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:

(a) The mass of sulphuric acid needed for the reaction.

(b) The volume of sulphur dioxide gas liberated at S.T.P.
[Atomic weight: Mg=24, H=1, S=32, O=16]

$\begin{matrix} \text{Mg} & + & 2\text{H}$2 \text{SO}4 & \longrightarrow & \text{MgSO}4 & + & 2\text{H}2\text{O} & + & \text{SO}_2 \ 1 \text{ mole} && 2 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \ 24 && 2[2(1) + 32 + 4(16)] & & 24 + 32 + 4(16) & & 2[2(1) + 16] && 32 + 2(16) \ = 24 \text{ g} & & = 196 \text{ g} & & = 120 \text{ g} & & 36 \text{ g} & & = 64\text{ g} \ \end{matrix}

(a) 24 g of Mg reacts with 196 g of H₂SO₄

∴ 60 g of Mg reacts with $\dfrac{196}{24}$ x 60 = 490 g of H₂SO₄

Hence, 490 g of H₂SO₄ is required.

(b) 24 g of Mg gives 22.4 L volume of SO₂

∴ 60 g of Mg gives $\dfrac{22.4}{24}$ x 60 = 56 L

Hence, the volume of sulphur dioxide gas liberated at S.T.P. is 56 L.

Thus, 490 g of sulphuric acid are needed and 56 L of SO₂ will be liberated at S.T.P.