Sure, let's solve this step-by-step:
1. Determine the molar mass of aluminum (Al) and aluminum oxide (Al₂O₃):
- Molar mass of Al = 26.98 g/mol
- For Al₂O₃:
- Molar mass of aluminum (Al) is 26.98 g/mol. Since there are two aluminum atoms in Al₂O₃:
[tex]\( 2 \times 26.98 = 53.96 \text{ g/mol} \)[/tex]
- Molar mass of oxygen (O) is 16.00 g/mol. Since there are three oxygen atoms in Al₂O₃:
[tex]\( 3 \times 16.00 = 48.00 \text{ g/mol} \)[/tex]
- Therefore, total molar mass of Al₂O₃:
[tex]\( 53.96 + 48.00 = 101.96 \text{ g/mol} \)[/tex]
2. Calculate the number of moles of Al in 375 grams:
[tex]\[
\text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} = \frac{375 \text{ g}}{26.98 \text{ g/mol}} \approx 13.899 \text{ moles}
\][/tex]
3. Use the stoichiometric ratio to find the moles of Al₂O₃ produced:
The balanced chemical equation is:
[tex]\[
4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3
\][/tex]
From this, 4 moles of Al produce 2 moles of Al₂O₃, giving a molar ratio of [tex]\( \frac{4}{2} = 2 \)[/tex].
Thus, the number of moles of Al₂O₃ produced from 13.899 moles of Al:
[tex]\[
\text{Moles of Al}_2\text{O}_3 = \frac{13.899}{2} \approx 6.950 \text{ moles}
\][/tex]
4. Calculate the mass of Al₂O₃ produced:
[tex]\[
\text{Mass of Al}_2\text{O}_3 = \text{moles} \times \text{molar mass} = 6.950 \text{ moles} \times 101.96 \text{ g/mol} \approx 708.580 \text{ grams}
\][/tex]
5. Express the answer to three significant figures:
[tex]\[
708.580 \text{ grams} \approx 708.58 \text{ grams}
\][/tex]
Therefore, the mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produced is:
[tex]\[
\boxed{708.58 \text{ grams}}
\][/tex]
