Final answer:
To produce 585 g of aluminum with 84.2% efficiency, approximately 1313 g of aluminum oxide and 232 g of carbon are required by the Hall-Heroult process.
Step-by-step explanation:
To prepare 585 g of aluminum (Al) using the Hall-Heroult process, we need to calculate the amount of aluminum oxide (Al2O3) and carbon (C) required according to the overall reaction:
2 Al2O3 (1) + 3 C(s) → 4 Al(1) + 3 CO2 (g)
First, we determine the molar mass of aluminum, which is approximately 26.98 g/mol. For 585 g of aluminum, the number of moles is:
585 g Al ÷ 26.98 g/mol = 21.7 mol Al
According to the balanced chemical reaction, 2 moles of Al2O3 produce 4 moles of Al. Therefore, for 21.7 moles of Al:
(21.7 mol Al) ÷ (4 mol Al) × (2 mol Al2O3) = 10.85 mol Al2O3
The molar mass of Al2O3 is about 101.96 g/mol, so the mass of Al2O3 needed is:
10.85 mol Al2O3 × 101.96 g/mol = 1106 g Al2O3
The chemical reaction also shows that 3 moles of carbon are needed for every 2 moles of Al2O3. Therefore, for 10.85 moles of Al2O3:
(10.85 mol Al2O3) ÷ (2 mol Al2O3) × (3 mol C) = 16.275 mol C
The molar mass of carbon is 12.01 g/mol, hence the mass of carbon needed is:
16.275 mol C × 12.01 g/mol = 195.4 g C
Considering the process has an efficiency of 84.2%, the actual amounts needed are:
Mass of Al2O3 = 1106 g ÷ 0.842 ≈ 1313 g Al2O3
Mass of C = 195.4 g ÷ 0.842 ≈ 232 g C
Thus, approximately 1313 g of aluminum oxide and 232 g of carbon are needed to produce 585 g of aluminum with 84.2% efficiency.