The ratio of equivalent masses of $\mathrm{NH}_3$ in the reactions is : \(i) $4 \mathrm{NH}_3+5 \ma

Question

The ratio of equivalent masses of $\mathrm{NH}_3$ in the reactions is : \(i) $4 \mathrm{NH}_3+5 \mathrm{O}_2 \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O}$, \(ii) $2 \mathrm{NH}_3 \longrightarrow \mathrm{~N}_2+3 \mathrm{H}_2$

Professor PC Thomas Classes · Answer

$ \begin{array}{ll} (\mathrm{d}): \mathrm{N}^{3-} \longrightarrow \mathrm{N}^{2+}+5 e^{-} & (	herefore E=M / 5) \ 6 e^{-}+2 \mathrm{~N}^{3-} \longrightarrow\left(\mathrm{N}^0ight)_2 & (	herefore E=M / 3) \end{array} $