The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

Question

Quizplus · Accepted Answer

To find the grams of NH3 needed, first calculate the theoretical yield of NO. The molar mass of NO is 30.01 g/mol, so 70.5 g of NO is 70.5 / 30.01 = 2.35 moles. Since the yield is 29.8%, the theoretical moles of NO needed is 2.35 / 0.298 = 7.89 moles. From the balanced equation, 4 moles of NH3 produce 4 moles of NO, so 7.89 moles of NO require 7.89 moles of NH3. The molar mass of NH3 is 17.03 g/mol, so 7.89 moles of NH3 is 7.89 * 17.03 = 134.3 g, which rounds to 134 g.

The Reaction for the Oxidation of NH3 Is Given As

The Reaction for the Oxidation of NH₃ Is Given As