Question

The reaction between concentrated sulphuric acid and magnesium can be represented by the equation given below:

\[\ce{Mg + 2H2SO4 -> MgSO4 + 2H2O + SO2}\]

If 60 g of magnesium is used in the reaction, calculate the following:

1. The mass of sulphuric acid needed for the reaction.
2. The volume of sulphur dioxide gas liberated at S.T.P.
   [Atomic weight: Mg = 24, H = 1, S = 32, O = 16]

Answer 1

\[\ce{Mg + 2H2SO4 -> MgSO4 + 2H2O + SO2}\]

a. Gram molecular mass of sulphuric acid:

= 2 × 1 + 1 × 32 + 4 × 16

= 2 + 32 + 64

= 98 g

Moles of Mg produced = `60/24`

= 2.5 moles

Now, 1 mole of Mg requires 2 moles of H₂SO₄.

Therefore, 2.5 moles of Mg will require = 2 × 2.5

= 5 moles of H₂SO₄

Now, mass of H₂SO₄ required = 5 × 98

= 490 g

b. 24 g of magnesium liberates 1 mole of sulphur dioxide at S.T.P.

Therefore, 1 g of magnesium liberates `22.4/24` L of sulphur dioxide at S.T.P.

60 g of magnesium liberates `22.4/24 xx 60 L` = 56 L of sulphur dioxide at S.T.P.

∴ The volume of sulphur dioxide gas liberated at S.T.P. is 56 L.