If one mole of aluminum chloride dissociates into one mole of aluminum and three moles of chloride ions, how many chloride ions will be produced when 5.22 moles of aluminum chloride dissociates?

The answer is: 15.66 moles of chloride ions.

Dissociation of aluminium chloride in water: AlCl₃(aq) → Al³⁺(aq) + 3Cl⁻(aq).

n(AlCl₃) = 5.22 mol; amount of aluminium chloride.

From chemical reaction: n(AlCl₃) : n(Cl⁻) = 1 : 3.

n(Cl⁻) = 3 · 5.22 mol.

n(Cl⁻) = 15.66 mol; amount of chloride ions.

N(Cl⁻) = n(Cl⁻) · Na.

N(Cl⁻) = 15.66 mol · 6.022·10²³ 1/mol.

N(Cl⁻) = 9.43·10²⁴; number of chloride ions.

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