When 2.50 mol of Al and 3.00 mol of Cl2 combine in the reaction 2Al(s) + 3Cl2(g) → 2AlCl3(s), how many moles of the excess reactant remain at the end of the reaction?

• A. 1.00 mol of Al
• B. 0.50 mol of Cl2
• C. 1.00 mol of Cl2
• D. 0.50 mol of Al

Answer: (D)

Verified step by step guidance

1. Identify the balanced chemical equation: 2Al(s) + 3Cl2(g) → 2AlCl3(s). This equation tells us that 2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3.
2. Determine the limiting reactant by comparing the mole ratio of the reactants provided to the mole ratio in the balanced equation. Start by calculating the mole ratio of Al to Cl2 from the given amounts: 2.50 mol Al and 3.00 mol Cl2.
3. Calculate the theoretical amount of Cl2 needed to react with 2.50 mol of Al using the stoichiometric ratio from the balanced equation: (3 mol Cl2 / 2 mol Al) * 2.50 mol Al = 3.75 mol Cl2. Since only 3.00 mol of Cl2 is available, Cl2 is the limiting reactant.
4. Calculate the amount of Al that reacts with the available Cl2: (2 mol Al / 3 mol Cl2) * 3.00 mol Cl2 = 2.00 mol Al. This is the amount of Al that will react completely with the available Cl2.
5. Determine the amount of excess Al remaining: Subtract the amount of Al that reacted (2.00 mol) from the initial amount (2.50 mol): 2.50 mol Al - 2.00 mol Al = 0.50 mol Al. Therefore, 0.50 mol of Al remains as the excess reactant.