Question

In a reaction, $4$ moles of electrons are transferred to $1$ mole of $HN{O}_3$ when acted as an oxidant. The possible reduction product is:

• A. $1/2$ mole of $N_2$
• B. $1/2$ mole of $N_{2}O$
• C. $1$ mole of $N{O}_2$
• D. $1$ mole of $N{H}_3$

Answer 1

The correct option is B $1/2$ mole of $N_{2}O$

Oxidation state of nitrogen in ${HNO}_3$ is +5. In $N_2$, oxidation state of nitrogen is 0. In $N_{2}O$, ${NO}_2$ and ${NH}_3$ the oxidation state of nitrogen is +1, +4 and -3 respectively.

(a) needs 5 moles (5 electrons per N, times 2 to get N2 and then 0.5 moles = 5.

(b), change in oxidation number is 4, 2 nitrogens to get N2O 0.5 moles = 4 moles of electrons.

(c) only needs one mole of electrons to change to +4

(d) would need 8 moles.

So, correct answer is option B.