Question

What is the possible product of the reaction in which $1mole$ of $HN{O}_3$ is reduced by absorption of $4moles$ electrons?

• A. $0.5mole$ of $N_2$
• B. $0.5mole$ of $N_{2}O$
• C. $1mole$ of $N{O}_3$
• D. $1mole$ of $N{H}_3$

Answer 1

Answer: (B)

Here, when nitric acid is converted into any of the nitrogen compound, the oxidation state of nitrogen is changed and for oxygen and hydrogen, it remains unchanged.

Let the oxidation state of Nitrogen in $HN{O}_3$ be $x.$

Then, $+1+x+3\left(-2\right)=0\Rightarrow x-5=0\Rightarrow x=+5$ .

Now, after absorption of $4moles$ of electrons, the oxidation state of nitrogen will become

$N^{5+}+4e^{-}\to N^{+}$

Out of the given compounds, the compound having the oxidation state of nitrogen as $+1$ is $N_{2}O$ .

Thus, here nitric acid is getting converted into dinitrogen oxide.

$1moleofHN{O}_3\stackrel{gives}{\to }xmolesof{N}_{2}O$

Applying principle of atom conservation for nitrogen,

$1\times molesofHN{O}_3=2\times molesof{N}_{2}O1\times 1=2xx=\frac{1}{2}=0.5mol$