1. Balanced Chemical Equation:
The balanced chemical equation is already provided:
Al₂O₃ + 3C → 2Al + 3CO
2. Calculate Moles of Reactants:
* Moles of Al₂O₃:
- Molar mass of Al₂O₃ = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol
- Moles of Al₂O₃ = (60.0 g) / (101.96 g/mol) = 0.588 mol
* Moles of C:
- Molar mass of C = 12.01 g/mol
- Moles of C = (30.0 g) / (12.01 g/mol) = 2.498 mol
3. Determine the Limiting Reactant:
* Compare the mole ratios: The balanced equation shows that 1 mole of Al₂O₃ reacts with 3 moles of C.
* Calculate the required moles of C: 0.588 mol Al₂O₃ * (3 mol C / 1 mol Al₂O₃) = 1.764 mol C
* Conclusion: We have more moles of C (2.498 mol) than needed (1.764 mol) for the reaction. Therefore, Al₂O₃ is the limiting reactant.
4. Calculate Theoretical Yield of Aluminum:
* Mole ratio: From the balanced equation, 1 mole of Al₂O₃ produces 2 moles of Al.
* Moles of Al: 0.588 mol Al₂O₃ * (2 mol Al / 1 mol Al₂O₃) = 1.176 mol Al
* Mass of Al: 1.176 mol Al * (26.98 g/mol) = 31.7 g Al
Therefore, the theoretical yield of aluminum that can be produced is 31.7 g.
