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?

(D) The balanced chemical equation for the reaction is: $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_{2}O_{(g)}$
From the stoichiometry,$68 \ g$ of $NH_{3}$ reacts with $160 \ g$ of $O_{2}$ to produce $120 \ g$ of $NO$.
To find the limiting reagent,calculate the $O_{2}$ required for $10 \ g$ of $NH_{3}$: $\frac{160 \ g \ O_{2}}{68 \ g \ NH_{3}} \times 10 \ g \ NH_{3} = 23.53 \ g \ O_{2}$.
Since only $20 \ g$ of $O_{2}$ is available,$O_{2}$ is the limiting reagent.
Using the limiting reagent to calculate the yield of $NO$: $\frac{120 \ g \ NO}{160 \ g \ O_{2}} \times 20 \ g \ O_{2} = 15 \ g \ NO$.
Thus,the maximum weight of nitric oxide obtained is $15 \ g$.