Question

$\ln$ the reduction of $HN{O}_3$ to $N_{2}O$ the no.of moles of electrons involved per mole of $HN{O}_3$ is :

• A. 8
• B. 4
• C. 3
• D. 6

Answer 1

The correct option is B 4

$HN{O}_3\to 0.5N_{2}O$
Oxidation number of nitrogen in $HN{O}_3$ and $N_{2}O$ is +5 and +1 respectively so number of moles of electrons per mole of $HN{O}_3$ is 4.