Answer :
To calculate the mass of Al₂O₃ (aluminum oxide) formed when 4.5 grams of Al (aluminum) completely react, follow these steps:
1. Determine the molar masses:
- Molar mass of Al (aluminum): 26.98 g/mol
- Molar mass of Al₂O₃ (aluminum oxide): 101.96 g/mol
2. Calculate the number of moles of aluminum (Al):
- Use the formula:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
- Plug in the values:
[tex]\[ \text{moles of Al} = \frac{4.5 \text{ g}}{26.98 \text{ g/mol}} ≈ 0.1668 \text{ moles} \][/tex]
3. Use the stoichiometric ratio to find the moles of Al₂O₃ (aluminum oxide):
- From the balanced equation, [tex]\(2 \text{ Al (s)} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2 \text{Fe (l)}\)[/tex], we can see that 2 moles of Al produce 1 mole of Al₂O₃.
- Therefore:
[tex]\[ \text{moles of Al}_2\text{O}_3 = \frac{\text{moles of Al}}{2} \approx \frac{0.1668 \text{ moles}}{2} ≈ 0.0834 \text{ moles} \][/tex]
4. Calculate the mass of Al₂O₃ formed:
- Use the formula:
[tex]\[ \text{mass of Al}_2\text{O}_3 = \text{moles of Al}_2\text{O}_3 \times \text{molar mass of Al}_2\text{O}_3 \][/tex]
- Plug in the values:
[tex]\[ \text{mass of Al}_2\text{O}_3 = 0.0834 \text{ moles} \times 101.96 \text{ g/mol} ≈ 8.503 \text{ g} \][/tex]
Therefore, when 4.5 grams of Al completely react, approximately 8.50 grams of Al₂O₃ are formed.
1. Determine the molar masses:
- Molar mass of Al (aluminum): 26.98 g/mol
- Molar mass of Al₂O₃ (aluminum oxide): 101.96 g/mol
2. Calculate the number of moles of aluminum (Al):
- Use the formula:
[tex]\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
- Plug in the values:
[tex]\[ \text{moles of Al} = \frac{4.5 \text{ g}}{26.98 \text{ g/mol}} ≈ 0.1668 \text{ moles} \][/tex]
3. Use the stoichiometric ratio to find the moles of Al₂O₃ (aluminum oxide):
- From the balanced equation, [tex]\(2 \text{ Al (s)} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2 \text{Fe (l)}\)[/tex], we can see that 2 moles of Al produce 1 mole of Al₂O₃.
- Therefore:
[tex]\[ \text{moles of Al}_2\text{O}_3 = \frac{\text{moles of Al}}{2} \approx \frac{0.1668 \text{ moles}}{2} ≈ 0.0834 \text{ moles} \][/tex]
4. Calculate the mass of Al₂O₃ formed:
- Use the formula:
[tex]\[ \text{mass of Al}_2\text{O}_3 = \text{moles of Al}_2\text{O}_3 \times \text{molar mass of Al}_2\text{O}_3 \][/tex]
- Plug in the values:
[tex]\[ \text{mass of Al}_2\text{O}_3 = 0.0834 \text{ moles} \times 101.96 \text{ g/mol} ≈ 8.503 \text{ g} \][/tex]
Therefore, when 4.5 grams of Al completely react, approximately 8.50 grams of Al₂O₃ are formed.
