5.3 g of sodium carbonate reacts with excess ethanoic acid: 2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2. What volume of carbon dioxide forms at room temperature and pressure? (C 12, O 16, Na 23; molar gas volume 24 dm3)
- Mr of Na2CO3 = (2 × 23) + 12 + (3 × 16) = 106.
- Moles of Na2CO3 = 5.3 ÷ 106 = 0.050 mol.
- 1 mol of Na2CO3 gives 1 mol of CO2, so moles of CO2 = 0.050 mol.
- Volume of CO2 = 0.050 × 24 = 1.2 dm3.
Answer: 1.2 dm3 of carbon dioxide