Question

In Ostwald’s process for the manufacturing of nitric acid, the first step involves the oxidation of ammonia gas to give nitric oxide gas and stream. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 gram of oxygen?

Answer 1

Hint: Ostwald process is an industrial process which involves the catalytic oxidation of ammonia. In the Ostwald process, ammonia is made to react with oxygen and water in the presence of platinum catalyst to yield nitric acid.

Complete step by step answer:
The Ostwald process refers to a chemical process that converts ammonia into nitric acid (\[HN{O_3}\]) in two steps. In the first step, ammonia is oxidized leading to the formation of nitric oxide and nitrogen dioxide. And in the second step, nitrogen dioxide is absorbed in water. The balanced chemical equations for the complete Ostwald process are written below:
\[4N{H_3} + 5{O_{2\;}} \to 4NO{\text{ + }}6{H_2}O{\text{ }} \\
2NO + {O_2} \to 2N{O_2} \\
2N{O_2} + {H_2}O \to HN{O_3} + HN{O_2}\]
From the first chemical equation stated above, we can clearly see that 4 moles of ammonia (\[N{H_3}\]) react with 5 moles of \[{O_2}\] to give 4 moles of nitric oxide (NO).
1 mole of \[NH_3\] is equivalent to 17 g of \[NH_3\]. (since the molecular weight of \[NH_3\] is 17 g/mol).
Therefore, 4 moles of \[N{H_3}\] is equivalent to $4 \times 17 = 68g$ of \[N{H_3}\]
1 mole of NO is equivalent to 30 g of NO (since molecular weight of NO is 30 g/mol)
Therefore, 4 moles of NO is equivalent to $4 \times 30 = 120g$of NO
Similarly, 1 mole of \[{O_2}\] is equivalent to 32 g of \[{O_2}\] (since molecular weight of \[{O_2}\] is 32 g/mol)
Therefore, 5 moles of \[{O_2}\] is equivalent to $5 \times 32 = 160g$ of \[{O_2}\]
5 moles of \[{O_2}\](160 g) react with 4 moles of \[N{H_3}\](68 g)
Therefore, 20 g of \[{O_2}\] react with $\dfrac{{68}}{{160}} \times 20 = 8.5g$ of \[N{H_3}\]
We can say \[N{H_3}\] is present in a limiting amount but oxygen is present in excess so we proceed by \[N{H_3}\].
68 g of \[N{H_3}\] gives 120 g of \[NO\] in balanced chemical reaction
So 1 gram of \[N{H_3}\] gives \[\dfrac{{120}}{{68}}\] = 1.76g of NO
And so 10 gram of \[N{H_3}\] gives $10 \times 1.76$ = 17.6 g of NO
Alternatively, if we proceed the reaction by \[{O_2}\].
160 g of \[{O_2}\] gives 120 g of NO in balanced chemical reaction
So 1 gram of \[{O_2}\] gives \[\dfrac{{120}}{{160}}\] = 0.75 g of NO
And so 20 gram of \[{O_2}\] gives $20 \times 0.75$ = 15 g of NO
If we start a reaction with 10.0 g of ammonia then we get a maximum 17.6 g of nitric oxide. And if we start a reaction by 20.0 g of oxygen then we get a maximum 15 g of nitric oxide.

Note: The Ostwald process is the significant process of the modern chemical industry, as it provides the main raw materials for the production of the most common type of fertilizers. The Ostwald process is closely related with the Haber process that provides the main requisite raw material for the Ostwald process i.e. ammonia.