Na₂ LU 1069/m01 2. If you had to make a 0.250M solution of CaCl2 from 100mL of solution, determine the mass of Calcium Chloride necessary. 250m 1000 me X100ML = ·025ML of
Na₂ LU 1069/m01 2. If you had to make a 0.250M solution of CaCl2 from 100mL of solution, determine the mass of Calcium Chloride necessary. 250m 1000 me X100ML = ·025ML of
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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Please help with question 2
![### Educational Text on Stoichiometry and Solution Preparation
#### 1. Calculating Mass for Sodium Carbonate Solution:
If you need to prepare a 0.750 M solution of Na2CO3 from 100 mL of solution, determine the mass of Sodium Carbonate necessary.
- Calculation:
- Using volume and concentration:
\[
\frac{100 \text{ mL}}{1000 \text{ mL/L}} \times 0.750 \text{ mol/L} = 0.075 \text{ mol}
\]
- Molar mass of Na2CO3 = 106 g/mol:
\[
\text{Mass of Na2CO3} = 106 \text{ g/mol} \times 0.075 \text{ mol} = 7.95 \text{ g}
\]
#### 2. Calculating Mass for Calcium Chloride Solution:
If you need to prepare a 0.250 M solution of CaCl2 from 100 mL of solution, determine the mass of Calcium Chloride necessary.
- Calculation:
- Using volume and concentration:
\[
\frac{0.250 \text{ mol/L} \times 100 \text{ mL}}{1000 \text{ mL/L}} = 0.025 \text{ mol}
\]
#### 3. Reaction of Sodium Carbonate with Calcium Chloride:
Using 10.0 mL of the Sodium Carbonate solution and an excess of Calcium Chloride, determine how many grams of Calcium Carbonate should be theoretically produced.
- Reaction equation:
\[
\text{Na}_2\text{CO}_3 + \text{CaCl}_2 \rightarrow \text{CaCO}_3 + 2\text{NaCl}
\]
- Calculation based on given molarities and volumes.
#### 4. Reaction of Calcium Chloride with Sodium Carbonate:
Using 25.0 mL of the Calcium Chloride solution and an excess of Sodium Carbonate, determine how many grams of Calcium Carbonate should be theoretically produced.
#### 5. Theoretical Yield Calculation:
Using information in problems 3 and 4, calculate the theoretical yield of 25.0 mL of a 0.250 M solution of CaCl2 when mixed with 10.0 mL of a 0.750](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c605e87-2c96-4aa2-99a1-2c5d12ea3556%2Fb0ac518f-b95e-407e-b052-ee9f93b2a262%2F8rhskfa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Text on Stoichiometry and Solution Preparation
#### 1. Calculating Mass for Sodium Carbonate Solution:
If you need to prepare a 0.750 M solution of Na2CO3 from 100 mL of solution, determine the mass of Sodium Carbonate necessary.
- Calculation:
- Using volume and concentration:
\[
\frac{100 \text{ mL}}{1000 \text{ mL/L}} \times 0.750 \text{ mol/L} = 0.075 \text{ mol}
\]
- Molar mass of Na2CO3 = 106 g/mol:
\[
\text{Mass of Na2CO3} = 106 \text{ g/mol} \times 0.075 \text{ mol} = 7.95 \text{ g}
\]
#### 2. Calculating Mass for Calcium Chloride Solution:
If you need to prepare a 0.250 M solution of CaCl2 from 100 mL of solution, determine the mass of Calcium Chloride necessary.
- Calculation:
- Using volume and concentration:
\[
\frac{0.250 \text{ mol/L} \times 100 \text{ mL}}{1000 \text{ mL/L}} = 0.025 \text{ mol}
\]
#### 3. Reaction of Sodium Carbonate with Calcium Chloride:
Using 10.0 mL of the Sodium Carbonate solution and an excess of Calcium Chloride, determine how many grams of Calcium Carbonate should be theoretically produced.
- Reaction equation:
\[
\text{Na}_2\text{CO}_3 + \text{CaCl}_2 \rightarrow \text{CaCO}_3 + 2\text{NaCl}
\]
- Calculation based on given molarities and volumes.
#### 4. Reaction of Calcium Chloride with Sodium Carbonate:
Using 25.0 mL of the Calcium Chloride solution and an excess of Sodium Carbonate, determine how many grams of Calcium Carbonate should be theoretically produced.
#### 5. Theoretical Yield Calculation:
Using information in problems 3 and 4, calculate the theoretical yield of 25.0 mL of a 0.250 M solution of CaCl2 when mixed with 10.0 mL of a 0.750
Expert Solution
Step 1
Molarity is a concentration term for a solution. The molarity of the given solution is defined as the number of moles of solute present in one liter of solution. One molar is the molarity of a solution where one mole of solute is dissolved in a liter of solution.
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