Question

Hydro-zine reacts with $KI{O}_3$ in presence of $HCl$ as $N_2{H}_4+I{O}_3^{-}+2H^{+}+C{l}^{-}\to ICl+N_2+3H_{2}O$
The equivalent masses of $N_2{H}_4$ and $KI{O}_3$:

• A. $8,87$
• B. $8,35.6$
• C. $16,53.5$
• D. $8,53.5$

Answer 1

The correct option is B $8,35.6$

$N_2{H}_4+I{O}_3{}^{-}+2H+C{l}^{-}⟶ICl+N_2+3H_{2}O$

Oxidation state of $N$ in $N_2{H}_4=-2$

Oxidation state of $N$ in $N_2$=$O$

$\therefore$ Change in oxidation of $N$ is $2$

& change in oxidation state of $N_2{H}_4=4$

$\therefore$ Equivalent weight of $N_2{H}_4=\frac{Molecularweight}{4}$

=$\frac{32}{4}$

=$8$

Oxidation state of $I$ in ${I{O}_3}^{-}=+5$

Oxidation state of $I$ in $ICl=-1$

$\therefore$ Change in oxidation state of $KI{O}_3=6$

$\therefore$ Equivalent weight of $KI{O}_3$=$\frac{Molecularweight}{6}$

=$\frac{214}{6}$

=$35.6$

$\therefore$ (B) is the correct answer.