1. **State the problem:** Calculate the maximum mass of aluminum chloride (AlCl₃) formed when 17.0 g of aluminum (Al) reacts with 22.0 g of chlorine gas (Cl₂) according to the reaction: $$2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)$$ 2. **Molar masses:** - Aluminum (Al): 26.98 g/mol - Chlorine gas (Cl₂): 70.90 g/mol (since Cl = 35.45 g/mol, Cl₂ = 2 \times 35.45) - Aluminum chloride (AlCl₃): 133.34 g/mol (Al = 26.98 + 3 \times 35.45) 3. **Calculate moles of each reactant:** $$\text{moles Al} = \frac{17.0}{26.98} = 0.630 \, \text{mol}$$ $$\text{moles Cl}_2 = \frac{22.0}{70.90} = 0.310 \, \text{mol}$$ 4. **Determine limiting reactant using stoichiometry:** From the balanced equation, 2 mol Al reacts with 3 mol Cl₂. Calculate moles of Al needed to react with 0.310 mol Cl₂: $$\text{moles Al needed} = 0.310 \times \frac{2}{3} = 0.207 \, \text{mol}$$ Since we have 0.630 mol Al available, which is more than 0.207 mol needed, Cl₂ is the limiting reactant. 5. **Calculate moles of AlCl₃ produced:** From the balanced equation, 3 mol Cl₂ produce 2 mol AlCl₃. $$\text{moles AlCl}_3 = 0.310 \times \frac{2}{3} = 0.207 \, \text{mol}$$ 6. **Calculate mass of AlCl₃ produced:** $$\text{mass AlCl}_3 = 0.207 \times 133.34 = 27.6 \, \text{g}$$ **Final answer:** The maximum mass of aluminum chloride formed is **27.6 g**.