Final answer:
Using stoichiometry based on the balanced chemical equation for the reaction of aluminum with chlorine gas, 33.0 g of Cl2 with excess aluminum will produce 0.31 moles of aluminum chloride (AlCl3).
Step-by-step explanation:
To determine how many moles of aluminum chloride (AlCl3) can be produced from 33.0 g of chlorine gas (Cl2) with excess aluminum, we use stoichiometry based on the balanced chemical equation:
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
First, we need to calculate the number of moles of Cl2 present in 33.0 g. The molar mass of Cl2 is 70.9 g/mol (35.45 g/mol for Cl, times 2 because Cl2 is diatomic):
moles Cl2 = 33.0 g Cl2 / 70.9 g/mol = 0.465 moles Cl2
From the stoichiometry of the equation, 3 moles of Cl2 produce 2 moles of AlCl3. Therefore, we can set up the following stoichiometric calculation:
moles AlCl3 = (0.465 moles Cl2) × (2 moles AlCl3 / 3 moles Cl2)
This simplifies to:
moles AlCl3 = 0.31 moles AlCl3
Assuming we have excess aluminum, the reaction will produce 0.31 moles of aluminum chloride.