4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.
4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
4c) is the tricky one must have confused myself and got it wrong but I got A) and B) correct as I will attach my work in a picture below as well thank you.
![### Chemical Reaction Problem Solving
**Problem Statement:**
A 50.0 mL of 0.250 M BaCl₂ solution were added to 50.0 mL of 0.160 M Al₂(SO₄)₃ solution.
**Reaction Equation:**
\[ 3 \, \text{BaCl}_2 (\text{aq}) + 1 \, \text{Al}_2(\text{SO}_4)_3 (\text{aq}) \rightarrow 3 \, \text{BaSO}_4 (\text{s}) + 2 \, \text{AlCl}_3 (\text{aq}) \]
**(a) Determining the Limiting Reactant**
- Calculate moles of BaCl₂:
\[
\text{moles of BaCl}_2 = M \times V = (0.250 \, \text{M})(50.0 \times 10^{-3} \, \text{L}) = 0.0125 \, \text{mol}
\]
- Calculate moles of Al₂(SO₄)₃:
\[
\text{moles of Al}_2(\text{SO}_4)_3 = M \times V = (0.160 \, \text{M})(50.0 \times 10^{-3} \, \text{L}) = 0.008 \, \text{mol}
\]
- Compare required moles:
\[
\frac{1 \, \text{mol Al}_2(\text{SO}_4)_3 }{3 \, \text{mol BaCl}_2} \Rightarrow 0.0125 \, \text{mol BaCl}_2 \Rightarrow 0.0042 \, \text{mol of Al}_2(\text{SO}_4)_3
\]
\[
\text{Excess of Al}_2(\text{SO}_4)_3: 0.008 \, \text{mol} - 0.0042 \, \text{mol} = 0.0038 \, \text{mol}
\]
- **Conclusion:** BaCl₂ is the limiting reactant.
**(b) Calculating Mass of BaSO₄ (Molar Mass = 233.4 g/mol)**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f4bfcc4-2f05-41cd-813d-7d76098e5301%2F3ecf6492-19d3-4f5d-81f9-cdf923aa0990%2Fehfy9ds_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction Problem Solving
**Problem Statement:**
A 50.0 mL of 0.250 M BaCl₂ solution were added to 50.0 mL of 0.160 M Al₂(SO₄)₃ solution.
**Reaction Equation:**
\[ 3 \, \text{BaCl}_2 (\text{aq}) + 1 \, \text{Al}_2(\text{SO}_4)_3 (\text{aq}) \rightarrow 3 \, \text{BaSO}_4 (\text{s}) + 2 \, \text{AlCl}_3 (\text{aq}) \]
**(a) Determining the Limiting Reactant**
- Calculate moles of BaCl₂:
\[
\text{moles of BaCl}_2 = M \times V = (0.250 \, \text{M})(50.0 \times 10^{-3} \, \text{L}) = 0.0125 \, \text{mol}
\]
- Calculate moles of Al₂(SO₄)₃:
\[
\text{moles of Al}_2(\text{SO}_4)_3 = M \times V = (0.160 \, \text{M})(50.0 \times 10^{-3} \, \text{L}) = 0.008 \, \text{mol}
\]
- Compare required moles:
\[
\frac{1 \, \text{mol Al}_2(\text{SO}_4)_3 }{3 \, \text{mol BaCl}_2} \Rightarrow 0.0125 \, \text{mol BaCl}_2 \Rightarrow 0.0042 \, \text{mol of Al}_2(\text{SO}_4)_3
\]
\[
\text{Excess of Al}_2(\text{SO}_4)_3: 0.008 \, \text{mol} - 0.0042 \, \text{mol} = 0.0038 \, \text{mol}
\]
- **Conclusion:** BaCl₂ is the limiting reactant.
**(b) Calculating Mass of BaSO₄ (Molar Mass = 233.4 g/mol)**
![**Transcription for Educational Website**
### Chemical Reaction Problem
A 50.0 mL of 0.250 M BaCl₂ solution was added to 50.0 mL of 0.160 M Al₂(SO₄)₃ solution.
#### The Reaction Equation:
\[ 3 \text{BaCl}_2 (\text{aq}) + \text{Al}_2(\text{SO}_4)_3 (\text{aq}) \rightarrow 3 \text{BaSO}_4 (\text{s}) + 2 \text{AlCl}_3 (\text{aq}) \]
**(a) What is the limiting reactant?**
**Answer:** _______________
**(b) Calculate the mass of BaSO₄ (233.4 g/mol).**
**Answer:** _______________
**(c) What are the molar concentrations (molarities) of the remaining ions in the solution?**
**Answer:** _______________
### Explanation
This exercise involves determining the limiting reactant when two solutions are mixed, calculating the mass of a precipitate formed, and finding the concentrations of ions remaining in the solution after the reaction. The reaction involves barium chloride and aluminum sulfate to form barium sulfate and aluminum chloride. Each question requires understanding stoichiometry and molarity concepts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f4bfcc4-2f05-41cd-813d-7d76098e5301%2F3ecf6492-19d3-4f5d-81f9-cdf923aa0990%2F6heve3_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
### Chemical Reaction Problem
A 50.0 mL of 0.250 M BaCl₂ solution was added to 50.0 mL of 0.160 M Al₂(SO₄)₃ solution.
#### The Reaction Equation:
\[ 3 \text{BaCl}_2 (\text{aq}) + \text{Al}_2(\text{SO}_4)_3 (\text{aq}) \rightarrow 3 \text{BaSO}_4 (\text{s}) + 2 \text{AlCl}_3 (\text{aq}) \]
**(a) What is the limiting reactant?**
**Answer:** _______________
**(b) Calculate the mass of BaSO₄ (233.4 g/mol).**
**Answer:** _______________
**(c) What are the molar concentrations (molarities) of the remaining ions in the solution?**
**Answer:** _______________
### Explanation
This exercise involves determining the limiting reactant when two solutions are mixed, calculating the mass of a precipitate formed, and finding the concentrations of ions remaining in the solution after the reaction. The reaction involves barium chloride and aluminum sulfate to form barium sulfate and aluminum chloride. Each question requires understanding stoichiometry and molarity concepts.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY