Calculate Mass of KClO₃ to Liberate 6.72 L of O₂

Question: Calculate the mass of potassium chlorate (KClO₃) required to liberate 6.72 litres of O₂ at STP.

Given: • Volume of O₂ to be liberated = 6.72 L (at STP) • Molar volume at STP = 22.4 L/mol

Chemical equation: 2KClO₃ → 2KCl + 3O₂

Step 1: Find moles of O₂ Moles of O₂ = Volume at STP / Molar volume = 6.72 / 22.4 = 0.3 mol of O₂

Step 2: Find moles of KClO₃ needed From the balanced equation: 2 mol KClO₃ produces 3 mol O₂ Therefore: mol KClO₃ = (2/3) × mol O₂ = (2/3) × 0.3 = 0.2 mol of KClO₃

Step 3: Find mass of KClO₃ Molar mass of KClO₃: K = 39, Cl = 35.5, O₃ = 48 Molar mass = 39 + 35.5 + 48 = 122.5 g/mol

Mass = Moles × Molar mass = 0.2 × 122.5 = 24.5 g

Answer: 24.5 g of KClO₃ is required to liberate 6.72 L of O₂ at STP.

Decomposition of Potassium Chlorate: 2KClO₃ →(MnO₂ catalyst, heat)→ 2KCl + 3O₂

Conditions: • Heat required (high temperature) • MnO₂ (manganese dioxide) acts as a catalyst • MnO₂ is not consumed in the reaction

Verification: • 0.2 mol KClO₃ decomposes to produce (3/2) × 0.2 = 0.3 mol O₂ • At STP, 0.3 mol O₂ = 0.3 × 22.4 = 6.72 L ✓

Molar mass calculation check: KClO₃: • K: 39 g/mol • Cl: 35.5 g/mol • O: 3 × 16 = 48 g/mol • Total: 39 + 35.5 + 48 = 122.5 g/mol ✓

General formula for this type of problem: Mass of KClO₃ = (Volume of O₂/22.4) × (2/3) × 122.5

Variation 1: Mass of KClO₃ for 3.36 L O₂ Mol O₂ = 3.36/22.4 = 0.15 mol Mol KClO₃ = (2/3) × 0.15 = 0.10 mol Mass = 0.10 × 122.5 = 12.25 g

Variation 2: Mass of KClO₃ for 1 mol O₂ Mol KClO₃ = (2/3) × 1 = 0.667 mol Mass = 0.667 × 122.5 = 81.67 g

Variation 3: What mass of KCl is produced along with 6.72 L O₂? Mol KCl = mol KClO₃ = 0.2 mol (2:2 ratio) Molar mass KCl = 39 + 35.5 = 74.5 g/mol Mass KCl = 0.2 × 74.5 = 14.9 g

Key mole concept formulas: • Moles = Mass / Molar mass • Moles of gas = Volume at STP / 22.4 • At STP: T = 273 K (0°C), P = 1 atm (101.325 kPa) • 1 mole of any gas at STP = 22.4 L