Final answer:
Using the stoichiometry of the balanced chemical equation and the molar masses of the reactants, we determine that chlorine is the limiting reactant. Calculations show that 0.601 moles of chlorine can produce 53.5 g of aluminum chloride.
Step-by-step explanation:
To determine the mass of aluminum chloride (AlCl3) formed from the reaction of aluminum (Al) with chlorine gas (Cl2), we must first identify the limiting reactant. We do this by comparing the moles of each reactant to their stoichiometric coefficients in the balanced chemical equation.
The balanced equation for the reaction of aluminum with chlorine gas to produce aluminum chloride is:
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
We need to calculate the number of moles of each reactant:
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl2 = 70.90 g/mol
- Number of moles of Al = 13.5 g / 26.98 g/mol = 0.500 moles
- Number of moles of Cl2 = 42.6 g / 70.90 g/mol = 0.601 moles
From the balanced equation, we see that 2 moles of Al reacts with 3 moles of Cl2, meaning that 1 mole of Al would require 1.5 moles of Cl2. Therefore:
- Aluminum requires 0.500 moles × 1.5 = 0.750 moles of Cl2
- We have 0.601 moles of Cl2, which is less than 0.750 moles
So, chlorine is the limiting reactant, and we use its moles to find the mass of AlCl3 that can be formed:
- From the balanced equation, 3 moles of Cl2 reacts to form 2 moles of AlCl3
- Therefore, 0.601 moles of Cl2 would produce (0.601 moles Cl2 / 3 moles Cl2 × 2 moles AlCl3) = 0.401 moles of AlCl3
- The molar mass of AlCl3 = 133.33 g/mol
- The mass of AlCl3 formed = 0.401 moles × 133.33 g/mol = 53.5 g