For the reaction shown, calculate how many grams of each product form when the following amount of reactant completely reacts to form products. Assume that there is more than enough of the other reactant.

[tex]\[2 \text{Al} (s) + \text{Fe}_2\text{O}_3 (s) \rightarrow \text{Al}_2\text{O}_3 (s) + 2 \text{Fe} (l)\][/tex]

Part A

Calculate the mass of [tex]\text{Al}_2\text{O}_3[/tex] formed when [tex]4.5 \, \text{g Al}[/tex] completely reacts.

Express your answer using two significant figures.



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.