Solid aluminum (Al)and chlorine (C1₂) gas react to form solid aluminum chloride (AIC13). Suppose you have 7.0 mol of Al and 1.0 mol of Cl₂ in a reactor. Calculate the largest amount of AlCl3 that could be produced. Round your answer to the nearest 0.1 mol. mol 1.0 X S ?

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## Chemical Reaction and Stoichiometry Problem

### Problem Statement:
Solid aluminum (Al) and chlorine (Cl₂) gas react to form solid aluminum chloride (AlCl₃). Suppose you have 7.0 mol of Al and 1.0 mol of Cl₂ in a reactor. 

Calculate the largest amount of AlCl₃ that could be produced. Round your answer to the nearest 0.1 mol.

### Input Fields:
- A text box labeled "mol" for entering the answer.

### Interactive Elements:
- A small input dialog box featuring a checkbox for some function (labeled with a "10^n" symbol, potentially indicating scientific notation).
- Buttons for multiplication (×), backspace (⟲), and a help button (?).

### Bottom Navigation:
- Buttons labeled "Explanation" and "Check" for navigating through the problem-solving process.

### Detailed Diagram/Calculation Instructions:

To solve the problem, we must use stoichiometry based on the balanced chemical equation:

\[ 2 Al(s) + 3 Cl_2(g) \rightarrow 2 AlCl_3(s) \]

From the balanced equation, we can deduce the mole ratios:
- 2 moles of Al reacts with 3 moles of Cl₂ to produce 2 moles of AlCl₃.

Given:
- 7.0 moles of Al
- 1.0 moles of Cl₂

**Step-by-Step Solution:**

1. **Determine the Limiting Reactant:**
   - Calculate the amount of Cl₂ needed to react with 7.0 moles of Al:
     \[ 7.0 \text{ moles Al} \times \frac{3 \text{ moles Cl}_2}{2 \text{ moles Al}} = 10.5 \text{ moles Cl}_2 \]
   - We only have 1.0 mole of Cl₂.

   Since we need 10.5 moles of Cl₂ but only have 1.0 mole, Cl₂ is the limiting reactant.

2. **Calculate the Amount of AlCl₃ Produced:**
   - Using the mole ratio between Cl₂ and AlCl₃ from the balanced equation:
     \[ 1.0 \text{ moles Cl}_2 \times \frac{2 \text{ moles AlCl}_3}{3 \text{ moles Cl
Transcribed Image Text:## Chemical Reaction and Stoichiometry Problem ### Problem Statement: Solid aluminum (Al) and chlorine (Cl₂) gas react to form solid aluminum chloride (AlCl₃). Suppose you have 7.0 mol of Al and 1.0 mol of Cl₂ in a reactor. Calculate the largest amount of AlCl₃ that could be produced. Round your answer to the nearest 0.1 mol. ### Input Fields: - A text box labeled "mol" for entering the answer. ### Interactive Elements: - A small input dialog box featuring a checkbox for some function (labeled with a "10^n" symbol, potentially indicating scientific notation). - Buttons for multiplication (×), backspace (⟲), and a help button (?). ### Bottom Navigation: - Buttons labeled "Explanation" and "Check" for navigating through the problem-solving process. ### Detailed Diagram/Calculation Instructions: To solve the problem, we must use stoichiometry based on the balanced chemical equation: \[ 2 Al(s) + 3 Cl_2(g) \rightarrow 2 AlCl_3(s) \] From the balanced equation, we can deduce the mole ratios: - 2 moles of Al reacts with 3 moles of Cl₂ to produce 2 moles of AlCl₃. Given: - 7.0 moles of Al - 1.0 moles of Cl₂ **Step-by-Step Solution:** 1. **Determine the Limiting Reactant:** - Calculate the amount of Cl₂ needed to react with 7.0 moles of Al: \[ 7.0 \text{ moles Al} \times \frac{3 \text{ moles Cl}_2}{2 \text{ moles Al}} = 10.5 \text{ moles Cl}_2 \] - We only have 1.0 mole of Cl₂. Since we need 10.5 moles of Cl₂ but only have 1.0 mole, Cl₂ is the limiting reactant. 2. **Calculate the Amount of AlCl₃ Produced:** - Using the mole ratio between Cl₂ and AlCl₃ from the balanced equation: \[ 1.0 \text{ moles Cl}_2 \times \frac{2 \text{ moles AlCl}_3}{3 \text{ moles Cl
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