To liberate 6.72 litres of oxygen (O₂) at STP, 24.5 g of potassium chlorate (KClO₃) is required. This is calculated using the decomposition reaction of potassium chlorate: 2KClO₃ → 2KCl + 3O₂. At STP, 1 mole of any gas occupies 22.4 litres. The calculation uses stoichiometry and the mole concept.
Reaction: 2KClO₃ → 2KCl + 3O₂ (with MnO₂ catalyst and heat).
Moles of O₂ = 6.72/22.4 = 0.3 mol.
Moles of KClO₃ = (2/3) × 0.3 = 0.2 mol.
Molar mass of KClO₃ = 39 + 35.5 + 48 = 122.5 g/mol.
Mass of KClO₃ = 0.2 × 122.5 = 24.5 g.
Answer: 24.5 g of KClO₃ required to liberate 6.72 L O₂ at STP.
At STP: 1 mole of gas = 22.4 litres.
MnO₂ acts as a catalyst — not consumed in the reaction.
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
Reaction: 2KClO₃ → 2KCl + 3O₂. Moles of O₂ = 6.72/22.4 = 0.3 mol. From stoichiometry: moles KClO₃ = (2/3) × 0.3 = 0.2 mol. Molar mass of KClO₃ = 122.5 g/mol. Mass = 0.2 × 122.5 = 24.5 g. Answer: 24.5 g of KClO₃ is needed.
2KClO₃ → 2KCl + 3O₂ (catalyst: MnO₂, condition: heat). Potassium chlorate decomposes to potassium chloride and oxygen gas. This reaction is used in laboratories to produce oxygen. MnO₂ acts as a catalyst and is not consumed.
Molar mass of KClO₃ = K + Cl + 3O = 39 + 35.5 + (3×16) = 39 + 35.5 + 48 = 122.5 g/mol.
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