To determine the limiting reactant in the balanced chemical equation:
[tex]\[
4 \text{AlF}_3 + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 + 6 \text{F}_2
\][/tex]
given 9 moles of [tex]\(\text{AlF}_3\)[/tex] and 12 moles of [tex]\(\text{O}_2\)[/tex], we need to follow these steps:
Step 1: Identify the stoichiometric coefficients.
From the balanced equation:
- For [tex]\(\text{AlF}_3\)[/tex]: 4 moles
- For [tex]\(\text{O}_2\)[/tex]: 3 moles
Step 2: Calculate how many moles of product each reactant would fully produce.
We determine the amount of product formed if each reactant completely reacts based on the stoichiometric coefficients:
For [tex]\(\text{AlF}_3\)[/tex]:
- Using the stoichiometric ratio, we see that 4 moles of [tex]\(\text{AlF}_3\)[/tex] produce a certain number of products. Therefore, if we have 9 moles of [tex]\(\text{AlF}_3\)[/tex]:
[tex]\[
\text{moles of product from } \text{AlF}_3 = \frac{9 \text{ moles AlF}_3}{4} = 2.25 \text{ moles of product}
\][/tex]
For [tex]\(\text{O}_2\)[/tex]:
- Similarly, we use the stoichiometric ratio to find that 3 moles of [tex]\(\text{O}_2\)[/tex] produce a certain number of products. With 12 moles of [tex]\(\text{O}_2\)[/tex]:
[tex]\[
\text{moles of product from } \text{O}_2 = \frac{12 \text{ moles O}_2}{3} = 4.0 \text{ moles of product}
\][/tex]
Step 3: Determine which reactant is limiting.
To find the limiting reactant, we compare the amounts of product each reactant can form:
- [tex]\(\text{AlF}_3\)[/tex] can produce 2.25 moles of product.
- [tex]\(\text{O}_2\)[/tex] can produce 4.0 moles of product.
The limiting reactant is the one that produces the least amount of product.
Conclusion:
Since [tex]\(\text{AlF}_3\)[/tex] only produces 2.25 moles of product compared to the 4.0 moles by [tex]\(\text{O}_2\)[/tex], [tex]\(\text{AlF}_3\)[/tex] is the limiting reactant in this reaction.
Thus, the limiting reactant is [tex]\(\text{AlF}_3\)[/tex].
