N_2H_4 + IO_3^- + 2H^+ + Cl^- to ICl + N_2 + | vedclass

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(C) In the reaction,the oxidation state of $N$ in $N_2H_4$ changes from $-2$ to $0$. The change per atom is $2$,and for $2$ atoms of $N$,the total cha

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$8$ and $53.5$

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(C) In the reaction,the oxidation state of $N$ in $N_2H_4$ changes from $-2$ to $0$. The change per atom is $2$,and for $2$ atoms of $N$,the total change in oxidation state is $4$. Thus,the $n$-factor for $N_2H_4$ is $4$.Equivalent mass of $N_2H_4 = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{32}{4} = 8$.The oxidation state of $I$ in $IO_3^-$ changes from $+5$ to $+1$ in $ICl$. The change in oxidation state is $5 - 1 = 4$. Thus,the $n$-factor for $KIO_3$ is $4$.Equivalent mass of $KIO_3 = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{214}{4} = 53.5$.

$N_2H_4 + IO_3^- + 2H^+ + Cl^- \to ICl + N_2 + 3H_2O$
The equivalent masses of $N_2H_4$ and $KIO_3$ respectively are:

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$N_2H_4 + IO_3^- + 2H^+ + Cl^- \to ICl + N_2 + 3H_2O$The equivalent masses of $N_2H_4$ and $KIO_3$ respectively are: