A) 2.8
B) 3.0
C) 1.4
D) 4.4
Correct Answer: A
Solution :
\[\underset{2.3g\text{ or }\left( \frac{1}{20}\text{mol} \right)}{\mathop{\text{HCOOH}}}\,\xrightarrow{\text{Conc}\text{.}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}\underset{\frac{1}{20}\text{mol}}{\mathop{\text{CO(g)}}}\,\text{+}{{\text{H}}_{\text{2}}}\text{O(l)}\] \[\begin{align} & \text{COOH}\xrightarrow{\text{Conc}\text{.}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}\underset{\frac{1}{20}\text{mol}}{\mathop{\text{CO(g)}}}\,\text{+}\underset{\frac{1}{20}\text{mol}}{\mathop{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{(g)+}{{\text{H}}_{\text{2}}}\text{O(l)} \\ & \underset{4.5g\,\,or\,\,\left( \frac{1}{20}\text{mol} \right)}{\mathop{\begin{align} & \text{ }\!\!|\!\!\text{ } \\ & \text{COOH} \\ \end{align}}}\, \\ \end{align}\]Gaseous mixture formed is \[\text{CO}\] and \[\text{C}{{\text{O}}_{\text{2}}}\]when it is passed through KOH, only \[\text{C}{{\text{O}}_{\text{2}}}\]is absorbed. So the remaining gas is CO. So, weight of remaining gaseous product CO is \[\frac{2}{20}\times 28=2.8g\] So, the correct option is
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