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?
The balanced chemical equation for the given reaction is given as follows with the amount of substance reacting :-
Hence, we can clearly see that 68 g of NH3 reacts with 160 g of O2.
Therefore, 10g of NH3 will react with = 160 x 10 / 68g = 23.53 g of O2.
But the available amount of O2 is 20 g.
Therefore, O2 is the limiting reagent.
Now, 160 g of O2 gives 120g of NO.
Hence, 20 g of O2 will give = 120 X 20 /160 = 15 g of NO.
Therefore, a maximum of 15 g of nitric oxide can be obtained.