A 10 g mixture of CaCO₃ and MgCO₃ was heated until constant mass. If the molar ratio of CaCO₃ to MgCO₃ was 1:2, how many liters of CO₂ were released at STP?
- 2.44 L
- 2.24 L
- 4.48 L
- 1.12 L
📝 Detailed explanation
Let x be the moles of CaCO₃ and 2x be the moles of MgCO₃.100x + 84(2x) = 10.
100x + 168x = 10 => 268x = 10 => x = 0.0373 mol.
Total moles CO₂ = x + 2x = 3x = 3 * 0.0373 = 0.1119 mol.
Volume CO₂ = 0.1119 * 22.4 ≈ 2.5 L.
(Option 2.44 is the closest precise value).