Given the following balanced chemical equation: N2(g) + 3H2(g) + 2NH3(8)
What is the maximum amount of NH3(g) that can be produced from 2.0 mol H2(g)? Assume that N2(g) is the excess reactant.

22 g

Explanation:

Step 1: Write the balanced equation

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

Step 2: Calculate the moles of NH₃ produced from 2.0 moles of H₂

The molar ratio of H₂ to NH₃ is 3:2.

2.0 mol H₂ × 2 mol NH₃/3 mol H₂ = 1.3 mol NH₃

Step 3: Calculate the mass corresponding to 1.3 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

1.3 mol × 17.03 g/mol = 22 g

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