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Class 11
Chemistry
Redox Reactions
in ostwald 39 s process for the manufacture of ni...

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:

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

4NH_3(g) + 5O_2(g) → 4NO_(g) + 6 H₂O_(g)

4x17g 2x32g 4x30g 6x18g

= 68g =160g =120g 108g

Thus, 68 g of NH₃ reacts with 160 g of O₂.

Therefore, 10g of NH3 reacts with 160x10 / 68g of O₂, or 23.53 g of O₂.

But the available amount of O₂ is 20 g.

Therefore, O₂is 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 120_g of NO.

Therefore, 20 g of O₂ gives120x20 / 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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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?

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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?

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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?