Final answer:
24.0 grams of aluminum will yield 0.889 moles of aluminum chloride, assuming there is excess chlorine, according to the stoichiometric ratio of the reaction.
Step-by-step explanation:
The question asks how many moles of aluminum chloride could be produced from 24.0 g of aluminum assuming there is excess chlorine. To solve this, we first need to convert the mass of aluminum to moles using aluminum's molar mass (26.98 g/mol). The procedure involves using the stoichiometry of the reaction between aluminum and chlorine gas, which is: 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s). This tells us that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.
So, for 24.0 g of aluminum, the moles of aluminum are calculated as:
(24.0 g Al) / (26.98 g/mol Al) = 0.889 moles of Al
Since we have excess chlorine, the amount of aluminum chloride produced will be based on the aluminum. According to the stoichiometry, 2 moles of Al yield 2 moles of AlCl3, thus:
(0.889 moles Al) × (1 mol AlCl3 / 1 mol Al) = 0.889 moles of AlCl3