A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction?
Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

Question

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Quizplus · Accepted Answer

The theoretical yield of NH₃ can be calculated using stoichiometry. The molar mass of N₂ is 28.0 g/mol, so 24.0 g of N₂ is 0.857 mol. According to the balanced equation, 1 mol of N₂ produces 2 mol of NH₃, so 0.857 mol of N₂ produces 1.714 mol of NH₃. The molar mass of NH₃ is 17.0 g/mol, so the theoretical yield is 1.714 mol × 17.0 g/mol = 29.14 g. The percent yield is (actual yield/theoretical yield) × 100% = (3.85 g / 29.14 g) × 100% = 13.2%.

A 240 G Sample of Nitrogen Gas Reacts with an Excess