How Many Grams of Aluminum Oxide Are Produced When Aluminum Reacts with Oxygen?

In chemical reactions, understanding the production of a desired product, such as aluminum oxide, is crucial. This article will guide you through a detailed step-by-step process to determine the grams of aluminum oxide (Al2O3) produced from a given amount of aluminum and oxygen. Let's begin with the balanced equation and the methodology to solve this problem.

Step 1: Writing the Balanced Equation

The balanced equation for the reaction between aluminum (Al) and oxygen (O2) to form aluminum oxide (Al2O3) is:

4Al 3O2 → 2Al2O3

Step 2: Calculating Molar Masses

Molar mass of Al: 26.98 g/mol Molar mass of O2: 32.00 g/mol Molar mass of Al2O3: (2 x 26.98) (3 x 16.00) 53.96 g/mol 48.00 g/mol 101.96 g/mol

Step 3: Converting Grams to Moles

Moles of Al: Moles of Al 23.4 g / 26.98 g/mol 0.867 mol Moles of O2: Moles of O2 34.2 g / 32.00 g/mol 1.069 mol

Step 4: Determining the Limiting Reactant

The balanced equation indicates that 4 moles of aluminum react with 3 moles of oxygen. To find the limiting reactant:

Required O2 for 0.867 mol Al: Required O2 0.867 mol Al × (3 mol O2 / 4 mol Al) 0.650 mol O2

Since we have 1.069 moles of oxygen available, aluminum is the limiting reactant.

Step 5: Calculating the Amount of Al2O3 Produced

From the balanced equation, 4 moles of aluminum produce 2 moles of aluminum oxide. Therefore, 0.867 moles of aluminum will produce:

Moles of Al2O3: Moles of Al2O3 0.867 mol Al × (2 mol Al2O3 / 4 mol Al) 0.434 mol Al2O3

Step 6: Converting Moles to Grams

Finally, convert the moles of aluminum oxide to grams:

Mass of Al2O3: Mass of Al2O3 0.434 mol × 101.96 g/mol 44.19 g

Conclusion

Thus, approximately 44.19 grams of aluminum oxide are produced when 23.4 g of aluminum reacts with 34.2 g of oxygen. This example demonstrates how to determine the amount of product formed in a chemical reaction by identifying the limiting reactant and using stoichiometry.