An important intermediate reaction in the industrial production process of nitric acid is shown.What is the theoretical yield of H₂O (18.02 g/mol) if 45.0 kg of NH₃ (17.04 g/mol) reacts with excess O₂ (32.00 g/mol) ?
4 NH₃ (g) +5O₂ (g) $\rightarrow$ 4 NO(g) +6 H₂ O(l)

Question

An important intermediate reaction in the industrial production process of nitric acid is shown.What is the theoretical yield of H₂O (18.02 g/mol) if 45.0 kg of NH₃ (17.04 g/mol) reacts with excess O₂ (32.00 g/mol) ?
4 NH₃ (g) +5O₂ (g)

\rightarrow

4 NO(g) +6 H₂ O(l)

Quizplus · Accepted Answer

The balanced equation shows that 4 moles of NH3 produce 6 moles of H2O. First, calculate the moles of NH3: 45,000 g / 17.04 g/mol = 2,640.85 mol. According to the stoichiometry, 2,640.85 mol NH3 will produce (2,640.85 mol * 6/4) = 3,961.28 mol H2O. The mass of H2O is then 3,961.28 mol * 18.02 g/mol = 71,384.3 g, or 71.4 kg.

An Important Intermediate Reaction in the Industrial Production Process of Nitric