During the thermal decomposition of 50 grams of an impure calcium carbonate sample, 8.96 liters of carbon dioxide (at STP) was collected. What is the percentage purity of the sample?
- 40%
- 80%
- 90%
- 100%
📝 Detailed explanation
CaCO₃ → CaO + CO₂.Moles of CO₂ = 8.96 / 22.4 = 0.4 mol.
Moles of pure CaCO₃ = 0.4 mol.
Mass of pure CaCO₃ = 0.4 × 100 = 40 g.
Purity = (40 / 50) × 100 = 80%.