What quantity in moles of precipitate will be formed when 176.0 mL of 0.200 M NaBr is reacted with excess Pb(NO3), in the following chemical reaction? 2 NaBr (aq) + Pb(NO₂)₂ (aq) → PbBr₂ (s) + 2 NaNO₂ (aq) 3
What quantity in moles of precipitate will be formed when 176.0 mL of 0.200 M NaBr is reacted with excess Pb(NO3), in the following chemical reaction? 2 NaBr (aq) + Pb(NO₂)₂ (aq) → PbBr₂ (s) + 2 NaNO₂ (aq) 3
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
![**Question:**
What quantity in moles of precipitate will be formed when 176.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?
\[ 2 \text{NaBr (aq)} + \text{Pb(NO₃)₂ (aq)} \rightarrow \text{PbBr₂ (s)} + 2 \text{NaNO₃ (aq)} \]
**Explanation:**
This chemical equation describes a precipitation reaction where aqueous sodium bromide (NaBr) reacts with lead(II) nitrate (Pb(NO₃)₂) to form solid lead(II) bromide (PbBr₂) as a precipitate and aqueous sodium nitrate (NaNO₃).
To find the number of moles of precipitate (PbBr₂) produced:
1. **Calculate Moles of NaBr:**
\[
\text{Moles of NaBr} = \text{Volume (L) of NaBr} \times \text{Molarity of NaBr}
\]
Given: 176.0 mL = 0.176 L, and 0.200 M NaBr
\[
\text{Moles of NaBr} = 0.176 \, \text{L} \times 0.200 \, \text{mol/L} = 0.0352 \, \text{mol}
\]
2. **Use Stoichiometry:**
According to the balanced chemical equation, 2 moles of NaBr react with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbBr₂.
\[
\text{Moles of PbBr₂} = \frac{0.0352 \, \text{mol NaBr}}{2} = 0.0176 \, \text{mol PbBr₂}
\]
Therefore, 0.0176 moles of lead(II) bromide precipitate will be formed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70274774-6f46-4808-8916-795560a9e0a1%2F5317ad79-e09d-447b-9798-cae46f036e74%2Fpuslq1d_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What quantity in moles of precipitate will be formed when 176.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?
\[ 2 \text{NaBr (aq)} + \text{Pb(NO₃)₂ (aq)} \rightarrow \text{PbBr₂ (s)} + 2 \text{NaNO₃ (aq)} \]
**Explanation:**
This chemical equation describes a precipitation reaction where aqueous sodium bromide (NaBr) reacts with lead(II) nitrate (Pb(NO₃)₂) to form solid lead(II) bromide (PbBr₂) as a precipitate and aqueous sodium nitrate (NaNO₃).
To find the number of moles of precipitate (PbBr₂) produced:
1. **Calculate Moles of NaBr:**
\[
\text{Moles of NaBr} = \text{Volume (L) of NaBr} \times \text{Molarity of NaBr}
\]
Given: 176.0 mL = 0.176 L, and 0.200 M NaBr
\[
\text{Moles of NaBr} = 0.176 \, \text{L} \times 0.200 \, \text{mol/L} = 0.0352 \, \text{mol}
\]
2. **Use Stoichiometry:**
According to the balanced chemical equation, 2 moles of NaBr react with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbBr₂.
\[
\text{Moles of PbBr₂} = \frac{0.0352 \, \text{mol NaBr}}{2} = 0.0176 \, \text{mol PbBr₂}
\]
Therefore, 0.0176 moles of lead(II) bromide precipitate will be formed.
![**Question:**
What volume (in mL) of 0.300 M HCl would be required to completely react with 4.65 g of Al in the following chemical reaction?
\[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 \text{(aq)} + 3 \text{H}_2 \text{(g)} \]
**Explanation:**
The given chemical reaction shows that aluminum (Al) reacts with hydrochloric acid (HCl) to produce aluminum chloride (AlCl₃) and hydrogen gas (H₂). The stoichiometry of the reaction indicates that 2 moles of Al react with 6 moles of HCl.
To determine the volume of HCl required:
1. **Calculate moles of aluminum (Al):**
- Molar mass of Al = 26.98 g/mol
- Moles of Al = mass of Al / molar mass of Al = 4.65 g / 26.98 g/mol
2. **Use stoichiometry to find moles of HCl:**
- According to the reaction, 2 moles of Al react with 6 moles of HCl.
- Therefore, calculate moles of HCl needed for the moles of Al calculated.
3. **Calculate the volume of HCl solution:**
- Molarity (M) of HCl = moles of solute / volume of solution in liters
- Rearrange to find volume: Volume = moles of HCl / Molarity of HCl
- Convert the volume from liters to milliliters.
By following this process, you will determine the required volume of 0.300 M HCl solution to react completely with 4.65 g of aluminum.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70274774-6f46-4808-8916-795560a9e0a1%2F5317ad79-e09d-447b-9798-cae46f036e74%2Fydom5rn_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What volume (in mL) of 0.300 M HCl would be required to completely react with 4.65 g of Al in the following chemical reaction?
\[ 2 \text{Al(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{AlCl}_3 \text{(aq)} + 3 \text{H}_2 \text{(g)} \]
**Explanation:**
The given chemical reaction shows that aluminum (Al) reacts with hydrochloric acid (HCl) to produce aluminum chloride (AlCl₃) and hydrogen gas (H₂). The stoichiometry of the reaction indicates that 2 moles of Al react with 6 moles of HCl.
To determine the volume of HCl required:
1. **Calculate moles of aluminum (Al):**
- Molar mass of Al = 26.98 g/mol
- Moles of Al = mass of Al / molar mass of Al = 4.65 g / 26.98 g/mol
2. **Use stoichiometry to find moles of HCl:**
- According to the reaction, 2 moles of Al react with 6 moles of HCl.
- Therefore, calculate moles of HCl needed for the moles of Al calculated.
3. **Calculate the volume of HCl solution:**
- Molarity (M) of HCl = moles of solute / volume of solution in liters
- Rearrange to find volume: Volume = moles of HCl / Molarity of HCl
- Convert the volume from liters to milliliters.
By following this process, you will determine the required volume of 0.300 M HCl solution to react completely with 4.65 g of aluminum.
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 3 steps with 3 images
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