Final answer:
Nitrogen is the limiting reactant, and the mass of ammonia produced is 4.1432 g, calculated by using the stoichiometry of the balanced chemical equation for the reaction.
Step-by-step explanation:
To determine the limiting reactant and the mass of ammonia produced, let's refer to the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia:
3H₂(g) + N₂(g) → 2NH₃(g)
This equation tells us that 3 moles of hydrogen reacts with 1 mole of nitrogen to produce 2 moles of ammonia. First, we need to convert the given masses of nitrogen and hydrogen to moles using their molecular weights (Molar mass of N₂ = 28.02 g/mol and H₂ = 2.016 g/mol):
- Moles of N₂ = 3.41 g / 28.02 g/mol = 0.1217 moles
- Moles of H₂ = 2.79 g / 2.016 g/mol = 1.384 moles
Using the stoichiometry of the balanced equation:
- For N₂ - 0.1217 moles N₂ × (3 moles H₂ / 1 mole N₂) = 0.3651 moles H₂ needed
- For H₂ - 1.384 moles H₂ × (1 mole N₂ / 3 moles H₂) = 0.4613 moles N₂ needed
Since we actually have more moles of hydrogen (1.384 moles) than the 0.3651 moles needed to react with all of the 0.1217 moles of nitrogen, we can see that nitrogen is the limiting reactant. Next, we calculate the amount of ammonia that can be produced from the 0.1217 moles of nitrogen:
- 0.1217 moles N₂ × (2 moles NH₃ / 1 mole N₂) = 0.2434 moles NH₃
- Mass of NH₃ = 0.2434 moles × 17.031 g/mol (molar mass of NH₃) = 4.1432 g
Therefore, nitrogen is the limiting reactant, and 4.1432 g of ammonia is produced.