Question

Consider the following reaction sequences:

$A\left(aq\right)+NaOH\left(aq\right)\to B\left(g\right)+NaCl\left(aq\right)$
$A\left(aq\right)+C\left(aq\right)\stackrel{\Delta }{\to }D\left(g\right)+NaCl\left(aq\right)+H_{2}O$
$FeS{O}_4\left(aq\right)+C{H}_{3}COOH\left(aq\right)+C\left(aq\right)\to \left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$

If gas B turns red limtus paper blue, then which of the following option(s) is/are not correct?

• A. C may be $NaN{O}_2$
• B. C may be $NaN{O}_3$
• C. Gas D may be $N_2$
• D. Gas D may be $N_{2}O$

Answer 1

Answer: (B), (D)

The correct options are
B C may be $NaN{O}_3$
D Gas D may be $N_{2}O$

B is $N{H}_3$, which means A must contain ammonium ions,
$\therefore \underset{\left(A\right)}{N{H}_{4}Cl\left(aq\right)}+NaOH\left(aq\right)\to \underset{\left(B\right)}{N{H}_3\left(g\right)}+NaCl\left(aq\right)+H_{2}O$

The brown ring test is given by both Nitrates and Nitrites, but for Nitrates, concentrated sulphuric acid must be used, therefore C can be $NaN{O}_2$ and not $NaN{O}_3$.

$\underset{\left(A\right)}{N{H}_{4}Cl}+\underset{\left(C\right)}{NaN{O}_2}\stackrel{\Delta }{\to }\underset{\left(D\right)}{N_2\left(g\right)}+NaCl\left(aq\right)+H_{2}O$