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

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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

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$+2$

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(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 $X$?

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