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Limiting reactants and excess explained

Syllabus VWO: Quantities and relationships

Compare amounts with the reaction ratio and calculate product and excess.

Compare on the same basis

For N₂ + 3 H₂ → 2 NH₃, three times as much H₂ as N₂ is needed, measured in mol. Divide each available amount by its coefficient. The smallest result determines the reaction extent.

Use only what reacts

Multiply the extent by each coefficient. Subtract the amount used from the amount available. Here we assume complete consumption of the limiting reactant; equilibrium is not modelled.

Worked example

1.00 mol N₂ and 2.40 mol H₂ are available.

  1. Compare 1.00 / 1 = 1.00 with 2.40 / 3 = 0.800.
  2. H₂ limits the reaction. 2 × 0.800 = 1.60 mol NH₃ forms.

1.00 − 0.800 = 0.20 mol N₂ remains.