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

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(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

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(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 $X$?

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