Final answer:
The limiting reactant is Cl2, and the mass of AlCl3 produced when 35.0 g of Al reacts with 28.0 g of Cl2 is 35.1 g.
Step-by-step explanation:
In order to determine the limiting reactant and the mass of AlCl3 produced when 35.0 g of Al reacts with 28.0 g of Cl2, the first step is to convert the mass of each reactant into moles using their molar masses.
Aluminum (Al) has a molar mass of ~26.98 g/mol, and chlorine gas (Cl2) has a molar mass of ~70.90 g/mol.
We can then use stoichiometry based on the balanced equation to see which reactant gives the fewer moles of the product, AlCl3.
The reaction given is 2Al(s) + 3Cl2(g) → 2AlCl3(s).
- For Al: 35.0 g Al × (1 mol Al / 26.98 g Al) = 1.297 moles of Al
- For Cl2: 28.0 g Cl2 × (1 mol Cl2 / 70.90 g Cl2) = 0.395 moles of Cl2
Using the stoichiometry from the given balanced equation, 1.297 moles of Al would require 1.297 × (3/2) = 1.9455 moles of Cl2 to fully react.
But, we only have 0.395 moles of Cl2, so Cl2 is the limiting reactant.
From the limiting reactant (Cl2), we calculate the amount of AlCl3 formed:
0.395 moles of Cl2 × (2 moles AlCl3 / 3 moles Cl2) = 0.2633 moles of AlCl3.
Finally, we convert moles of AlCl3 to grams:
0.2633 moles of AlCl3 × (133.33 g AlCl3 / 1 mol AlCl3) = 35.1 g of AlCl3.
Therefore, the limiting reactant is Cl2, and the mass of AlCl3 produced is 35.1 g.