To solve this problem, we need to use the stoichiometry of the given chemical reaction:
[tex]\[
2 \text{Al}_2\text{O}_3 \rightarrow 4 \text{Al} + 3 \text{O}_2
\][/tex]
This equation tells us that 4 moles of aluminum ([tex]\( \text{Al} \)[/tex]) are produced for every 3 moles of oxygen ([tex]\( \text{O}_2 \)[/tex]).
Here's a step-by-step method to determine the number of moles of oxygen produced if 11.0 moles of aluminum are produced:
1. Identify the molar ratio:
The balanced chemical equation shows that 4 moles of [tex]\( \text{Al} \)[/tex] correspond to 3 moles of [tex]\( \text{O}_2 \)[/tex].
So, the molar ratio of [tex]\( \text{Al} \)[/tex] to [tex]\( \text{O}_2 \)[/tex] is:
[tex]\[
\frac{4 \text{ moles Al}}{3 \text{ moles } O_2}
\][/tex]
2. Calculate the moles of O_2 produced using the molar ratio:
Given that 11.0 moles of [tex]\( \text{Al} \)[/tex] are produced, calculate the moles of [tex]\( \text{O}_2 \)[/tex] produced by dividing the moles of [tex]\( \text{Al} \)[/tex] by this ratio:
[tex]\[
\text{Moles of } O_2 = \frac{11.0 \text{ moles } \text{Al}}{\frac{4}{3}} = 11.0 \text{ moles } \text{Al} \times \frac{3}{4}
\][/tex]
[tex]\[
\text{Moles of } O_2 = 11.0 \times 0.75 = 8.25 \text{ moles of } O_2
\][/tex]
3. Conclude the result:
Thus, the number of moles of oxygen produced when 11.0 moles of aluminum are produced is 8.25 moles.
Among the given options:
- 8.25 mol
- 11.0 mol
- 14.7 mol
- 16.5 mol
The correct answer is:
[tex]\[
\boxed{8.25 \text{ mol}}
\][/tex]
