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Mass of Calcium Carbonate (Precipitate, day 2) Percent Yield

Mass of Calcium Carbonate (Precipitate, day 2) Percent Yield

Chemistry
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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### Educational Transcription of Scientific Experiment #### Data Table This table presents the results of two trials conducted to determine the yield of Calcium Carbonate in a chemical reaction. The experiment involved the mixing of Sodium Carbonate (Na₂CO₃) and Calcium Chloride (CaCl₂). | Data Entry | Trial 1 | Trial 2 | |----------------------------------------|---------------------------|---------------------------| | **Quantities Used** | 30.0 mL of 0.250M CaCl₂, | 50.0 mL of 0.250M CaCl₂, | | | 15.0 mL of 0.750M Na₂CO₃ | 10.0 mL of 0.750M Na₂CO₃ | | **Theoretical Yield of Precipitate** | 0.75 g | 0.75 g | | **Mass of Filter Paper BEFORE experiment** | 0.53 g | 0.53 g | | **Mass of DRY Filter Paper and Precipitate (day 2)** | 1.33 g | 1.33 g | | **Mass of Calcium Carbonate (Precipitate, day 2)** | | | | **Percent Yield** | | | #### Calculations Below are the calculations performed to determine the moles of reactants and theoretical yield of Calcium Carbonate: - **Trial 1**: - \( \text{CaCl}_2: (0.250 \times 30) / 1000 = 0.0075 \text{ moles} \) - \( \text{Na}_2\text{CO}_3: 0.750 \times 15 / 1000 = 0.0112 \text{ moles} \) - \(\text{Theoretical Yield: 0.0075} \times 160.2 = \text{0.75 g} \) - **Trial 2**: - \( \text{CaCl}_2: (0.250 \times 50) / 1000 = 0.0125 \text{ moles} \) - \( \text{Na}_2\text{CO}_3: 0.750 \times 10 / 100
Transcribed Image Text:### Educational Transcription of Scientific Experiment #### Data Table This table presents the results of two trials conducted to determine the yield of Calcium Carbonate in a chemical reaction. The experiment involved the mixing of Sodium Carbonate (Na₂CO₃) and Calcium Chloride (CaCl₂). | Data Entry | Trial 1 | Trial 2 | |----------------------------------------|---------------------------|---------------------------| | **Quantities Used** | 30.0 mL of 0.250M CaCl₂, | 50.0 mL of 0.250M CaCl₂, | | | 15.0 mL of 0.750M Na₂CO₃ | 10.0 mL of 0.750M Na₂CO₃ | | **Theoretical Yield of Precipitate** | 0.75 g | 0.75 g | | **Mass of Filter Paper BEFORE experiment** | 0.53 g | 0.53 g | | **Mass of DRY Filter Paper and Precipitate (day 2)** | 1.33 g | 1.33 g | | **Mass of Calcium Carbonate (Precipitate, day 2)** | | | | **Percent Yield** | | | #### Calculations Below are the calculations performed to determine the moles of reactants and theoretical yield of Calcium Carbonate: - **Trial 1**: - \( \text{CaCl}_2: (0.250 \times 30) / 1000 = 0.0075 \text{ moles} \) - \( \text{Na}_2\text{CO}_3: 0.750 \times 15 / 1000 = 0.0112 \text{ moles} \) - \(\text{Theoretical Yield: 0.0075} \times 160.2 = \text{0.75 g} \) - **Trial 2**: - \( \text{CaCl}_2: (0.250 \times 50) / 1000 = 0.0125 \text{ moles} \) - \( \text{Na}_2\text{CO}_3: 0.750 \times 10 / 100
Expert Solution
Step 1

Mass of dry filter paper = 0.75g

Mass of dry filter paper + precipitate = 1.33g

So, for, trial 1 and 2:

Mass of precipitate (Calcium Carbonate) = 1.33g - Mass of dry filter paper

Mass of precipitate (Calcium Carbonate) = 1.33g - 0.75g

Mass of precipitate (Calcium Carbonate) = 0.58g

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