Answer :
The correct answer is option D. 56.3 g
To solve this problem, we need to determine the limiting reactant and then calculate the maximum amount of ammonia (NH₃) that can be produced based on the given quantities of H₂ and N₂.
Given information:
- Reaction equation: N₂ + 3H₂ → 2NH₃
- Mass of H₂ = 10.0 g
- Mass of N₂ = 80.0 g
Step 1: Convert the given masses to moles.
Molar mass of H₂ = 2.016 g/mol
Moles of H₂ = 10.0 g / 2.016 g/mol = 4.96 mol
Molar mass of N₂ = 28.014 g/mol
Moles of N₂ = 80.0 g / 28.014 g/mol = 2.85 mol
Step 2: Determine the limiting reactant.
According to the balanced equation, 3 moles of H₂ react with 1 mole of N₂.
Moles of H₂ required for 2.85 mol of N₂ = 2.85 mol × 3 = 8.55 mol
Moles of N₂ required for 4.96 mol of H₂ = 4.96 mol / 3 = 1.65 mol
Since there are only 4.96 mol of H₂ available, but 8.55 mol is required, H₂ is the limiting reactant.
Step 3: Calculate the maximum amount of NH₃ that can be produced.
According to the balanced equation, 2 moles of NH₃ are produced for every 3 moles of H₂ consumed.
Moles of NH₃ produced = (4.96 mol H₂ / 3) × 2 = 3.31 mol
Step 4: Convert moles of NH₃ to grams.
Molar mass of NH₃ = 17.031 g/mol
Mass of NH₃ = 3.31 mol × 17.031 g/mol = 56.3 g
Therefore, the maximum amount of ammonia (NH₃) that can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂ is 56.3 g.
The correct answer is option D. 56.3 g
