2.43 g of magnesium is added to 50.0 cm3 of 2.00 mol dm−3 HCl: Mg + 2HCl → MgCl2 + H2. What volume of hydrogen forms at STP? (Ar: Mg 24.31; molar volume 22.7 dm3 mol−1)
- Moles of Mg = 2.43 ÷ 24.31 = 0.100 mol. Moles of HCl = 2.00 × 0.0500 = 0.100 mol.
- 0.100 mol of Mg needs 0.200 mol of HCl, but only 0.100 mol is present, so HCl is limiting.
- Moles of H2 = 0.100 ÷ 2 = 0.0500 mol.
- Volume = 0.0500 × 22.7 = 1.135 dm3.
Answer: 1.14 dm3 of hydrogen (HCl is the limiting reactant)