Solutions containing 5.07 g of aluminum chloride and 2.71 g of silver nitrate are combined to form 1.00 g of solid product. What is the percent yield in this reaction? Note: to receive full credit, your work must also show the limiting reactant and theoretical yield for the reaction, clearly labeled

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**Determining the Percent Yield of a Chemical Reaction**

**Problem Statement:**

Solutions containing 5.07 g of aluminum chloride and 2.71 g of silver nitrate are combined to form 1.00 g of solid product. What is the percent yield in this reaction?

**Note:** To receive full credit, your work must also show the limiting reactant and theoretical yield for the reaction, clearly labeled.

---

### Explanation

To solve this problem, take the following steps:

1. **Identify the Reactants and Products:**
    - Aluminum chloride (AlCl₃)
    - Silver nitrate (AgNO₃)
    - Solid product (assume it is either AgCl or another solid formed in this reaction)

2. **Determine the Balanced Chemical Equation:**
    - Write the balanced chemical reaction that might occur when aluminum chloride reacts with silver nitrate.

3. **Calculate Moles of Each Reactant:**
    - Calculate the number of moles for each reactant using the given masses and their molar masses.

4. **Identify the Limiting Reactant:**
    - Determine which reactant is the limiting reagent by using stoichiometry.

5. **Calculate the Theoretical Yield:**
    - Use the limiting reactant to calculate the theoretical yield of the product.

6. **Calculate the Percent Yield:**
    - Use the following formula to determine the percent yield:
      \[
      \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
      \]

### Steps in Detail:

1. **Balanced Chemical Equation:**
    
    AlCl₃ + 3AgNO₃ → 3AgCl + Al(NO₃)₃

2. **Moles Calculation:**
    - Molar mass of AlCl₃ = 133.34 g/mol
    - Moles of AlCl₃ = 5.07 g / 133.34 g/mol
    - Molar mass of AgNO₃ = 169.87 g/mol
    - Moles of AgNO₃ = 2.71 g / 169.87 g/mol

3. **Limiting Reactant:**
    - Use the stoichiometric coefficients from the balanced equation to identify the limiting reactant.

4. **Theoretical Yield:**
    - Find the theoretical yield of
Transcribed Image Text:--- **Determining the Percent Yield of a Chemical Reaction** **Problem Statement:** Solutions containing 5.07 g of aluminum chloride and 2.71 g of silver nitrate are combined to form 1.00 g of solid product. What is the percent yield in this reaction? **Note:** To receive full credit, your work must also show the limiting reactant and theoretical yield for the reaction, clearly labeled. --- ### Explanation To solve this problem, take the following steps: 1. **Identify the Reactants and Products:** - Aluminum chloride (AlCl₃) - Silver nitrate (AgNO₃) - Solid product (assume it is either AgCl or another solid formed in this reaction) 2. **Determine the Balanced Chemical Equation:** - Write the balanced chemical reaction that might occur when aluminum chloride reacts with silver nitrate. 3. **Calculate Moles of Each Reactant:** - Calculate the number of moles for each reactant using the given masses and their molar masses. 4. **Identify the Limiting Reactant:** - Determine which reactant is the limiting reagent by using stoichiometry. 5. **Calculate the Theoretical Yield:** - Use the limiting reactant to calculate the theoretical yield of the product. 6. **Calculate the Percent Yield:** - Use the following formula to determine the percent yield: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] ### Steps in Detail: 1. **Balanced Chemical Equation:** AlCl₃ + 3AgNO₃ → 3AgCl + Al(NO₃)₃ 2. **Moles Calculation:** - Molar mass of AlCl₃ = 133.34 g/mol - Moles of AlCl₃ = 5.07 g / 133.34 g/mol - Molar mass of AgNO₃ = 169.87 g/mol - Moles of AgNO₃ = 2.71 g / 169.87 g/mol 3. **Limiting Reactant:** - Use the stoichiometric coefficients from the balanced equation to identify the limiting reactant. 4. **Theoretical Yield:** - Find the theoretical yield of
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