1. Write the balanced chemical equation:

2 Al + 3 Cl₂ → 2 AlCl₃

2. Calculate the moles of aluminum:

* Molar mass of aluminum (Al) = 26.98 g/mol

* Moles of Al = (34.0 g) / (26.98 g/mol) = 1.26 mol

3. Determine the limiting reactant:

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

* To react completely with 1.26 moles of Al, you would need (1.26 mol Al) * (3 mol Cl₂ / 2 mol Al) = 1.89 mol Cl₂.

* Since you only have 39.0 g of Cl₂, let's calculate how many moles that is:

* Molar mass of chlorine gas (Cl₂) = 70.90 g/mol

* Moles of Cl₂ = (39.0 g) / (70.90 g/mol) = 0.55 mol Cl₂

* Chlorine gas (Cl₂) is the limiting reactant because you have less of it than you need to react with all of the aluminum.

4. Calculate the moles of aluminum chloride (AlCl₃) produced:

* The balanced equation shows that 3 moles of Cl₂ produce 2 moles of AlCl₃.

* Moles of AlCl₃ = (0.55 mol Cl₂) * (2 mol AlCl₃ / 3 mol Cl₂) = 0.37 mol AlCl₃

Therefore, you could produce 0.37 moles of aluminum chloride (AlCl₃) from the given amounts of aluminum and chlorine gas.