Question

In a reaction, $4$ moles of electrons are transferred to $1$ mole of ${HNO}_3$. The possible product obtained due to reduction is :

• A. $0.4$ mol of $N_2$
• B. $0.5$ mol of $N_{2}O$
• C. $1$ mol of $N{O}_2$
• D. $1$ mol of ${NH}_3$

Answer 1

Answer: (B)

$0.5$ mol of $N_{2}O$

$H{NO}_3+4e^{-}⟶Product$
+5 +1
In $N_{2}O$, oxidation number of $N=+1$.
2 moles of $HN{O}_3$ produces 1 mole of $N_{2}O$.
So, $0.5$ mol of $N_{2}O$ are formed by reduction of $1$ mol of $H{NO}_3$ with $4$ mol of electrons.