1. Write the balanced chemical equation:

Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(l)

2. Calculate the moles of aluminum hydroxide:

* Molar mass of Al(OH)₃ = 78 g/mol

* Moles of Al(OH)₃ = (328 g) / (78 g/mol) = 4.21 mol

3. Determine the mole ratio of aluminum hydroxide to chloride ions:

* From the balanced equation, 1 mole of Al(OH)₃ produces 3 moles of Cl⁻ ions.

4. Calculate the moles of chloride ions:

* Moles of Cl⁻ = 4.21 mol Al(OH)₃ * (3 mol Cl⁻ / 1 mol Al(OH)₃) = 12.63 mol

5. Calculate the number of chloride ions:

* Avogadro's number = 6.022 x 10²³ ions/mol

* Number of Cl⁻ ions = 12.63 mol * (6.022 x 10²³ ions/mol) = 7.61 x 10²⁴ chloride ions

Therefore, 328 grams of aluminum hydroxide reacting with excess hydrochloric acid can form 7.61 x 10²⁴ chloride ions.