Final answer:
A total of 59.06 grams of aluminum chloride can be produced from the given amounts of aluminum and chlorine gas, with chlorine being the limiting reagent.
Step-by-step explanation:
Stoichiometry of Aluminum Chloride Formation
To determine how many grams of aluminum chloride could be produced from 22.77 grams of aluminum and 47.20 grams of chlorine gas, it is necessary to first convert these amounts into moles using their respective molar masses. The molar mass of aluminum (Al) is approximately 26.98 g/mol, and the molar mass of chlorine gas (Cl₂) is approximately 70.90 g/mol.
The balanced chemical equation for the reaction is:
2 Al (s) + 3 Cl₂ (g) → 2 AlCl₃ (s)
Next, we analyze which reactant is the limiting reagent, which will determine the maximum amount of aluminum chloride (AlCl₃) that can be produced.
From the given masses:
- 22.77 grams of Al is equivalent to 0.843 moles (22.77 g ÷ 26.98 g/mol).
- 47.20 grams of Cl₂ is equivalent to 0.665 moles (47.20 g ÷ 70.90 g/mol).
Using the stoichiometry from the balanced equation, we know that 2 moles of Al react with 3 moles of Cl₂ to form 2 moles of AlCl₃. So, we can set up the ratio:
0.843 moles Al × (2 moles AlCl₃ / 2 moles Al) = 0.843 moles of AlCl₃ can be formed from Al
0.665 moles Cl₂ × (2 moles AlCl₃ / 3 moles Cl₂) = 0.443 moles of AlCl₃ can be formed from Cl₂
The chlorine is the limiting reagent because it produces fewer moles of aluminum chloride (AlCl₃). Since 0.443 moles of AlCl₃ could be formed by the chlorine, and the molar mass of AlCl₃ is approximately 133.33 g/mol, the mass of AlCl₃ produced is:
- 0.443 moles × 133.33 g/mol = 59.06 grams of AlCl₃ (rounded to 2 decimal places)