Sure, let's solve the problem step-by-step.
We are given the balanced chemical equation:
[tex]\[2 Al_2O_3 \rightarrow 4 Al + 3 O_2 \][/tex]
We need to calculate how many moles of aluminum [tex]\((Al)\)[/tex] can be produced from 5.00 moles of aluminum oxide [tex]\((Al_2O_3)\)[/tex].
1. Understanding the stoichiometry: The balanced equation shows that 2 moles of [tex]\(Al_2O_3\)[/tex] produce 4 moles of [tex]\(Al\)[/tex]. This gives us a mole ratio of:
[tex]\[\frac{4 \text{ moles } Al}{2 \text{ moles } Al_2O_3} = 2 \][/tex]
2. Applying the ratio to the given amount: We have 5.00 moles of [tex]\(Al_2O_3\)[/tex]. Using the mole ratio from the equation, we can set up the following relationship:
[tex]\[ \text{moles of } Al = \text{moles of } Al_2O_3 \times \left( \frac{4 \, \text{moles } Al}{2 \text{ moles } Al_2O_3} \right) \][/tex]
[tex]\[ \text{moles of } Al = 5.00 \, \text{moles } Al_2O_3 \times 2 \][/tex]
3. Calculation:
[tex]\[ \text{moles of } Al = 5.00 \times 2 = 10.0 \text{ moles } Al \][/tex]
Therefore, 10.0 moles of aluminum are produced from 5.00 moles of aluminum oxide.
The correct answer is:
[tex]\[ \boxed{10 \text{ mol}} \][/tex]
