If 24.0 g of NaOH is added to 0.650 L of 1.00 M Cu(NOs)2, how many grams of Cu(OH)2 will be formed in the following precipitation reaction? 2 N2OH(aq) + Cu(NO:)2(aq) → Cu(OH)2 (s) + 2 NANO:(aq)
If 24.0 g of NaOH is added to 0.650 L of 1.00 M Cu(NOs)2, how many grams of Cu(OH)2 will be formed in the following precipitation reaction? 2 N2OH(aq) + Cu(NO:)2(aq) → Cu(OH)2 (s) + 2 NANO:(aq)
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...
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![## Precipitation Reaction Problem
**Question:**
If 24.0 g of NaOH is added to 0.650 L of 1.00 M Cu(NO₃)₂, how many grams of Cu(OH)₂ will be formed in the following precipitation reaction?
\[ 2 \text{ NaOH(aq) } + \text{ Cu(NO}_3\text{)}_2\text{(aq) } \rightarrow \text{ Cu(OH)}_2\text{(s) } + 2 \text{ NaNO}_3\text{(aq)} \]
**Details:**
- *NaOH* (Sodium Hydroxide): 24.0 grams
- *Cu(NO₃)₂* (Copper(II) Nitrate): 0.650 liters of 1.00 M solution
### Steps for Solving the Problem:
1. **Determine Moles of Reactants:**
- Calculate the moles of NaOH using its molar mass.
- Calculate the moles of Cu(NO₃)₂ using its molarity and volume.
2. **Identify the Limiting Reagent:**
- Use stoichiometry based on the balanced chemical equation.
3. **Calculate Moles of Product (Cu(OH)₂):**
- Determine moles of Cu(OH)₂ formed using the limiting reagent.
4. **Convert Moles of Cu(OH)₂ to Grams:**
- Use the molar mass of Cu(OH)₂ to find the grams produced.
### Visual Aid:
The page includes a numeric calculator interface for entering and calculating values. The calculator displays as:
- A number pad with digits (1-9, 0) and function keys (+, -, C for clear)
This interface assists in performing calculations directly while viewing the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea791450-d01e-4088-949f-ede525bcff36%2Ffef44ce4-fbcc-4871-aa2c-4900a5f33143%2Fplz0r6q_processed.png&w=3840&q=75)
Transcribed Image Text:## Precipitation Reaction Problem
**Question:**
If 24.0 g of NaOH is added to 0.650 L of 1.00 M Cu(NO₃)₂, how many grams of Cu(OH)₂ will be formed in the following precipitation reaction?
\[ 2 \text{ NaOH(aq) } + \text{ Cu(NO}_3\text{)}_2\text{(aq) } \rightarrow \text{ Cu(OH)}_2\text{(s) } + 2 \text{ NaNO}_3\text{(aq)} \]
**Details:**
- *NaOH* (Sodium Hydroxide): 24.0 grams
- *Cu(NO₃)₂* (Copper(II) Nitrate): 0.650 liters of 1.00 M solution
### Steps for Solving the Problem:
1. **Determine Moles of Reactants:**
- Calculate the moles of NaOH using its molar mass.
- Calculate the moles of Cu(NO₃)₂ using its molarity and volume.
2. **Identify the Limiting Reagent:**
- Use stoichiometry based on the balanced chemical equation.
3. **Calculate Moles of Product (Cu(OH)₂):**
- Determine moles of Cu(OH)₂ formed using the limiting reagent.
4. **Convert Moles of Cu(OH)₂ to Grams:**
- Use the molar mass of Cu(OH)₂ to find the grams produced.
### Visual Aid:
The page includes a numeric calculator interface for entering and calculating values. The calculator displays as:
- A number pad with digits (1-9, 0) and function keys (+, -, C for clear)
This interface assists in performing calculations directly while viewing the problem.
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