Final answer:
The theoretical yield of AlCl₃ using 2.5 g Al and 2.5 g Cl₂ is determined by identifying Cl₂ as the limiting reactant. Applying stoichiometry and molar masses, we find that the yield is closest to 13.0 g (option A).
Step-by-step explanation:
To calculate the theoretical yield of AlCl₃ in grams from the initial amounts of reactants, we first determine which reactant is the limiting reactant. The balanced chemical equation 2 Al(s) + 3 Cl₂ (g) → 2 AlCl₃ (s) shows that 2 moles of Al will react with 3 moles of Cl₂ to produce 2 moles of AlCl₃.
The molar mass of Al is approximately 26.98 g/mol, which means 2.5 g Al is equal to 2.5 g / 26.98 g/mol ≈ 0.0927 moles of Al. The molar mass of Cl₂ is approximately 70.90 g/mol, which means 2.5 g Cl₂ is equal to 2.5 g / 70.90 g/mol ≈ 0.0353 moles of Cl₂. Since the stoichiometry of the reaction is 2 mol Al : 3 mol Cl₂, we can see that Cl₂ is the limiting reactant because we have less than the required 0.1386 moles of Cl₂ (0.0927 mol Al * 3/2).
Using the limiting reactant, the theoretical yield of AlCl₃ can be found by multiplying the moles of Cl₂ by the molar mass of AlCl₃ (133.33 g/mol) and using the ratio from the balanced equation: 0.0353 moles Cl₂ * 2 moles AlCl₃ / 3 moles Cl₂ * 133.33 g/mol AlCl₃ ≈ 15.7 g AlCl₃.
The closest answer to 15.7 g is 13.0g, so the theoretical yield of AlCl₃ is 13.0g (option A).