How many moles of hydrazine (N_{2}H_{4}) can | vedclass

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(B) The reduction half-reaction is: $BrO_{3}^{-} 	o Br^{-}$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor is $6$.For

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(B) The reduction half-reaction is: $BrO_{3}^{-} 	o Br^{-}$. The change in oxidation state of $Br$ is from $+5$ to $-1$,so the $n$-factor is $6$.For $\frac{2}{3}$ mole of $BrO_{3}^{-}$,the total electrons gained $= \frac{2}{3} 	imes 6 = 4 \, e^{-}$.The oxidation half-reaction is: $N_{2}H_{4} 	o N_{2}$. The change in oxidation state of $N$ is from $-2$ to $0$. Since there are two $N$ atoms,the $n$-factor is $2 	imes 2 = 4$.Let $n$ be the moles of $N_{2}H_{4}$. Total electrons lost $= n 	imes 4$.Equating total electrons lost and gained: $n 	imes 4 = 4$.Therefore,$n = 1 \, 	ext{mole}$.

How many moles of hydrazine $(N_{2}H_{4})$ can be oxidized into $N_{2}$ using $\frac{2}{3}$ mole of bromate ions $(BrO_{3}^{-} \to Br^{-})$ :-

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