Question

When HCOOH acid is heated with conc.${{H}_{2}}S{{O}_{4}}$ , the gas evolved is:
[A] Only $C{{O}_{2}}$
[B] $CO+C{{O}_{2}}$
[C] $S{{O}_{2}}+C{{O}_{2}}$
[D] $CO$

Answer 1

Hint: In this reaction, one acid acts as an oxidizing agent and the other acts as a reducing agent. The acid acting as a reducing agent gets hydrated as it is a strong dehydrating agent too and a gas is evolved. The gas evolved here is toxic.

Complete step by step answer:
 In the question we are given HCOOH which is methanoic acid and we commonly know it as formic acid. Formic acid is reacted with concentrated sulphuric acid which is also an acid.
We know that acids generally react with bases to give us salt and water but acids do not react with other acids readily. However, there are several acid and acid reactions where one acid is a strong oxidizing agent and the other acid is a strong reducing agent. These reactions generally take place at higher temperatures with concentrated acids.
Here, we have formic acid which will donate an electron pair and act as a reducing agent and the concentrated sulphuric acid will act as an oxidizing agent by accepting an electron pair. We can write the reaction as-
     \[HCOOH+{{H}_{2}}S{{O}_{4}}(conc.)\to CO+{{H}_{2}}S{{O}_{4}}\cdot {{H}_{2}}O\]

Here, hydrated sulphuric acid is formed as we know it is a strong dehydrating agent too. The gas evolved here is carbon monoxide which is highly toxic.
So, the correct answer is “Option D”.

Note: Here if we had hydrochloric acid and hydrobromic acid, there would be no reaction. Sulphuric acid is a strong dehydrating agent and thus might react with weaker acids when used concentrated sulphuric acid. It also dehydrates nitric acid-
     \[{{H}_{2}}S{{O}_{4}}+HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}^{-}+N{{O}_{2}}^{+}+{{H}_{2}}O\]