Blue vitriol (CuSO₄.5H₂O) is often added to swimming pools to kill algae. It is prepared by the reaction between copper metal and hot sulphuric acid to give CuSO₄ (aq) and SO₂(g). If one mole of copper is reacted with one mole of sulphuric acid, then the molecules of SO₂(g) obtained will be

$$\begin{gathered} \mathrm{Cu}+2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)⟶{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2\left(\mathrm{g}\right) \\ 1 \mathrm{mole} 2 \mathrm{mole} 1 \mathrm{mole} \end{gathered}$$

From the above equation we find that 1 mole of Cu reacts with 2 mole of H₂SO₄. Since we have only 1 mole of H₂SO₄ so it is the limiting reagent.

Now Since 2 moles of H₂SO₄ react with Cu to produce SO₂ = 1 mole

$\therefore$ 1 mole of H₂SO₄ reacts with Cu to produce SO₂ $=\frac{1}{2}\text{ mole }$

Now one mole of SO₂ = 6.023 × 10²³ molecules of SO₂ 

$\therefore \frac{1}{2}\text{ mole of }{\mathrm{SO}}_2=\frac{6.023\times {10}^{23}}{2}\text{ molecules of }{\mathrm{SO}}_2$

= 3 × 10²³ molecules of SO₂