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17. 55.0 g of potassium iodide and 185.5 g of lead(II) nitrate react and 35.0 g of lead(II) iodide is produced. What is the percent yield of the reaction? KI(aq) + Pb (NO3)2 (aq) → Pbl₂ (s) + KNO3(aq) Basara A: 45.8%

17. 55.0 g of potassium iodide and 185.5 g of lead(II) nitrate react and 35.0 g of lead(II) iodide is produced. What is the percent yield of the reaction? KI(aq) + Pb (NO3)2 (aq) → Pbl₂ (s) + KNO3(aq) Basara A: 45.8%

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Problem 17: Reaction Yield Calculation** In this problem, we are given the masses of reactants and products in a chemical reaction: - **Reactants:** - 55.0 grams of potassium iodide (KI) - 185.5 grams of lead(II) nitrate [Pb(NO₃)₂] - **Product:** - 35.0 grams of lead(II) iodide (PbI₂) is produced The reaction is represented by the following equation: \[ \text{KI}_{(aq)} + \text{Pb(NO}_3\text{)}_2\,(aq) \rightarrow \text{PbI}_2\,(s) + \text{KNO}_3\,(aq) \] We are asked to calculate the percent yield of this reaction. **Solution:** - The percent yield is given by the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] - In this problem, the actual yield is provided as 35.0 grams of \(\text{PbI}_2\). - The theoretical yield needs to be calculated based on the balanced chemical equation and the stoichiometry between reactants and products. The final percent yield, provided at the end of the problem, is: - **Answer: 45.8%**
Transcribed Image Text:**Problem 17: Reaction Yield Calculation** In this problem, we are given the masses of reactants and products in a chemical reaction: - **Reactants:** - 55.0 grams of potassium iodide (KI) - 185.5 grams of lead(II) nitrate [Pb(NO₃)₂] - **Product:** - 35.0 grams of lead(II) iodide (PbI₂) is produced The reaction is represented by the following equation: \[ \text{KI}_{(aq)} + \text{Pb(NO}_3\text{)}_2\,(aq) \rightarrow \text{PbI}_2\,(s) + \text{KNO}_3\,(aq) \] We are asked to calculate the percent yield of this reaction. **Solution:** - The percent yield is given by the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] - In this problem, the actual yield is provided as 35.0 grams of \(\text{PbI}_2\). - The theoretical yield needs to be calculated based on the balanced chemical equation and the stoichiometry between reactants and products. The final percent yield, provided at the end of the problem, is: - **Answer: 45.8%**
Expert Solution
Step 1

Given,

2KI(aq) +  Pb(NO3)2(aq)  → PbI2(s) + 2KNO3(aq)

mass of potassium iodide (KI) react = 55.0 g

mass of lead(II) nitrate [Pb(NO3)2] = 185.5 g

mass of lead(II) iodide [PbI2] = actual yield of PbI2 = 35.0 g

Percent yield of the reaction = ?

Note:

molar mass of KI = 166.0 g/mol

molar mass of Pb(NO3)2 = 331.2 g/mol

molar mass of PbI2 = 461.01 g/mol

 

 

 

 

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