Final answer:
Form 80.1 g of AlCl₃, the calculated mass of Cl₂ needed is 63.96 g. The closest answer choice to this value is 60.1 g, which indicates that 60.1 g of Cl₂ are needed for the reaction.
Step-by-step explanation:
To find out how many grams of Cl₂ are needed to form 80.1 g of AlCl₃, we first need to establish the molar mass of AlCl₃ which is 133.33 g/mol. Next, we calculate the number of moles of AlCl₃ formed by dividing 80.1 g by the molar mass of AlCl₃:
80.1 g AlCl₃ × (1 mol AlCl₃ / 133.33 g AlCl₃) = 0.601 moles of AlCl₃
Now, using the stoichiometry of the balanced chemical equation 2Al(s) + 3Cl₂(g) → 2AlCl₃(s), we see that there is a 3:2 mole ratio between Cl₂ and AlCl₃. To find the moles of Cl₂ needed, we multiply the moles of AlCl₃ by the ratio:
0.601 moles of AlCl₃ × (3 moles Cl₂ / 2 moles AlCl₃) = 0.9015 moles of Cl₂
Lastly, to find the mass of Cl₂, we multiply the moles of Cl₂ by the molar mass of Cl₂:
0.9015 moles of Cl₂ × (70.90 g/mol) = 63.96 g of Cl₂
From the choices provided, the closest amount to the calculated mass of Cl₂ needed is 60.1 g.