2.5 mol of hydrazine (N_2H_4) loses 25 mol | vedclass

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

Question

(C) The reaction for the oxidation of hydrazine is: $N_2H_4 \rightarrow N_2^X + n e^-$.In $N_2H_4$,the oxidation state of $N$ is $-2$.Let the oxidatio

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The reaction for the oxidation of hydrazine is: $N_2H_4 ightarrow N_2^X + n e^-$.In $N_2H_4$,the oxidation state of $N$ is $-2$.Let the oxidation state of $N$ in compound $X$ be $X$.The change in oxidation state per mole of $N_2H_4$ is: $n = 2 	imes (X - (-2)) = 2(X + 2)$.Total electrons lost by $2.5 \ mol$ of $N_2H_4$ is: $2.5 	imes 2(X + 2) = 5(X + 2)$.Given that the total electrons lost is $25 \ mol$,we have: $5(X + 2) = 25$.$X + 2 = 5$.$X = +3$.Thus,the oxidation state of $N$ in compound $X$ is $+3$.

$2.5 \ mol$ of hydrazine $(N_2H_4)$ loses $25 \ mol$ of electrons to form a new compound $X$. Assuming all lost electrons are contained in the new compound,what is the oxidation state of nitrogen in compound $X$?

Similar Questions

In which of the following acids is the phosphorus atom present in its lowest oxidation state?

Arrange the following substances in the increasing order of the oxidation number of $Mn$: $MnCl_2, MnO_2, Mn(OH)_3, KMnO_4$.

$A$ compound contains atoms of $x, y$ and $z$ elements. Oxidation numbers of $x, y$ and $z$ are $+3, +5$ and $-2$ respectively. Identify the possible formula of this compound?

Calculate the correct increasing order of average oxidation number of sulphur in $S_8, S_2O_3^{2-}, S_2O_7^{2-}, S_4O_6^{2-}, SO_3^{2-}$

The difference between the oxidation numbers of the two types of sulphur atoms in $Na_2S_4O_6$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module