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

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(B) In the reaction,$N_2H_4$ acts as the reductant.The oxidation state of $N$ in $N_2H_4$ is $-2$,and in $N_2$ it is $0$.The change in oxidation state

Vedclass · Accepted Answer

$8$

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(B) In the reaction,$N_2H_4$ acts as the reductant.The oxidation state of $N$ in $N_2H_4$ is $-2$,and in $N_2$ it is $0$.The change in oxidation state per nitrogen atom is $0 - (-2) = 2$.Since there are $2$ nitrogen atoms in $N_2H_4$,the total $n$-factor is $2 	imes 2 = 4$.The molar mass of $N_2H_4$ is $2 	imes 14 + 4 	imes 1 = 32 \ g/mol$.The equivalent weight $(EW)$ is calculated as:$EW = \frac{	ext{Molar Mass}}{n	ext{-factor}} = \frac{32}{4} = 8$.

Column $I$	Column $II$
$(A)$ $Cu + \text{dil } HNO_3$	$(p)$ $NO$
$(B)$ $Cu + \text{conc } HNO_3$	$(q)$ $NO_2$
$(C)$ $Zn + \text{dil } HNO_3$	$(r)$ $N_2O$
$(D)$ $Zn + \text{conc } HNO_3$	$(s)$ $Cu(NO_3)_2$
	$(t)$ $Zn(NO_3)_2$

$N_2H_4 + IO_3^- + 2H^+ + Cl^- \to ICl + N_2 + 3H_2O$. What is the equivalent weight of the reductant in the above reaction?

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