Consider the reaction 4 HCl(g) + O2(g) → 2H2O(l) + 2Cl2(g) . When 63.1 g of HCl react with 17.2 g of O2, 49.3 g of Cl2 are collected.?

Molar mass of HCl = (1.0 + 35.5) g/mol = 36.5 g/mol
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Molar mass of Cl₂ = 35.5 × 2 g/mol = 71.0 g/mol
(The molar masses learned in the lessons should be used instead.)

Initial number of moles of HCl = (63.1 g) / (36.5 g/mol) = 1.729 mol
Initial number of moles of O₂ = (17.2 g) / (32.0 g/mol) = 0.5375 mol

4HCl(g) + O₂(g) → 2H₂O(l) + 2Cl₂(g)
Mole ratio HCl : O₂ : Cl₂ = 4 : 1 : 2

If 1.729 mol HCl completely reacts, O₂ needed = (1.729 mol) × (1/4) = 0.4323 mol < 0.5375 mol
Hence, O₂ is in excess, and HCl completely reacts (the limiting reactant/reagent).

No. of moles of HCl reacted = 1.729 mol
Maximum of moles of Cl₂ formed = (1.729 mol) × (1/2) = 0.8645 mol
Theoretic yield of Cl₂ = (0.8645 mol) × (71.0 g/mol) = 61.4 g
Percent yield of Cl₂ = (49.3/61.4) × 100% = 80.3%