The reaction between concentrated sulphuric acid and magnesium can be represented by the equation given below: Mg + 2H₂SO₄ ⟶ MgSO₄ + 2H₂O + SO₂ If 60 g of magnesium is used in the reaction, calculate the following:
The mass of sulphuric acid needed for the reaction.
Topic: Mole concept, gas volumes and stoichiometry
Answer
Mg + 2H₂SO₄ ⟶ MgSO₄ + 2H₂O + SO₂
[Mg = 24, H = 1, S = 32, O = 16]
Molar masses: Mg = 24 g mol⁻¹ H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g mol⁻¹
From the equation, 1 mol Mg reacts with 2 mol H₂SO₄: 24 g of Mg reacts with 2 × 98 = 196 g of H₂SO₄
By proportion, for 60 g of Mg: mass of H₂SO₄ = (196 ÷ 24) × 60 = 8.1667 × 60 = 490 g
Mass of sulphuric acid needed = 490 g
Set the equation up as a proportion — "24 g of magnesium needs 196 g of acid, so 60 g needs how much?" — and the arithmetic takes one line.
The coefficient 2 in front of H₂SO₄ is the mark-loser. It is 2 × 98 = 196, not 98. Using 98 gives 245 g, exactly half the right answer. Whenever a formula in a mass calculation has a number in front of it in the balanced equation, that number multiplies its molar mass.