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 78.0 g of each reactant? 

4NH₃(g)+5O₂(g)⟶4NO(g)+6H₂O(g)

Answer 1

Given reaction:

4 NH3 (g ) + 5O2 ---- > 4 NO (g) + 6 H2O (g)

Given that 78.0 g NH3

78.0 g O2

Number of moles = amount in g / molar mass

NH3: 78.0 g / 17 g/ mole

= 4.58 moles NH3

O2 : 78.0 g / 32 g/ mole

2.44 moles O2

Mole of NH3 = 2.44 moles o2 *4/5

= 1.952 mole NH3

Thus oxygen is limiting agent

Mole of H2O = 2.44 moles O2*6 moel H2O / 5 mole O2

= 2.928 mole H2O

Amount in g = number of moles * molar mass

2.928 mole H2O *18.02 g/ mole

= 52.8 g H2O