Solid aluminum  Al and chlorine  Cl2  gas react to form solid aluminum chloride  AlCl3 . Suppose you have 5.0 mol of Al and 2.0 mol of Cl2 in a reactor.   Suppose as much as possible of the Al  reacts. How much will be left? Round your answer to the nearest  0.1   mol .

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Solid aluminum

 Al

and chlorine

 Cl2 

gas react to form solid aluminum chloride

 AlCl3

. Suppose you have

5.0
mol

of

Al

and

2.0
mol

of

Cl2

in a reactor.

 

Suppose as much as possible of the

Al

 reacts. How much will be left? Round your answer to the nearest 

0.1

 

mol

.

 

 

 

Expert Solution
Step 1

Aluminium and chlorine react to form aluminium chloride. The equation form the balanced chemical reaction is as follows:

2Als + 3Cl2g  2AlCl3s

The moles of Al is = 5.0 mole

The mole of chlorine is = 2.0 mole

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Follow-up Question
**Chemical Reaction of Aluminum and Chlorine**

---

Solid aluminum (Al) and chlorine (Cl₂) gas react to form solid aluminum chloride (AlCl₃). Suppose you have 2.0 mol of Al and 1.0 mol of Cl₂ in a reactor.

**Problem Statement:**

What would be the limiting reactant? Enter its chemical formula below.

---

Explanation:

- **Reactants:**
  - Aluminum (Al) - 2.0 mol
  - Chlorine (Cl₂) - 1.0 mol

- **Product:**
  - Aluminum Chloride (AlCl₃)

To solve this, you will need to use the stoichiometry of the balanced chemical equation:

\[ 2Al + 3Cl₂ \rightarrow 2AlCl₃ \]

From the equation, you can see that 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride. By comparing the mole ratio of the reactants you have to the ratio in the balanced equation, you can determine which reactant will be consumed first, thereby identifying the limiting reactant.

Given that only 1.0 mole of Cl₂ is available and the reaction requires 3 moles of Cl₂ for every 2 moles of Al:

Calculate:

- Moles of Cl₂ required for 2 moles of Al: \( (2 \text{ moles Al}) \times \left(\frac{3 \text{ moles Cl₂}}{2 \text{ moles Al}}\right) = 3 \text{ moles Cl₂} \)

Since only 1.0 mole of Cl₂ is available (which is less than the required 3 moles), Cl₂ is the limiting reactant.

---

**Interaction with the Diagram:**

The diagram appears to have an input box accompanied by options for submission, which might include typical operations like 'Check Answer', 'Reset', and 'Help'. The chemical formula of the limiting reactant (in this case, Cl₂) should be input into the input box provided in the diagram for validation.

---
Transcribed Image Text:**Chemical Reaction of Aluminum and Chlorine** --- Solid aluminum (Al) and chlorine (Cl₂) gas react to form solid aluminum chloride (AlCl₃). Suppose you have 2.0 mol of Al and 1.0 mol of Cl₂ in a reactor. **Problem Statement:** What would be the limiting reactant? Enter its chemical formula below. --- Explanation: - **Reactants:** - Aluminum (Al) - 2.0 mol - Chlorine (Cl₂) - 1.0 mol - **Product:** - Aluminum Chloride (AlCl₃) To solve this, you will need to use the stoichiometry of the balanced chemical equation: \[ 2Al + 3Cl₂ \rightarrow 2AlCl₃ \] From the equation, you can see that 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride. By comparing the mole ratio of the reactants you have to the ratio in the balanced equation, you can determine which reactant will be consumed first, thereby identifying the limiting reactant. Given that only 1.0 mole of Cl₂ is available and the reaction requires 3 moles of Cl₂ for every 2 moles of Al: Calculate: - Moles of Cl₂ required for 2 moles of Al: \( (2 \text{ moles Al}) \times \left(\frac{3 \text{ moles Cl₂}}{2 \text{ moles Al}}\right) = 3 \text{ moles Cl₂} \) Since only 1.0 mole of Cl₂ is available (which is less than the required 3 moles), Cl₂ is the limiting reactant. --- **Interaction with the Diagram:** The diagram appears to have an input box accompanied by options for submission, which might include typical operations like 'Check Answer', 'Reset', and 'Help'. The chemical formula of the limiting reactant (in this case, Cl₂) should be input into the input box provided in the diagram for validation. ---
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