of CO formed from both reactions
$ =\frac{1}{20}+\frac{1}{20}=\frac{1}{10} $
Left mass of $CO=$ moles $\times$ molar mass
$ =\frac{1}{10} \times 28=2.8 g $
