Limiting and Excess Reactants

An excess reactant is supplied in greater quantity than required and remains after the reaction.

ZnS ≈ 1 mol and O₂ = 1.5 mol. Stoichiometric mixture. ZnO = 1 mol = 81 g.

N₂ = 0.5 mol. H₂ required = 1.5 mol but available = 4 mol. Excess H₂ = 2.5 mol = 5 g.

The limiting reactant determines the maximum yield regardless of excess reactant.

O₂ is limiting. H₂O formed = (2/3) × 2 = 1.33 mol.

Stoichiometric reaction produces 4 mol CO₂ = 89.6 dm³.

0.1 mol NaOH is insufficient to react completely with 0.1 mol H₂SO₄.

O₂ is limiting. One mole NO (30 g) is formed.

The excess reactant remains after the limiting reactant is exhausted.

O₂ is limiting. Four moles CO₂ are formed and 1 mol CO remains.

The reactants are in the exact stoichiometric ratio. Four moles NH₃ = 68 g are produced.

Moles of N₂ = 56/28 = 2 mol. Moles of H₂ = 12/2 = 6 mol. The mixture is stoichiometric. NH₃ formed = 4 mol. Mass = 4 × 17 = 68 g.

5 mol Al requires 3.75 mol O₂. Since 4 mol O₂ are available, oxygen is in excess. Aluminium is limiting.

Al = 0.5 mol, S = 0.5 mol. Sulphur is limiting. Al₂S₃ formed = 0.5/3 = 0.1667 mol. Mass = 0.1667 × 150 = 25 g.

Zn = 0.1 mol, HCl = 0.2 mol. Stoichiometric mixture. H₂ = 0.1 mol. Volume = 0.1 × 22.4 = 2.24 dm³.

The limiting reactant determines the maximum amount of product that can be formed.

Fe = 2 mol, H₂O = 2 mol. Water is limiting. Fe₃O₄ = 0.5 mol. Mass = 0.5 × 232 = 116 g.

P₄ = 0.5 mol and Cl₂ = 3 mol. Required Cl₂ = 3 mol. Both react completely.

Stoichiometric mixture produces 3 mol S = 96 g.