Ncert Exercises & Solutions

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?

NEETprep Answer:

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

4 {NH}_{3 (g)} + 5 O_{2 (g)} \to 4 {NO}_{(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

{NH}_{3}

reacts with 160g of

O_{2}

Therefore, 10g of

{NH}_{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, or 15 g of NO.
Hence, a maximum of 15 g of nitric oxide can be obtained.

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