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
![**Title: Calculating Moles of Aluminum in a Chemical Reaction**
**Description:**
In this example, we explore how to determine the number of moles of aluminum (Al) required to produce 86.0 grams of aluminum bromide (AlBr₃) in a chemical reaction. The reaction is given by the equation:
\[ 2 \text{Al(s)} + 3 \text{Br}_2(\ell) \rightarrow 2 \text{AlBr}_3 \]
**Explanation:**
The balanced chemical equation shows that 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide. The stoichiometry of the reaction indicates the necessary proportions of reactants and products.
To solve for the moles of aluminum required:
1. **Calculate the Molar Mass of AlBr₃:**
- Aluminum (Al): 26.98 g/mol
- Bromine (Br): 79.904 g/mol, but since there are three bromines in AlBr₃, multiply by 3.
- Molar mass of AlBr₃ = 26.98 + (79.904 * 3) = 266.69 g/mol
2. **Find the Moles of AlBr₃:**
- Use the given mass of AlBr₃ to find moles:
\[ \text{Moles of AlBr}_3 = \frac{86.0 \text{g}}{266.69 \text{g/mol}} \]
3. **Use Stoichiometry to Find Moles of Al:**
- The balanced equation tells us that 2 moles of Al produce 2 moles of AlBr₃.
- Therefore, the moles of Al required is equal to the moles of AlBr₃ calculated.
**Conclusion:**
By applying stoichiometry and understanding the balanced chemical equation, we can determine the exact amount of reactants needed in a chemical reaction to produce the desired product. This method is crucial in fields involving chemical manufacturing and laboratory work.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd862c25f-a636-4347-831e-84b629db5903%2F565e398e-7e2d-4da9-a3c7-ca575eb254c6%2Fi8mygkk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating Moles of Aluminum in a Chemical Reaction**
**Description:**
In this example, we explore how to determine the number of moles of aluminum (Al) required to produce 86.0 grams of aluminum bromide (AlBr₃) in a chemical reaction. The reaction is given by the equation:
\[ 2 \text{Al(s)} + 3 \text{Br}_2(\ell) \rightarrow 2 \text{AlBr}_3 \]
**Explanation:**
The balanced chemical equation shows that 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide. The stoichiometry of the reaction indicates the necessary proportions of reactants and products.
To solve for the moles of aluminum required:
1. **Calculate the Molar Mass of AlBr₃:**
- Aluminum (Al): 26.98 g/mol
- Bromine (Br): 79.904 g/mol, but since there are three bromines in AlBr₃, multiply by 3.
- Molar mass of AlBr₃ = 26.98 + (79.904 * 3) = 266.69 g/mol
2. **Find the Moles of AlBr₃:**
- Use the given mass of AlBr₃ to find moles:
\[ \text{Moles of AlBr}_3 = \frac{86.0 \text{g}}{266.69 \text{g/mol}} \]
3. **Use Stoichiometry to Find Moles of Al:**
- The balanced equation tells us that 2 moles of Al produce 2 moles of AlBr₃.
- Therefore, the moles of Al required is equal to the moles of AlBr₃ calculated.
**Conclusion:**
By applying stoichiometry and understanding the balanced chemical equation, we can determine the exact amount of reactants needed in a chemical reaction to produce the desired product. This method is crucial in fields involving chemical manufacturing and laboratory work.
![**Problem Statement:**
What mass of precipitate (in g) is formed when 250.0 mL of 0.610 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
\[ \text{CuCl}_2(aq) + 2 \text{KOH}(aq) \rightarrow \text{Cu(OH)}_2(s) + 2 \text{KCl}(aq) \]
**Explanation:**
This problem involves a precipitation reaction where copper(II) chloride reacts with potassium hydroxide to form copper(II) hydroxide as a precipitate and potassium chloride in solution. The task is to calculate the mass of Cu(OH)₂ formed.
**Approach:**
1. **Determine the Number of Moles of CuCl₂:**
- Convert the volume of CuCl₂ solution to liters.
- Use the molarity to find the moles of CuCl₂.
2. **Use Stoichiometry:**
- According to the balanced equation, 1 mole of CuCl₂ reacts with 2 moles of KOH to form 1 mole of Cu(OH)₂.
- Calculate the moles of Cu(OH)₂ formed, which will be equal to the moles of CuCl₂ because they react in a 1:1 ratio.
3. **Calculate the Mass of Precipitate:**
- Use the molar mass of Cu(OH)₂ to find the mass.
This process will yield the mass of the Cu(OH)₂ precipitate formed in grams.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd862c25f-a636-4347-831e-84b629db5903%2F565e398e-7e2d-4da9-a3c7-ca575eb254c6%2Fewg71y2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
What mass of precipitate (in g) is formed when 250.0 mL of 0.610 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
\[ \text{CuCl}_2(aq) + 2 \text{KOH}(aq) \rightarrow \text{Cu(OH)}_2(s) + 2 \text{KCl}(aq) \]
**Explanation:**
This problem involves a precipitation reaction where copper(II) chloride reacts with potassium hydroxide to form copper(II) hydroxide as a precipitate and potassium chloride in solution. The task is to calculate the mass of Cu(OH)₂ formed.
**Approach:**
1. **Determine the Number of Moles of CuCl₂:**
- Convert the volume of CuCl₂ solution to liters.
- Use the molarity to find the moles of CuCl₂.
2. **Use Stoichiometry:**
- According to the balanced equation, 1 mole of CuCl₂ reacts with 2 moles of KOH to form 1 mole of Cu(OH)₂.
- Calculate the moles of Cu(OH)₂ formed, which will be equal to the moles of CuCl₂ because they react in a 1:1 ratio.
3. **Calculate the Mass of Precipitate:**
- Use the molar mass of Cu(OH)₂ to find the mass.
This process will yield the mass of the Cu(OH)₂ precipitate formed in grams.
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