Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor as follows: 4NH₃ + 5O₂→ 4NO + 6H₂O
What is the maximum amount of water that may be produced if 40.0 g NH₃ and 50.0 g O₂ are mixed and allowed to react?

Question

Quizplus · Accepted Answer

First, calculate the molar mass of NH3 (17.03 g/mol) and O2 (32.00 g/mol). Then, find the moles of each reactant: 40.0 g NH3 / 17.03 g/mol = 2.35 mol NH3, and 50.0 g O2 / 32.00 g/mol = 1.56 mol O2. The reaction ratio is 4NH3:5O2, so O2 is the limiting reactant. From the balanced equation, 5 mol O2 produces 6 mol H2O, so 1.56 mol O2 will produce (1.56 mol O2 * 6 mol H2O / 5 mol O2) = 1.87 mol H2O, which rounds to 1.88 mol, making option B correct.

Ammonia Reacts with Diatomic Oxygen to Form Nitric Oxide and Water