Question

There is only an Experiment 1 in this lab.

Experiment 1: Double Replacement Precipitation Reaction between Calcium Chloride and Sodium Carbonate

In this experiment you will precipitate calcium carbonate from the reaction between sodium carbonate and calcium chloride. The reaction is:

Procedure

Place a weigh boat on the scale. Once you have the mass of your weigh boat, press the button on the right hand side (0/T). Your scale should now read 0.0 g.Use your metal spatula to weigh out 2.0 g of CaCl2 in the weigh boat (the total mass should be 2.0 g). Record the exact mass of the powder in Table 9.Be sure to wipe off your spatula, gloves, and scale on a paper towel between chemical samples to avoid cross-contamination.Add the 2.0 g of CaCl2 to the 250 mL beaker. Use the 100 mL graduated cylinder to add 50 mL of distilled water to the beaker and mix with the glass stir rod until all CaCl2 has dissolved.

Note: This is an exothermic process, so the beaker may become warm.

Place a 50 mL beaker on the scale. Once you have the mass of your beaker, press the button on the right hand side (0/T). Your scale should now read 0.0 g.Use your metal spatula to weigh out 2.5 g of Na2CO3 in the 50 mL beaker. Record the exact mass of the powder in Table 9.Remove the beaker from the scale. Use your 100 mL graduated cylinder to add 25 mL of distilled water to the 50 mL beaker and mix with the glass stir rod until all Na2CO3 has dissolved. Add all of the Na2CO3 solution to the beaker containing the CaCl2 solution. It is important that all of the Na2CO3 is added. To ensure this, rinse the 50 mL beaker with up to 5 mL distilled water, and pour the rinse into the CaCl2 solution.Stir the solution for three minutes. Then, allow it to sit for 15 minutes. This gives sufficient time for all CaCO3 to precipitate. Record your observations in Table 9.While the solution is sitting, set up the gravity filtration apparatus (Figure 4). Place a funnel in the 250 mL Erlenmeyer flask, such that the bottom of the funnel is also inside the mouth of the flask. Obtain a piece of filter paper. Use the scale to weigh the filter paper and record the mass in Table 9.Prepare a filtering funnel as shown in Figure 5. Fold a piece of filter paper in half twice to make quarters, and place the paper in the funnel so that three quarters are open on one side and one quarter is on the opposite side. Place the paper into the funnel and seat with a small amount of distilled water (this will prevent the filter paper from rising up).  Slowly filter the solution from the beaker. Additional distilled water may also be used to transfer any remaining solid into the filtration apparatus. After all the solution has been filtered, use the pipette to rinse the filter paper with approximately 5 mL of isopropyl alcohol to aid the drying process. Allow the isopropyl alcohol to completely drip through the filter before removing filter paper from the funnel. Carefully remove the filter paper from the funnel and put it precipitate-side up in a safe place to dry overnight. Air drying will take anywhere from 5 to 18 hours, depending on the humidity of your region. Allow the product to dry, undisturbed, and determine the mass of the product recovered (Experimental Yield) by re-weighing the system and subtracting the weight of the filter paper. Record the data in Table 9.

Lab questions

Item	Mass (g)
Calcium Chloride CaCl2	2.0g
Sodium Carbonate Na2CO3	2.6g
Filter Paper	3.9g
Calcium Carbonate, CaCO3 (Experimental Yield)	1.3g

1. Which is the limiting reagent, CaCl2 or Na2CO3? What is the theoretical yield of the product CaCO3? Show all work to prove your answers. Use the correctly balanced equation given in the Procedure

2. What happens to the excess reactant after the reaction is complete?

3. Use the following equation to find the percent yield of CaCO3. Show all work. The Experimental Yield is found in Table 9 and the Theoretical Yield is from question 1 above.

4. Explain why an experimental yield from an experiment (not necessarily this one) could be LOWER than the theoretical yield (i.e. the percent yield is less than 100%). HUMAN ERROR and CALCULATION ERROR are unacceptable reasons. It is expected that you follow all procedures to the letter, and that you know how to do the calculations properly (or find appropriate help to get the calculations right).

5. Explain why an experimental yield from an experiment (not necessarily this one) could be HIGHER than the theoretical yield (i.e. the percent yield is more than 100%). HUMAN ERROR and CALCULATION ERROR are unacceptable reasons. It is expected that you follow all procedures to the letter, and that you know how to do the calculations properly (or find appropriate help to get the calculations right).
6. What is the theoretical yield of CaCO3 if 6.0 g of CaCl2 (instead of 2.0 g) is used for the reaction while the amount of Na2CO3 remains the same at 2.5 g? Support your answer with calculations and numerical values. Show all work. Again, re-read the Introduction the lab Moodle shell and the example before the Post Lab Questions.

Answer 1

Given that-

Calcium Chloride CaCl2	2.0g
Sodium Carbonate Na2CO3	2.6g
Filter Paper	3.9g
Calcium Carbonate, CaCO3 (Experimental Yield)	1.3g

mol of CaCl₂ = mass/ m.wt.

Mol of CaCl₂ = 2.0 / 110.98 = 0.018021265 mol

mol of Na₂CO₃ = 2.6 / 105.98

mol of Na₂CO₃ = 0.02453293 mol

The balanced reaction is - Na₂CO₃ + CaCl₂ ---> CaCO₃ + 2NaCl

Thus you can see both reactants inovolve their one mole to form one mole of CaCO3.

1- From the above calculations of moles it is known that CaCl₂is present in less amount thus this is your limiting reagent here.

2- Excess Na₂CO₃ remains unreacted after the reaction is complete and is removed via filteration process into the filtrate.

3 - As per balanced reaction the mole of CaCO₃ produced = moles of limiting reagent =  0.018021265 mol

Theoretical mass produced = mol * m.wt. of CaCO₃

Theoretical Mass = 0.018021265 * 100.0869 g

Theoretical Mass = 1.8 g

Obtained mass = 1.3 g

% y = 1.3 *100 / 1.8

% y = 72.22 %

4 - Two major factor could be for less yield -

(I) - We can not provide the proper reaction condition in our synthesis lab such as temperature pressure requirements.

(II)- Some of the products may remain stuck with filter paper after filteration.

5 - For higher %Y than 100% is mainly possible due to impurities or may be if some of byproduct of the reaction is also present in the product that will lead to give a higher mass thus higher %Y.

6 - For mass of CaCl₂ = 6.0 g

mole of CaCl₂ = 6.0 / 110.98

mole of CaCl₂ = 0.05406379 mole

mol of Na₂CO₃ remains same = 0.02453293 mol

Here your limiting reagent will be now Na₂CO₃ thus theoretical moles of product =  0.02453293 mol

Theoretical Mass = 0.02453293 * 100.0869 g

Theoretical Mass = 2.4554249  g