o AgNO:(aq) + KBr(aq)→AgBr(s)+ KNO:(aq) a. Which reactant is limiting? Which reactant is in excess? b. What is the theoretical amount of AgBr(s) produced given the limiting reactant. c. If a student reported 15.5grams of the AgBr(s) being produced, what is the percent yield?

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Chapter1: Chemical Foundations
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**Problem 15: Limiting Reactant and Percent Yield Calculations**

A reaction combines 94.6 grams of silver nitrate with 126.4 grams of potassium bromide in the unbalanced equation:

\[ \text{AgNO}_3(aq) + \text{KBr}(aq) \rightarrow \text{AgBr}(s) + \text{KNO}_3(aq) \]

a. **Which reactant is limiting? Which reactant is in excess?**  
To determine the limiting reactant, calculate the moles of each reactant and use stoichiometry based on the balanced equation to find which one produces less of the product.

b. **What is the theoretical amount of AgBr(s) produced given the limiting reactant?**  
Use the moles of the limiting reactant to calculate the theoretical yield of silver bromide (AgBr) using stoichiometry.

c. **If a student reported 15.5 grams of the AgBr(s) being produced, what is the percent yield?**  
Percent yield is calculated using the formula:  
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]

**Instructions:**  
1. Balance the chemical equation given.
2. Calculate the molar mass of each reactant.
3. Determine moles of each reactant using their respective masses.
4. Identify the limiting reactant by comparing the mole ratio from the balanced equation.
5. Calculate the theoretical yield of AgBr using the moles of the limiting reactant.
6. Determine the percent yield with the provided actual yield of AgBr.

_Key Concepts:_ 
- Stoichiometry
- Mole-to-mass calculations
- Limiting reactants
- Theoretical vs. actual yield
- Percent yield calculation
Transcribed Image Text:**Problem 15: Limiting Reactant and Percent Yield Calculations** A reaction combines 94.6 grams of silver nitrate with 126.4 grams of potassium bromide in the unbalanced equation: \[ \text{AgNO}_3(aq) + \text{KBr}(aq) \rightarrow \text{AgBr}(s) + \text{KNO}_3(aq) \] a. **Which reactant is limiting? Which reactant is in excess?** To determine the limiting reactant, calculate the moles of each reactant and use stoichiometry based on the balanced equation to find which one produces less of the product. b. **What is the theoretical amount of AgBr(s) produced given the limiting reactant?** Use the moles of the limiting reactant to calculate the theoretical yield of silver bromide (AgBr) using stoichiometry. c. **If a student reported 15.5 grams of the AgBr(s) being produced, what is the percent yield?** Percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] **Instructions:** 1. Balance the chemical equation given. 2. Calculate the molar mass of each reactant. 3. Determine moles of each reactant using their respective masses. 4. Identify the limiting reactant by comparing the mole ratio from the balanced equation. 5. Calculate the theoretical yield of AgBr using the moles of the limiting reactant. 6. Determine the percent yield with the provided actual yield of AgBr. _Key Concepts:_ - Stoichiometry - Mole-to-mass calculations - Limiting reactants - Theoretical vs. actual yield - Percent yield calculation
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