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When the concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas(A) was evolved. This gas intensified when copper turning was added to this test tube. On cooling, the gas (A) changed into a colorless solid (B). identify (A) and (B).

Answer Verified Verified
Hint: Nitrate salt of alkali metals on reaction with conc. Sulphur acid results in the formation of brown gas and the brown gas at low temperature its dimers is and gives a colorless solid. Solve it.

Complete answer:
The above given data states that the salt is $NO{}_{3}^{-}$ of metal which with conc. sulphuric acid gives $HN{{O}_{3}}$ which decomposes to give brown gas i.e.${{O}_{2}}$ (A) because it a reddish brown gas having pungent smell ${{O}_{3}}$ is highly reactive and easily decomposed i.e. breaks into the brown gas i.e. $N{{O}_{2}}$ and oxygen and water along with it. Now, we have to take that nitrate of that element which on reaction with conc sulphuric acid gives nitric acid because if the resultant product is not nitric acid then there would be no release of brown gas. So now on recalling the reaction, we know that the sodium nitrate gives nitric acid on reaction with conc sulphuric acid so, therefore, sodium nitrate salt is taken.

$2NaN{{O}_ {3}} +{{H}_ {2}} S{{O}_ {4}} \to N{{a}_ {2}} S{{O}_ {4}} +2HN{{O}_ {3}} $
            $4HN{{O}_ {3}} \to \text{4N} {{\text{O}} _ {2}} \text {+2} {{\text{H}} _ {2}} \text{O+} {{\text{O}} _ {2}} $
                               Brown gas
                                (A)
The gas is intensified when copper turning is added to it and so due to reduction of $HN{{O}_ {3}} $ by Cu tuning.
               $\begin{align}
  & Cu+4HN{{O}_ {3}} \to Cu{{(N{{O}_ {3}})} _ {2}} +2{{H}_ {2}} O+2N{{O}_ {2}} \\
 & \text{}2HN{{O}_ {3}} \to 2N{{O}_ {2}} +{{H}_ {2}} O+O \\
 & \frac{Cu+2HN{{O}_{3}}+O\to Cu{{(N{{O}_{3}})}_{2}}+{{H}_{2}}O}{Cu+2HN{{O}_{3}}\to Cu{{(N{{O}_{3}})}_{2}}+2N{{O}_{2}}+2{{H}_{2}}O} \\
\end{align}$
At low temperature i.e. on cooling the $N{{O}_{2}}$, brown colored gas dimers is ( two molecules of $N{{O}_{2}}$joins to form one molecule) to give ${{N}_{2}}{{O}_{4}}$.
                 $N{{O}_ {2}} +N{{O}_ {2}} \to {{N}_ {2}} {{O}_ {4}} $
So, thus on cooling gas(A) $N{{O}_ {2}} $, changes to colorless gas B (${{N}_ {2}} {{O}_ {4}} $)

                $\begin{align}
  & 2N{{O}_ {2}} \text {} \rightleftharpoons {{N}_ {2}} {{O}_ {4}} \text {} \\
 & brown\text {} gas(A)\text {colorless gas(B)} \\
\end{align}$

Hence, the gas (A) is $N{{O}_ {2}} $ and the (B) is ${{N}_ {2}} {{O}_ {4}} $.

Note:
Salt of magnesium nitrate cannot be taken because on reaction with conc sulphuric acid, it gives $Mg{{(NO)} _ {3}} $and there is no release of brown gas i.e. $HN{{O}_ {3}} $.