1. Write the balanced chemical equation:

2 Al + 3 Cl₂ → 2 AlCl₃

2. Determine the limiting reactant:

* Calculate moles of aluminum:

* Moles of Al = (mass of Al) / (molar mass of Al) = 13.0 g / 26.98 g/mol = 0.482 mol

* Calculate moles of chlorine:

* Moles of Cl₂ = (mass of Cl₂) / (molar mass of Cl₂) = 18.0 g / 70.90 g/mol = 0.254 mol

* Determine the limiting reactant:

* From the balanced equation, 2 moles of Al react with 3 moles of Cl₂.

* The mole ratio of Al to Cl₂ is 2:3.

* Since we have 0.482 moles of Al, we would need 0.723 moles of Cl₂ (0.482 mol Al * (3 mol Cl₂ / 2 mol Al) = 0.723 mol Cl₂) to react completely.

* We only have 0.254 moles of Cl₂, so chlorine is the limiting reactant.

3. Calculate the mass of aluminum chloride formed:

* Use the mole ratio from the balanced equation:

* 3 moles of Cl₂ produce 2 moles of AlCl₃.

* Calculate moles of AlCl₃ formed:

* Moles of AlCl₃ = (0.254 mol Cl₂) * (2 mol AlCl₃ / 3 mol Cl₂) = 0.169 mol AlCl₃

* Calculate the mass of AlCl₃:

* Mass of AlCl₃ = (moles of AlCl₃) * (molar mass of AlCl₃) = 0.169 mol * 133.34 g/mol = 22.5 g

Therefore, the maximum mass of aluminum chloride that can be formed is 22.5 grams.