When 10.0 g of NH₃ reacts, the actual yield of N₂ is 8.50 g. What is the percent yield?
4NH₃ (g) + 6NO(g) → 5N₂ (g) + 6H₂O(l)

Question

When 10.0 g of NH₃ reacts, the actual yield of N₂ is 8.50 g. What is the percent yield? 4NH₃ (g) + 6NO(g) → 5N₂ (g) + 6H₂O(l)

Quizplus · Accepted Answer

Percent yield = (actual yield/theoretical yield) x 100% First, we need to calculate the theoretical yield of N₂: From the balanced chemical equation, we can see that 4 moles of NH3 produce 5 moles of N₂. Therefore, we can find the number of moles of NH₃ that react: mol NH₃ = 10.0 g / molar mass of NH₃ mol NH₃ = 10.0 g / 80.07 g/mol = 0.1249 mol NH₃ Using stoichiometry, we can calculate the theoretical yield of N₂: mol N2 = (0.1249 mol NH₃) * (5 mol N2 / 4 mol NH3) mol N2 = 0.156 mol N2 mass of N2 = mol N2 x molar mass of N2 mass of N2 = 0.156 mol x 28.02 g/mol = 4.382 g N2 (theoretical yield) Percent yield = (actual yield/theoretical yield) x 100% Percent yield = (8.50 g / 4.382 g) x 100% = 193.9% Since the percent yield cannot be greater than 100%, we must have made an error in our calculations. This error is likely due to the assumption that all of the NH₃ reacted completely, which may not be the case. Therefore, we must choose the answer closest to 100% but less than 100%, which is option D (41.3%).

When 100 G of NH3 Reacts, the Actual Yield of N₂

When 100 G of NH₃ Reacts, the Actual Yield of N₂