Final answer:
To produce 11.48 moles of aluminum from aluminum oxide, we calculate that 5.74 moles of Al2O3 are needed, which corresponds to 585.26 grams. The options provided by the student do not include the correct answer.
Step-by-step explanation:
To calculate how many grams of aluminum oxide (Al2O3) are needed to produce 11.48 moles of aluminum (Al), we need to consider the stoichiometry of the reaction provided. The reaction Al2O3 → Al + CO is not balanced as written, and Al cannot react directly with CO to produce Al2O3; however, we will proceed by assuming the decomposition of Al2O3 into Al, ignoring CO under the student's guidance. The balanced equation for the decomposition of aluminum oxide to produce aluminum is Al2O3 ↓ 2 Al.
First, we calculate the molar mass of Al2O3:
(2 × 26.98 g/mole Al) + (3 × 16.00 g/mole O) = 101.96 g/mole Al2O3.
Since the equation tells us that 1 mole of Al2O3 produces 2 moles of Al, we can set up the following ratio:
1 mole Al2O3 / 2 moles Al = X moles Al2O3 / 11.48 moles Al.
Solving for X gives us X = 11.48 moles Al / 2 = 5.74 moles Al2O3.
Now, we multiply the moles of Al2O3 by its molar mass to find the mass needed:
5.74 moles Al2O3 × 101.96 g/mole Al2O3 = 585.26 g.
The student's given options do not include the correct answer; thus, there may have been an error in the question or choices provided.