2 mole of N_2H_4 loses 16 mole of electrons a | vedclass

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

Question

(C) In $N_2H_4$,the oxidation state of $N$ is $-2$.Total moles of $N$ atoms in $2$ moles of $N_2H_4$ is $2 	imes 2 = 4$ moles of $N$.Total electrons

Vedclass · Accepted Answer

$+2$

Vedclass · Answer

(C) In $N_2H_4$,the oxidation state of $N$ is $-2$.Total moles of $N$ atoms in $2$ moles of $N_2H_4$ is $2 	imes 2 = 4$ moles of $N$.Total electrons lost = $16$ moles.Electrons lost per mole of $N$ atom = $\frac{16 	ext{ moles of } e^-}{4 	ext{ moles of } N} = 4$ moles of $e^-$.Since oxidation involves an increase in oxidation state,the new oxidation state of $N$ = (Initial oxidation state) + (Electrons lost per $N$ atom) = $-2 + 4 = +2$.

$2$ mole of $N_2H_4$ loses $16$ mole of electrons and is converted to a new compound $X$. Assuming that all of the $N$ atoms appear in the new compound,what is the oxidation state of $N$ in $X$?

Similar Questions

The equivalent weight of $KMnO_4$ in an alkaline medium is .......

The oxidation states of sulphur in $S_8$,$S_2F_2$,and $H_2S$ respectively,are:

The oxidation number of $S$ in $S^{2-}$ is:

In which substance does nitrogen exhibit the lowest oxidation state?

In which of the following compounds is the oxidation state of nitrogen correctly represented?

Vedclass Test Series

Exam Paper Generator

Online Exam Module