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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?
Solution
The balanced chemical equation for the given reaction is given as:
| 4NH3(g) | + | 5O2(g) | → | 4NO(g) | + | 6H2O(g) |
| 4 × 17 g = 68 g |
5 × 32g = 160 g |
4 × 30g = 120 g |
6 × 18g = 108 g |
Thus, 68 g of NH3 reacts with 160 g of O2.
Therefore, 10g of NH3 reacts with `(160xx10)/68` g of O2 or 23.53 g of O2.
But the available amount of O2 is 20 g.
Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).
Now, 160 g of O2 gives 120g of NO.
Therefore, 20 g of O2 gives `(120xx20)/160` g of N, 15 g of NO
Hence, a maximum of 15 g of nitric oxide can be obtained
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