Alright, let's solve this step-by-step.
Step 1: Determine the molar masses
- The molar mass of Aluminum (Al) is 26.98 grams per mole.
- The molar mass of Aluminum Oxide (Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex]) is calculated as follows:
[tex]\[
\text{Molar mass of } Al_2O_3 = 2 \times 26.98 \text{ (for Al)} + 3 \times 16 \text{ (for O)} = 53.96 + 48 = 101.96 \text{ g/mol}
\][/tex]
Step 2: Calculate the moles of Aluminum (Al)
- Given mass of Aluminum is 375 grams.
- To find the moles of Al, we use the formula:
[tex]\[
\text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{375 \text{ grams}}{26.98 \text{ g/mol}} \approx 13.899 \text{ moles}
\][/tex]
Step 3: Determine the moles of Aluminum Oxide (Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex]) produced
- According to the balanced chemical equation:
[tex]\[
4 \text{ Al} \rightarrow 2 \text{ Al}_2\text{O}_3
\][/tex]
- This reaction implies that 4 moles of Al produces 2 moles of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex].
- Therefore, the moles of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] produced will be half the moles of Al, as given by the ratio (2/4):
[tex]\[
\text{Moles of Al}_2\text{O}_3 = 13.899 \text{ moles of Al} \times \frac{2}{4} = 6.95 \text{ moles of Al}_2\text{O}_3
\][/tex]
Step 4: Convert the moles of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] to grams
- Using the molar mass of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex] (101.96 g/mol), we find the mass of Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex]:
[tex]\[
\text{Mass of Al}_2\text{O}_3 = \text{moles of Al}_2\text{O}_3 \times \text{molar mass of Al}_2\text{O}_3 = 6.95 \text{ moles} \times 101.96 \text{ g/mol} \approx 708.58 \text{ grams}
\][/tex]
Final Result:
The mass of Aluminum Oxide (Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex]) produced when 375 grams of Aluminum (Al) react completely is:
[tex]\[
\boxed{708.58} \text{ grams of Al}_2\text{O}_3
\][/tex]
