3 mole of N_2H_4 loses 30 mole of electrons | vedclass

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(D) The initial oxidation state of $N$ in $N_2H_4$ is $x$,where $2x + 4(+1) = 0$,so $x = -2$.Total electrons lost by $3 \ mole$ of $N_2H_4$ is $30 \ m

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$3$

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(D) The initial oxidation state of $N$ in $N_2H_4$ is $x$,where $2x + 4(+1) = 0$,so $x = -2$.Total electrons lost by $3 \ mole$ of $N_2H_4$ is $30 \ mole$,so electrons lost per mole of $N_2H_4$ is $30 / 3 = 10 \ mole$.Let the new oxidation state of $N$ be $y$. The change in oxidation state per mole of $N_2H_4$ is $2(y - (-2)) = 10$.$2(y + 2) = 10 \implies y + 2 = 5 \implies y = +3$.Thus,the oxidation state of $N$ in the new compound is $+3$.

$3 \ mole$ of $N_2H_4$ loses $30 \ mole$ of electrons to form a new compound. Calculate the oxidation state of $N$ in the new compound,assuming the oxidation state of $H$ remains unchanged.

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