A 10 g mixture of magnesium and magnesium carbonate was treated with excess hydrochloric acid. The volume of gas released was 5.6 L at STP. Determine the mass of magnesium in the initial mixture.
- 4 g
- 6 g
- 2.4 g
- 7.6 g
📝 Detailed explanation
Mg + 2HCl → MgCl₂ + H₂MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
Total moles of gas = 5.6 / 22.4 = 0.25 mol.
Let x = mass Mg, y = mass MgCO₃.
x + y = 10.
(x / 24) + (y / 84) = 0.25.
Substitute y = 10 - x: (x / 24) + (10 - x) / 84 = 0.25.
Multiply by 168: 7x + 2(10 - x) = 42.
7x + 20 - 2x = 42.
5x = 22, so x = 4.4 g.
Closest option is 4 g.