A 185.0 mL sample of 1.200 M Pb(NO3)2 is mixed with 123.50 mL of 1.500 M NaCI, and the PbCl2 precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr, precipitate is also collected and dried. What is the mass (in grams) of the PbBr2 precipitate, assuming the yield in each precipitation step is 100%?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Precipitation Reaction Problem**

**Question:**

A 185.0 mL sample of 1.200 M Pb(NO₃)₂ is mixed with 123.50 mL of 1.500 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass (in grams) of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%?

**Analysis:**

1. **Reaction 1: Formation of PbCl₂**
   - Solutions mixed: 185.0 mL of 1.200 M Pb(NO₃)₂ and 123.50 mL of 1.500 M NaCl.
   - Precipitate formed: PbCl₂.
   
2. **Reaction 2: Formation of PbBr₂**
   - Remaining solution: After filtration of PbCl₂.
   - Added solution: 200.0 mL of 3.000 M NaBr.
   - Precipitate formed: PbBr₂.
   
**Objective:**
Calculate the mass of the PbBr₂ precipitate.

**Key Assumptions:**
- 100% yield for each precipitation step. 

**Calculation Steps:**
- Determine moles of each reactant.
- Identify the limiting reactant for each reaction.
- Calculate moles of PbBr₂ formed.
- Convert moles of PbBr₂ to grams using its molar mass.
Transcribed Image Text:**Precipitation Reaction Problem** **Question:** A 185.0 mL sample of 1.200 M Pb(NO₃)₂ is mixed with 123.50 mL of 1.500 M NaCl, and the PbCl₂ precipitate is filtered from the solution. Then 200.0 mL of 3.000 M NaBr is added to the remaining solution, and the PbBr₂ precipitate is also collected and dried. What is the mass (in grams) of the PbBr₂ precipitate, assuming the yield in each precipitation step is 100%? **Analysis:** 1. **Reaction 1: Formation of PbCl₂** - Solutions mixed: 185.0 mL of 1.200 M Pb(NO₃)₂ and 123.50 mL of 1.500 M NaCl. - Precipitate formed: PbCl₂. 2. **Reaction 2: Formation of PbBr₂** - Remaining solution: After filtration of PbCl₂. - Added solution: 200.0 mL of 3.000 M NaBr. - Precipitate formed: PbBr₂. **Objective:** Calculate the mass of the PbBr₂ precipitate. **Key Assumptions:** - 100% yield for each precipitation step. **Calculation Steps:** - Determine moles of each reactant. - Identify the limiting reactant for each reaction. - Calculate moles of PbBr₂ formed. - Convert moles of PbBr₂ to grams using its molar mass.
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