For ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$ rate of disappearance of H₂ is 0.01 M.min^-1 The amount of NH₃ formed at that instant will be

$$\begin{gathered} \mathrm{For} \mathrm{the} \mathrm{reaction}: {\mathrm{N}}_2 + 3{\mathrm{H}}_2 \to 2{\mathrm{NH}}_3 \\ \mathrm{Rate} \mathrm{of} \mathrm{reaction} = -\frac{ \mathrm{d}\left[{\mathrm{N}}_2\right] }{\mathrm{dt}}= - \frac{1\mathrm{d}\left[{\mathrm{H}}_2\right]}{3 \mathrm{dt}}= \frac{1\mathrm{d}\left[{\mathrm{NH}}_3\right]}{2 \mathrm{dt}} \\ \therefore \frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{ \mathrm{dt}} = \frac{2}{3}\times \frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{dt}} = \frac{2}{3}\times 0.01\times 17=\frac{0.02}{3}\times 17=0.1132\mathrm{g} \end{gathered}$$