What is the theoretical yield in grams for this reaction under the given conditions?

Method 1:

Molar mass of H₂ = 1.0×2 g/mol = 2.0 g/mol
Molar mass of N₂ = 14.0×2 g/mol = 28.0 g/mol
Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol

Initial moles of H₂ = (1.49 g) / (2.0 g/mol) = 0.745 mol
Initial moles of N₂ = (9.64 g) / (17.0 g/mol) = 0.567 mol

Balanced equation for the reaction:
N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ = 1 : 3
If H₂ completely reacts, N₂ needed = (0.745 mol) × 1/3 = 0.248 mol < 0.567 mol
Hence, N₂ is in excess. The limiting reactant (limiting reagent) is H₂

According to the above equation, mole ratio H₂ : NH₃ = 3 : 2
Maximum moles of NH₃ produced = (0.745 mol) × 2/3 = 0.4967 mol
Theoretical yield of NH₃ = (0.4967 mol) × (17.0 g/mol) = 8.44 g

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Method 2:

If H₂ is the limiting reactant:
(1.49 g H₂) × (1 mol H₂ / 2.0 g H₂) × (2 mol NH₃ / 3 mol H₂) × (17.0 g NH₃ / 1 mol NH₃)

= 8.44 g NH₃

If N₂ is the limiting reactant:
(9.64 g N₂) × (1 mol N₂ / 28.0 g N₂) × (2 mol NH₃ / 1 mol N₂) × (17.0 g NH₃ / 1 mol NH₃)

= 11.7 g NH₃ > 8.44 g NH₃

Hence, H₂ is the limiting reactant, and the theoretical yield is 8.44 g.