Final answer:
When 300 grams of chlorine is the limiting reactant, it yields approximately 2.842 moles of aluminum chloride. Multiplying this by the molar mass of AlCl3 (approximately 133.33 g/mol) gives us 379 grams of aluminum chloride produced, though this result does not match any of the multiple-choice options provided.
Step-by-step explanation:
To determine the mass of aluminum chloride produced when aluminum and chlorine react, we must find out which of the two reactants is the limiting reactant. The balanced chemical equation for the reaction is:
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
Using provided information, we know that from 6.0 grams of aluminum, we can produce 0.22 moles of AlCl3, and from 3.8 grams of chlorine, we can produce 0.036 moles of AlCl3. Clearly, the chlorine is the limiting reactant and determines the amount of product formed. Since we have 300 grams of chlorine and we know that 3.8 grams yields 0.036 moles, we can set up a proportion to find the moles of AlCl3 that 300 grams of chlorine would yield.
Moles of AlCl3 from 300 grams of Cl2 = (0.036 mol AlCl3 / 3.8 g Cl2) * 300 g Cl2 = 2.842 moles of AlCl3
To find grams of AlCl3, we multiply the moles by the molar mass of AlCl3, which is approximately 133.33 g/mol:
2.842 moles * 133.33 g/mol = 379 grams of AlCl3
So, the total mass of aluminum chloride produced will be 379 grams, which does not match any of the options provided in the question, suggesting a possible discrepancy in the question or choices given.