The following series of reactions were carried out. PbCO:(s) + 2HN03(aq) → Pb(NO3)2(aq) + H2O(1) + CO2(g) Pb(NO:)2(aq) + 2HCI(aq) → 2HNO:(aq) + PbCl2(s) (a) If a student starts with 2.871 g of lead(II) carbonate for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride solid? b) If the student isolates 2.385 g of lead(II) chloride, what is the percent yield?
The following series of reactions were carried out. PbCO:(s) + 2HN03(aq) → Pb(NO3)2(aq) + H2O(1) + CO2(g) Pb(NO:)2(aq) + 2HCI(aq) → 2HNO:(aq) + PbCl2(s) (a) If a student starts with 2.871 g of lead(II) carbonate for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride solid? b) If the student isolates 2.385 g of lead(II) chloride, what is the percent yield?
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section5.8: Product- Or Reactant-favored Reactions And Thermodynamics
Problem 2.1ACP
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![**Chemical Reactions and Yield Calculation**
The following series of reactions were carried out:
1. \( \text{PbCO}_3(s) + 2\text{HNO}_3(aq) \rightarrow \text{Pb(NO}_3)_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \)
2. \( \text{Pb(NO}_3)_2(aq) + 2\text{HCl}(aq) \rightarrow 2\text{HNO}_3(aq) + \text{PbCl}_2(s) \)
---
**Problem (a):**
If a student starts with 2.871 g of lead(II) carbonate (\( \text{PbCO}_3 \)) for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride (\( \text{PbCl}_2 \)) solid?
**Problem (b):**
If the student isolates 2.385 g of lead(II) chloride, what is the percent yield?
---
### Explanation
To solve these problems, you will need to convert between moles and grams using molar masses and apply stoichiometry to determine theoretical and percent yields.
For **Problem (a)**, calculate the moles of \( \text{PbCO}_3 \) and use the stoichiometry of the given reactions to find the moles and then the grams of \( \text{PbCl}_2 \). This gives you the theoretical yield.
For **Problem (b)**, use the actual yield (2.385 g of \( \text{PbCl}_2 \)) and the theoretical yield from Problem (a) to determine the percent yield using the formula:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa8d3fce-d16d-4f4d-b492-1dee716a707b%2F1bce7c45-2a11-41b1-bfb6-35845907a5da%2Filtefck_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reactions and Yield Calculation**
The following series of reactions were carried out:
1. \( \text{PbCO}_3(s) + 2\text{HNO}_3(aq) \rightarrow \text{Pb(NO}_3)_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \)
2. \( \text{Pb(NO}_3)_2(aq) + 2\text{HCl}(aq) \rightarrow 2\text{HNO}_3(aq) + \text{PbCl}_2(s) \)
---
**Problem (a):**
If a student starts with 2.871 g of lead(II) carbonate (\( \text{PbCO}_3 \)) for the first reaction and all other reagents are added in excess, what is the theoretical yield of lead(II) chloride (\( \text{PbCl}_2 \)) solid?
**Problem (b):**
If the student isolates 2.385 g of lead(II) chloride, what is the percent yield?
---
### Explanation
To solve these problems, you will need to convert between moles and grams using molar masses and apply stoichiometry to determine theoretical and percent yields.
For **Problem (a)**, calculate the moles of \( \text{PbCO}_3 \) and use the stoichiometry of the given reactions to find the moles and then the grams of \( \text{PbCl}_2 \). This gives you the theoretical yield.
For **Problem (b)**, use the actual yield (2.385 g of \( \text{PbCl}_2 \)) and the theoretical yield from Problem (a) to determine the percent yield using the formula:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
Expert Solution
Moles of PbCl2
* Moles of Lead (ll) chloride produced = 0.0107 mol
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