8.25. In Ostwald’s process for the manufacture of nitric add, 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 10.0 g of ammonia and 20.0 g of oxygen?
8.25. In Ostwald’s process for the manufacture of nitric add, 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 10.0 g of ammonia and 20.0 g of oxygen?
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1 Answer
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The balanced equation for the reaction is:
4NH3? (g)+5O2? (g)→4NO (g)+6H2? O (g)
The molar masses of ammonia and oxygen are 17 g/mol and 32 g/mol respectively.
68 g of NH3 will react with O2 = 160 g.
Therefore, 10 g of NH3 will react with O2 = 160/68 x 10 g = 23.6 g
But the amount of O2 which is actually available is 20.0 g which is less than the amount which is needed. Therefore, O2 is the limiting reagent and hence calculations must be based upon the amount of O2 taken and not on the amount of NH3 taken. From the equation,
160 g of O2 produce NO = 120 g
Therefore, 20 g of O2 will produce NO =120/160 x 20 = 15 g
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