N2 + 3H2 → 2NH3. What is the theoretical yield, the limiting reactant, and the percent yield? Please show work.

Question

It has 3 moles of N2 that react with 3 moles of H2 and a actual of 31.63g

Answer 1

3 mol N₂ * (2 mol NH₃ / 1 mol N₂) = 6 mol NH₃.

3 mol H₂ * (2 mol NH₃ / 3 mol H₂) = 2 mol NH₃. So H₂ is the limiting reactant.

From here, 2 mol NH₃ * (17.02g NH₃ /1 mol NH₃) = 34.04g NH₃, the theoretical yield.

Percent yield = 100 * actual/theoretical = 100 * 31.63/34.04 = 92.9%.