0.0300 mol of ethanoic acid reacts completely with excess magnesium. What volume of hydrogen forms at room conditions? (1 mol of gas occupies 24.0 dm3.)
- Equation: 2CH3COOH + Mg → (CH3COO)2Mg + H2, so 2 mol of acid give 1 mol of H2.
- Moles of H2 = 0.0300 ÷ 2 = 0.0150 mol.
- Volume = 0.0150 × 24.0 = 0.360 dm3.
Answer: 0.360 dm3 of hydrogen.