Percent Yield Calculations For each trial, show the theoretical yield of Calcium Carbonate produced based on your limiting reactant and amounts of each reactant used. Trial 1 Trial 2 od
![**Chemical Reaction Experiment Data Analysis**
**Data Table**
| **Data Entry** | **Trial 1** | **Trial 2** |
|---------------------------------|--------------------------------------------|--------------------------------------------|
| **Quantities Used** | 30.0 mL of 0.250M CaCl₂, 15.0 mL of 0.750M Na₂CO₃ | 50.0 mL of 0.250M CaCl₂, 10.0 mL of 0.750M Na₂CO₃ |
| **Theoretical Yield of Precipitate** | [Space for calculations] | [Space for calculations] |
| **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)** | [Space for calculations] | [Space for calculations] |
| **Percent Yield** | [Space for calculations] | [Space for calculations] |
**Calculations**
For each trial, show the theoretical yield of Calcium Carbonate produced based on your limiting reactant and amounts of each reactant used.
- **Trial 1**
- [Space for the theoretical yield calculation]
- **Trial 2**
- [Space for the theoretical yield calculation]
This table outlines the data collected from two experimental trials involving the reaction of Calcium Chloride (CaCl₂) and Sodium Carbonate (Na₂CO₃). The procedure involves measuring specified volumes of reactants, recording the mass of filter paper before and after the experiment, and determining the theoretical yield and percent yield of Calcium Carbonate precipitate.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c605e87-2c96-4aa2-99a1-2c5d12ea3556%2F663ecb77-3cbe-49dc-a8ec-e97353a24783%2Flpa14lr_processed.jpeg&w=3840&q=75)
CaCl2 + Na2CO3 = 2NaCl + CaCO3
Trial 1:
Molarity of CaCl2, M1 = 0.250 M
Volume of CaCl2, V1 = 30 ml = 30/1000 = 0.03 lts (1000 ml = 1 lt)
Number of moles of CaCl2, n = Molarity x Volume in liters = 0.250 mol/lt x 0.03 lts = 0.0075 moles
Molarity of Na2CO3, M2 = 0.750 M
Volume of Na2CO3, V2 = 15 ml = 15/1000 = 0.015 lts (1000 ml = 1 lt)
Number of moles of Na2CO3, n = Molarity x Volume in liters = 0.750 mol/lt x 0.015 lts = 0.01125 moles
Molar ratio of CaCl2 and Na2CO3 = 1:1
So, 1mole of CaCl2 require 1 mole of Na2CO3
0.0075 moles of CaCl2 require === 0.0075 x 1/1 = 0.0075 moles of Na2CO3 but the moles of Na2CO3 present is 0.01125 moles which are excess in number.
This shows that CaCl2 is the limiting reactant and Na2CO3 is the excess reactant
Molar ratio of CaCl2 and CaCO3 is 1:1
So, theoretical moles of CaCO3 are 0.0075 moles
Molar mass of CaCO3 = 100 g/mol (1x Ca + 1 x C +3 x O = 40 +12 +48 = 100 g/mol)
Theoretical yield of CaCO3 = Theoretical moles x molar mass = 0.0075 moles x 100 g/mol = 0.75 g
Practical Mass of CaCO3 = Mass of CaCO3 + filter paper after drying day 2) - (Mass of filter paper) = 1.33 - 0.53 g = 0.8
Percent yield of CaCO3 = [Experimental yield/theoretical yield] x 100 = [0.8/0.75] x 100 = 106.67 %
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