What is the maximum number of grams of ammonia, NH₃, that can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂?
N₂ + 3H₂ $\rarr$ 2NH₃

Question

What is the maximum number of grams of ammonia, NH₃, that can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂?
N₂ + 3H₂

\rarr

2NH₃

Quizplus · Accepted Answer

To solve this problem, we need to use stoichiometry to determine the limiting reactant and the theoretical yield of ammonia that can be produced from the given amounts of reactants. The balanced chemical equation tells us that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, we can set up a ratio to determine the moles of ammonia that can be produced: 10.0 g H₂ * 1 mol H₂/34.08 g H₂ * 1 mol NH₃/3 mol H₂ = 0.0989 mol NH₃ 80.0 g N₂ * 1 mol N₂/80.06 g N₂ * 2 mol NH₃/1 mol N₂ = 1.998 mol NH₃ Since the stoichiometric ratio is 3 moles of H₂ to 2 moles of NH₃, and we have less than 3 moles of H₂ available, H₂ is the limiting reactant. Using the amount of limiting reactant, we can calculate the theoretical yield of ammonia: 0.0989 mol NH₃ * 17.03 g NH₃/1 mol NH₃ = 1.68 g NH₃ Therefore, the maximum number of grams of ammonia that can be obtained is 1.68 g. Choice C, 56.7 g, is incorrect because it is much larger than the theoretical yield. This suggests that a calculation error was made, perhaps by assuming that N₂ was the limiting reactant.

What Is the Maximum Number of Grams of Ammonia, NH3 →\rarr→

What Is the Maximum Number of Grams of Ammonia, NH₃ $\rarr$