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

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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### 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
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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