Final answer:
The production of AlBr3 is limited by the 9 moles of Br2, allowing for the creation of 6 moles of AlBr3. Al is in excess and Br2 is the limiting reagent in this reaction.
Step-by-step explanation:
To determine the number of moles of AlBr3 that can be produced from 8 moles of Al and 9 moles of Br2, we first need to write the balanced chemical equation for the reaction:
2 Al + 3 Br2 → 2 AlBr3
From the balanced equation, we can see that 2 moles of Al react with 3 moles of Br2 to produce 2 moles of AlBr3. Thus, the mole ratio of Al to AlBr3 is 1:1 and the mole ratio of Br2 to AlBr3 is 1.5:1.
Given the initial amounts, Al is present in excess because we have 8 moles of Al, and we would need 13.5 moles of Br2 to react completely with all the Al (since 9 moles of Al would need 13.5 moles of Br2 based on the 1.5:1 ratio). Since we only have 9 moles of Br2, this is the limiting reagent.
Here's how the math works out:
- For the limiting reagent Br2: 9 moles Br2 × (2 moles AlBr3 / 3 moles Br2) = 6 moles AlBr3
Therefore, we will be able to produce 6 moles of AlBr3 from 9 moles of Br2, with some Al remaining unreacted. The correct answer is not listed among the options provided, but the closest correct answer based on the available choices would be option b) 8 moles.