Question

In the 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 $N_{2}O$

Oxidation number of $HN{O}_3$ is $+5$

Oxidation number of $N_2$ is $O$, $N_{2}O$ is $+1$, $N{O}_2$ is $+4$ and $N{H}_3$ is $-3$.

Therefore, each change in oxidation number need $1$ mole of electrons for mole of product.

(a) Needs $4$ moles ($5$ electrons for $N$, times $2$ to get $N_2$ and then $0.4$ moles=$4$)

(b) Change in oxidation state=$4$, $2$ Nitrogen to get $N_{2}O$, $0.5$ moles=$4$ moles of electron.

(c) Only needs one mole of electron to change to $+4$ oxidation state

(d) It would need $8$ moles of electrons.