How many grams of Mg is required to completel | vedclass

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(A) The balanced redox reaction is:$NO_3^{-} + 4Mg + 10H^{+} \longrightarrow 4Mg^{2+} + NH_4^{+} + 3H_2O$ (or similar reduction to $NH_3$ in basic med

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$0.96$

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(A) The balanced redox reaction is:$NO_3^{-} + 4Mg + 10H^{+} \longrightarrow 4Mg^{2+} + NH_4^{+} + 3H_2O$ (or similar reduction to $NH_3$ in basic medium).In the reduction of $NO_3^{-}$ to $NH_3$,the oxidation state of $N$ changes from $+5$ to $-3$,so the $n$-factor for $NO_3^{-}$ is $8$.The oxidation of $Mg$ from $0$ to $+2$ gives an $n$-factor of $2$.According to the Law of Equivalents:$Eq. 	ext{ of } NO_3^{-} = Eq. 	ext{ of } Mg$$(n	ext{-factor}) 	imes M 	imes V(L) = (n	ext{-factor}) 	imes \frac{	ext{Mass of } Mg}{	ext{Molar mass of } Mg}$$8 	imes 0.1 	imes 0.1 = 2 	imes \frac{	ext{Mass}}{24}$$0.8 = \frac{	ext{Mass}}{12}$$	ext{Mass} = 0.8 	imes 1.2 = 0.96 \ g$

How many grams of $Mg$ is required to completely reduce $100 \ mL$ of $0.1 \ M$ $NO_3^{-}$ solution using the following reaction:
$NO_3^{-} + Mg \longrightarrow Mg^{2+} + NH_3$

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How many grams of $Mg$ is required to completely reduce $100 \ mL$ of $0.1 \ M$ $NO_3^{-}$ solution using the following reaction:$NO_3^{-} + Mg \longrightarrow Mg^{2+} + NH_3$