Nitrous acid was disproportionated to form water, $\mathrm{HNO}_3$ and $X$. In another reaction, sodium…
Nitrous acid was disproportionated to form water, $\mathrm{HNO}_3$ and $X$. In another reaction, sodium nitrite was reacted with $\mathrm{H}_2 \mathrm{SO}_4$ to form $\mathrm{NaHSO}_4, \mathrm{HNO}_3$, water and $Y$. What are $X$ and $Y$ respectively?
$\mathrm{NO}, \mathrm{N}_2 \mathrm{O}_3$
NO, NO
$\mathrm{N}_2 \mathrm{O}, \mathrm{NO}_2$
$\mathrm{NO}_2, \mathrm{~N}_2 \mathrm{O}_5$
Solution
Nitrous acid was disproportionated to form water, $\mathrm{HNO}_3$ and $\mathrm{NO}$. In another reaction, sodium nitrite was reacted with $\mathrm{H}_2 \mathrm{SO}_4$ to form $\mathrm{NaHSO}_4$, $\mathrm{HNO}_3$, water and NO.
$
\begin{aligned}
& 3 \mathrm{HNO}_2 \longrightarrow \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \\
& 2 \mathrm{NaNO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{NaHSO}_4+\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{NO}
\end{aligned}
$