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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 $(BrO_{3}^{-} \to Br^{-})$ :-

Similar Questions

In the redox reaction $xMnO + yPbO_2 + zHNO_3 \to HMnO_4 + Pb(NO_3)_2 + H_2O$,determine the values of $x, y,$ and $z$.

What is the equivalent weight of $IO_4^-$ when it is converted to $I_2$ in an acidic medium?

In the neutral or faintly alkaline medium,$KMnO_4$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from?

Assign $A, B, C, D$ from the given type of reaction.
$CuSO_4 + 2KCN \longrightarrow CuCN \downarrow + (CN)_2 \uparrow + K_2SO_4$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module