Aluminum Oxide Reduction to Aluminum: A Detailed Analysis and Yield Calculation

In the production of aluminum, one crucial process involves the reduction of aluminum oxide (Al2O3) to metallic aluminum (Al) through the application of electric current. This process is instrumental in the aluminum industry and must be understood in terms of its chemical principles and practical applications. In this article, we will explore a detailed calculation of how much aluminum is obtained from a given amount of aluminum oxide, with a yield of 97.5%.

Chemical Reaction and Principles

The chemical reaction involved in the reduction of aluminum oxide to aluminum can be represented as follows:

2 Al2O3 (aluminum oxide) 3 C (carbon) → 4 Al (aluminum) 3 CO2 (carbon dioxide)

This reaction is known as the Hall-Héroult process and is widely used in the industrial production of aluminum. The yield of this process is typically 97.5%, meaning that 97.5% of the aluminum oxide input is converted to metallic aluminum.

Calculation of Aluminum Obtained from Aluminum Oxide

To determine the amount of aluminum produced, we need to use stoichiometry and the given yield. Here is a detailed step-by-step calculation:

Step 1: Define the Given Values and Reagents

Given:

Amount of aluminum oxide (Al2O3) 80,000 kg Molar mass of Al2O3 102 g/mol Moles of aluminum in 102 g of Al2O3 54 g Moles of Al2O3 80000000 g / 102 g/mol Percentage yield 97.5%

Step 2: Convert Kilograms to Grams

To simplify the calculation, we convert kilograms to grams:

80,000 kg 80,000,000 g

Step 3: Calculate Moles of Aluminum Oxide

Moles of Al2O3 80,000,000 g / 102 g/mol 784,313.73 moles

Step 4: Apply the Mole Ratio from the Balanced Equation

From the balanced equation, the mole ratio of Al2O3 to Al is 2:4, meaning 2 moles of Al2O3 produce 4 moles of Al.

Moles of Al produced 784,313.73 moles × 4 / 2 1,568,627.46 moles

Step 5: Calculate the Mass of Aluminum

Moles of Al 1,568,627.46 moles

Molecular weight of Al 27 g/mol

Mass of Al obtained 1,568,627.46 moles × 27 g/mol × 97.5% 42,352,941 g

Converting grams to kilograms:

Mass of Al obtained 42,352,941 g 42,352.941 kg 4.24 × 104 kg

Conclusion

The detailed process of reduction of aluminum oxide to aluminum involves complex stoichiometric calculations. By following the steps outlined above, we can accurately predict the amount of aluminum produced from a given amount of aluminum oxide. The industrial yield of 97.5% ensures efficient and effective production, making it a cornerstone of the aluminum industry.