what is the maximum mass of ammonia that can be produced from a mixture of 1.20x 10^3g N2 and 5.60x 10^2g H2?

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

Initial moles of N₂ = (1.20 × 10³ g) / (28.0 g/mol) = 42.86 mol
Initial moles of H₂ = (5.60 × 10² g) / (2.0 g/mol) = 280.0 mol

Balanced equation for the reaction:
N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ = 1 : 3
If N₂ completely reacts, H₂ needed = (42.86 mol) × 3 = 128.6 mol < 280.0 mol
Hence, H₂ is in excess, and N₂ is the limiting reactant.

According to the above equation, mole ratio N₂ : NH₃ = 1 : 2
Moles of N₂ reacted = 42.86 mol
Maximum moles of NH₃ produced = (42.86 mol) × 2 = 85.72 mol
Maximum mass of NH₃ produced = (85.72 mol) × (17.0 g/mol) = 1.46 × 10³ g

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OR:

If N₂ is the limiting reactant:
(1.20 × 10³ g N₂) × (1 mol N₂ / 28.0 g N₂) × (Max. 2 mol NH₃ / 1 mol N₂) × (17.0 g NH₃ / 1 mol N₂)
= Max. 1.46 × 10³ g NH₃

If H₂ is the limiting reactant:
(5.60 × 10² g H₂) × (1 mol H₂ / 2.0 g H₂) × (Max. 2 mol NH₃ / 3 mol H₂) × (17.0 g NH₃ / 1 mol N₂)
= Max. 3.17 × 10³ g NH₃

Since N₂ produced the smaller yield of NH₃, N₂ is the limiting reactant.
Hence, maximum mass of NH₃ produced is 1.46 × 10³ g.