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: [Atomic weight: Mg=24, H=1, S=32, O=16]
The volume of sulphur dioxide gas liberated at S.T.P.
Topic: Mole concept, gas volumes and stoichiometry
Answer
Mg + 2H₂SO₄ ⟶ MgSO₄ + 2H₂O + SO₂
From the equation, 1 mol Mg gives 1 mol SO₂.
Moles of Mg = 60 ÷ 24 = 2.5 mol Moles of SO₂ = 2.5 mol (ratio 1 : 1)
Volume at STP = moles × 22.4 = 2.5 × 22.4 = 56 litres
Volume of sulphur dioxide at STP = 56 litres
The magnesium to sulphur dioxide ratio is 1 : 1, so the moles carry straight across and only the final conversion to volume is left.
Do not go via the acid. The 490 g from part (a) is not needed here and dragging it in only creates room for error — go back to the magnesium, which is the substance the question actually gives you. And convert moles to litres with 22.4, not to grams with 64; the question asks for a volume.