In Ostwald’s process for the manufacture of nitric acid,

The balanced chemical equation for the given reaction is given as:

Thus, $68 \mathrm{~g}$ of $\mathrm{NH}_{3}$ reacts with $160 \mathrm{~g}$ of $\mathrm{O}_{2}$.

Therefore, $10 \mathrm{~g}$ of $\mathrm{NH}_{3}$ reacts with $\frac{160 \times 10}{68} \mathrm{~g}$ of $\mathrm{O}_{2}$, or $23.53 \mathrm{~g}$ of $\mathrm{O}_{2}$.

But the available amount of $\mathrm{O}_{2}$ is $20 \mathrm{~g}$.

Therefore, $\mathrm{O}_{2}$ is the limiting reagent (we have considered the amount of $\mathrm{O}_{2}$ to calculate the weight of nitric oxide obtained in the reaction).

Now, $160 \mathrm{~g}$ of $\mathrm{O}_{2}$ gives $120 \mathrm{~g}$ of $\mathrm{NO}$.

Therefore, $20 \mathrm{~g}$ of $\mathrm{O}_{2}$ gives $\frac{120 \times 20}{160} \mathrm{~g}$ of $\mathrm{N}$, or $15 \mathrm{~g}$ of $\mathrm{NO}$.

Hence, a maximum of $15 \mathrm{~g}$ of nitric oxide can be obtained.