Answer to Question #70726 in General Chemistry for Brooke

Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 92.8 g of each reactant?

Answer 1

4NH 3 + 5O 2  4NO + 6H 2 O
That means that 4 mol of ammonia reacts with 5 mol of oxygen producing 6 mol of water.
Let’s define the amounts of each reactant to know what is the limiting reagent.
The amount of ammonia is 92,8 g / 17 g/mol = 5,46 mol. It reacts with = 6.825 mol of oxygen.
The amount of oxygen is 92.8 g / 32 g/mol = 2,9 mol.
So there is less oxygen than ammonia requires, that’s why oxygen is limiting reagent.
From 2,9 mol oxygen we’ll obtain = 3.48 mol of water.
3,48 mol of water is 3,48×18 = 62.64 g.
Answer
62.64 g of water is the maximum mass of H 2 O that can be produced by combining 92.8 g of
each reactant.