Search any question & find its solution
Question:
Answered & Verified by Expert
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?
Options:
Solution:
2776 Upvotes
Verified Answer
The correct answer is:
NO, NO
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.
$$
\begin{aligned}
& 3 \mathrm{HNO}_2 \longrightarrow \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \\
& 2 \mathrm{NaNO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{NaHSO}_4+\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{NO}
\end{aligned}
$$
$$
\begin{aligned}
& 3 \mathrm{HNO}_2 \longrightarrow \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+2 \mathrm{NO} \\
& 2 \mathrm{NaNO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{NaHSO}_4+\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{NO}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.