This is a limiting reagent problem based on the reaction between aluminum (Al) and chlorine (Cl$_2$) to form aluminum chloride (AlCl$_3$).

The balanced chemical equation for the reaction is: $2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3$

Step 1: Determine the mole ratio

From the equation:

• 2 moles of Al react with 3 moles of Cl$_2$ to produce 2 moles of AlCl$_3$.

Step 2: Find the limiting reagent

You have:

• 9.0 moles of Al
• 2.0 moles of Cl$_2$

Now, calculate how much Cl$_2$ is required to react with 9.0 moles of Al: $\text{Moles of Cl}_2 \, \text{needed} = \frac{3}{2} \times 9.0 = 13.5 \, \text{moles}$ But you only have 2.0 moles of Cl$_2$, so Cl$_2$ is the limiting reagent.

Step 3: Determine the maximum amount of AlCl$_3$ that can be produced

Using the limiting reagent (2.0 moles of Cl$_2$), calculate the moles of AlCl$_3$ formed: $\text{Moles of AlCl}_3 = \frac{2}{3} \times 2.0 = 1.33 \, \text{moles}$

Step 4: Round the answer

The largest amount of AlCl$_3$ that could be produced is 1.3 moles.

Let me know if you want further clarifications or details! Here are some related questions to consider:

1. What is the concept of a limiting reagent, and why is it important in stoichiometry?
2. How does rounding affect precision in chemical calculations?
3. Can you calculate how much of the excess reactant (Al) remains after the reaction?
4. What are the real-world implications of limiting reagents in industrial chemistry?
5. How can you verify this type of calculation experimentally?

Tip: Always balance the chemical equation first; it's the foundation for accurate stoichiometric calculations!