In Ostwals'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. ammonia and 20.00 g of oxygen?

Open in App

Solution

The balanced chemical equation for the given reaction is given as:
$4 N H_{3 (g)} + 5 O_{2 (g)} ⟶ 4 N O_{(g)} + 6 H_{2} O_{(g)}$
$4 \times 17 g 5 \times 32 g 4 \times 30 g 6 \times 18 g$
$= 68 g = 160 g = 120 g = 108 g$
Thus, 68g of $N H_{3}$ reacts with 160 g of $O_{2}$
Therefore, 10g of $N H_{3}$ reacts with $\frac{160 \times 10}{68} g$ of $O_{2}$ or 23.53g 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 120g of NO.
Therefore, 20 g of $O_{2}$ gives $\frac{120 \times 20}{160} g$ of N, 15g of NO.
Hence, a maximum of 15g of nitric oxide can be obtained.

Suggest Corrections

Similar questions

H_{2} O

N O

20.3 g

5.89 g

The first step in the manufacture of nitric acid is the catalytic oxidation of ammonia. What is name of the catalyst?

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program