Question

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?

Answer 1

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

$4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(g)$
$4×17\,g$ $5×32\,g$ $4×30\,g$ $6×18\,g$
$=68\,g$ $=160\,g$ $=120\,g$ $=108 \,g$
Thus, $68 \,g$ of $NH_3$ reacts with $160 \,g$ of $O_2$.

Therefore, $10\,g$ of $NH_3$ reacts with $\frac{160\times10}{68}$ $g$ of $O_2$, or $23.53$ $g$ of $O_2$.
But the available amount of $O_2$ is $20$ $g$.
Therefore, $O_2$ is the limiting reagent (we have considered the amount of $O_2$ to calculate the weight of nitric oxide obtained in the reaction).

Now, $160$ $g$ of $O_2$ gives $120$ $g$ of $NO$.

Therefore, $20$ $g$ of $O_2$ gives $\frac{120\times20}{160}$ $g$ of $N$, or $15$ $g$ of $NO$.

Hence, a maximum of $15 \,g$ of nitric oxide can be obtained.