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 75.9 g of each reactant? 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

Answer 1

51.1 g of H2O

Step-by-step explanation:

The equation of the reaction is;

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

We decide on the limiting reactant;

Number of moles of NH3 = 75.9 g/17 g/mol = 4.46 moles

If 4 moles of NH3 produced 6 moles of H2O

4.46 moles of NH3 will produce 4.46 * 6/4 = 6.69 moles of H2O

For O2

Number of moles of O2 = 75.9 g/32g/mol = 2.37 moles

If 5 moles of O2 produced 6 moles of H2O

2.37 moles of O2 will produce 2.37 * 6/5 = 2.84 moles of H2O

Hence O2 is the limiting reactant

Mass of water produced = 2.84 moles of H2O * 18 g/mol = 51.1 g of H2O