Practice Questions

Na = 2 mol, H₂O = 3 mol. Na is limiting. H₂ = 1 mol = 2 g.

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

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

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

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

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

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

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.

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

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.