How many grams of Mg is required to completely reduce 100 ml ,   0 . 1 M   NO 3 - solution using…

Given reaction

$\underset{+5}{{\mathrm{NO}}_3^{-}}+\underset{0}{\mathrm{Mg}}⟶\underset{+2}{{\mathrm{Mg}}^{2+}}+\underset{-3}{\mathrm{N}}{\mathrm{H}}_3$

In this redox reaction oxidation state of N is changing from $+5 \mathrm{to} -3$. So, n-factor for the ${\mathrm{NO}}_3^{-}$ will be $8.$ And for the $\mathrm{Mg}$ it will be $2$.

Now according to Law of equivalent, Number of equivalent of both should be equal.

$$\begin{gathered} \mathrm{Eq}. \mathrm{of} {\mathrm{NO}}_3^{-} = \mathrm{Eq}. \mathrm{of} \mathrm{Mg} \\ \mathrm{n}-\mathrm{factor}\times \mathrm{M}\times \mathrm{V} = \mathrm{n}-\mathrm{factor}\times \frac{\mathrm{Mass} \mathrm{of} \mathrm{Mg}}{\mathrm{Molar} \mathrm{mass}} \\ 8\times 0.1\times 0.1\mathrm{L} = 2\times \frac{\mathrm{Mass} \mathrm{of} \mathrm{Mg}}{24} \\ \mathrm{Mass} \mathrm{of} \mathrm{Mg} = 0.08\times 12 = 0.96 \mathrm{g} \end{gathered}$$