Final answer:
The question asked about the amount of NH₃ produced when 5.4 moles of N₂ react with 3.2 moles of H₂. H₂ is the limiting reactant, and the stoichiometry of the reaction indicates that 2.1333... moles of NH₃ are produced, which does not match the given options.
Step-by-step explanation:
When 5.4 moles of N₂ react with 3.2 moles of H₂, we can determine how many moles of NH₃ will be formed by looking at the stoichiometry of the balanced chemical equation N₂ + 3H₂ → 2NH₃. The equation tells us that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. To solve the problem, we first need to identify the limiting reactant, which is the reactant that will be completely consumed first, thus limiting the amount of product formed.
In this case, 1 mole of N₂ would require 3 moles of H₂ to fully react. Since we have 5.4 moles of N₂, it would require 5.4 x 3 = 16.2 moles of H₂. However, we only have 3.2 moles of H₂ available, which means H₂ is the limiting reactant.
Now, using the mole ratio, we can calculate the amount of NH₃ produced from the 3.2 moles of H₂. Since the ratio of H₂ to NH₃ is 3:2, the moles of NH₃ produced will be 3.2 moles H₂ x (2 moles NH₃ / 3 moles H₂) = 2.1333... moles of NH₃. Given the options provided, none match this calculation, which indicates that there may be an error in the question.