Final answer:
0.110 mol of aluminum reacting with an excess of chlorine gas will produce 0.110 mol of aluminum chloride according to the balanced equation, given the one-to-one stoichiometry of aluminum to aluminum chloride.
Step-by-step explanation:
When 0.110 mol of aluminum reacts with an excess of chlorine gas, Cl2, the reaction to form aluminum chloride (AlCl3) is expressed by the balanced chemical equation 2Al + 3Cl2 → 2AlCl3. According to the stoichiometry of the reaction, 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. Therefore, the amount of aluminum chloride produced is directly proportional to the amount of aluminum used when chlorine is in excess.
Given that there is an excess of chlorine, the amount of aluminum present will determine the amount of aluminum chloride produced. Since the stoichiometry is one-to-one for aluminum to aluminum chloride, 0.110 mol of aluminum will yield 0.110 mol of aluminum chloride (AlCl3).