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?

(C) The balanced chemical equation for the reaction is:
$4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_{2}O_{(g)}$
From the stoichiometry:
$4 \times 17 \, g$ of $NH_{3}$ $(68 \, g)$ reacts with $5 \times 32 \, g$ of $O_{2}$ $(160 \, g)$ to produce $4 \times 30 \, g$ of $NO$ $(120 \, g)$.
Calculate the limiting reagent:
For $10.00 \, g$ of $NH_{3}$,the required $O_{2} = \frac{160 \, g \, O_{2}}{68 \, g \, NH_{3}} \times 10.00 \, g \, NH_{3} \approx 23.53 \, g \, O_{2}$.
Since only $20.00 \, g$ of $O_{2}$ is available,$O_{2}$ is the limiting reagent.
Calculate the yield of $NO$ based on the limiting reagent $(O_{2})$:
$160 \, g$ of $O_{2}$ produces $120 \, g$ of $NO$.
Therefore,$20.00 \, g$ of $O_{2}$ produces $\frac{120 \, g \, NO}{160 \, g \, O_{2}} \times 20.00 \, g \, O_{2} = 15.00 \, g$ of $NO$.