8.25 :  In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Thus, 68g of ${\mathrm{NH}}_{3}$ reacts with 160g of ${\mathrm{O}}_{2}$.
Therefore, 10g of ${\mathrm{NH}}_{3}$ reacts with $\frac{160×10}{68}$g of ${\mathrm{O}}_{2}$, or 23.53g of ${\mathrm{O}}_{2}$
But the available amount of ${\mathrm{O}}_{2}$ is 20 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 g of ${\mathrm{O}}_{2}$ gives 120g of NO.
Therefore, 20 g of ${\mathrm{O}}_{2}$ gives $\frac{120×20}{160}$ g of N, or 15 g of NO.