To calculate the percent yield for the production of aluminum (Al) in this reaction, we need to follow the steps methodically:
1. Balanced Chemical Equation:
[tex]\[
\text{Al}_2\text{O}_3(s) + 3\text{C}(s) \rightarrow 2\text{Al}(s) + 3\text{CO}(g)
\][/tex]
2. Molar Masses (Given):
- Molar Mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] = 101.96 g/mol
- Molar Mass of Al = 26.98 g/mol
- Molar Mass of C = 12.01 g/mol
3. Given Data:
- Mass of [tex]\(\text{Al}\)[/tex] produced = 4.26 g
- Mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] reacted = 1.500 \times 10 = 15.00 g
- Mass of C reacted = 6.00 g
4. Calculating Moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex] Reacted:
[tex]\[
\text{Moles of }\text{Al}_2\text{O}_3\text{ reacted} = \frac{\text{Mass of }\text{Al}_2\text{O}_3}{\text{Molar Mass of }\text{Al}_2\text{O}_3} = \frac{15.00 \text{ g}}{101.96 \text{ g/mol}} \approx 0.1471 \text{ mol}
\][/tex]
5. Stoichiometry of the Reaction:
According to the balanced equation, 1 mol of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produces 2 mols of Al.
6. Calculating Theoretical Moles of Al:
[tex]\[
\text{Moles of Al produced (theoretically)} = 2 \times \text{Moles of }\text{Al}_2\text{O}_3\text{ reacted} = 2 \times 0.1471 \text{ mol} = 0.2942 \text{ mol}
\][/tex]
7. Convert Theoretical Moles of Al to Grams:
[tex]\[
\text{Mass of Al produced (theoretically)} = \text{Moles of Al produced (theoretically)} \times \text{Molar Mass of Al} = 0.2942 \text{ mol} \times 26.98 \text{ g/mol} \approx 7.9384 \text{ g}
\][/tex]
8. Percent Yield Calculation:
[tex]\[
\text{Percent Yield} = \left( \frac{\text{Actual Mass of Al produced}}{\text{Theoretical Mass of Al produced}} \right) \times 100 = \left( \frac{4.26 \text{ g}}{7.9384 \text{ g}} \right) \times 100 \approx 53.7\%
\][/tex]
So, the percent yield for the production of Al is [tex]\( 53.7\% \)[/tex] which matches one of the given answer choices. Thus, the correct answer is:
[tex]\[ 53.7\% \][/tex]
