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

What is the oxidation state of $Cr$ when the $pH$ of the aqueous solution of potassium dichromate changes from acidic to basic?

The oxidation number of $S$ in $Na_2S_4O_6$ is ...........

In the reaction of potassium dichromate,potassium chloride and sulfuric acid (conc.),the oxidation state of the chromium in the product is $(+)$ . . . . . . .

Which one of the following does not have an underlined atom with a fractional oxidation state?

The formal oxidation numbers of $Cr$ and $Cl$ in the ions $Cr_2O_7^{2-}$ and $ClO_3^{-}$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module