CALCULATIONS WITH A CHEMICAL REACTION 6B Objectives: 1. To observe the reaction between solutions of calcium chloride and sodium carbonate, forming insoluble calcium carbonate 2. To calculate the number of moles of each of the starting materials to present in the solutions 3. To determine the reactant that is in excess 4. To determine the theoretical amount of calcium carbonate that could be produced 5. To compare the theoretical amount to the actual amount of calcium carbonate and calculate the percent yield - Ring stand and ring support Safety goggles Lab apron Drying oven 0.60M sodium carbonate solution, Na2CO3 0.40M calcium chloride solution, CaCl2 Distilled water is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m Materials: - Centigram balance - Graduated cylinder (100mL) - Funnel - Filter paper - Wash bottle - 2 beakers (250mL) - Stirring rod with rubber scraper - Post Lab Discussion: Percent yield = actual amount (g) theoretical amount (g) = _1.04g_ x 100 = 104% 1.00g Data and Observations: Volume of Na2CO3(aq) Volume of CaCl2(aq) Describe what happens after mixing the two solutions Mass of dry filter paper Mass of filter paper + dry solid x100 35.0 mL 25.0 mL Turns instantaneously white, like curdled milk. Then starts to become clearer on the top 0.82 g 1.86 g sh Th Questions and Calculations: 1. Calculate the following values: a. Moles of sodium carbonate used: 35.0mL Na2CO3 x __1L___ x 0.60mol Na2CO3 = 0.021mol Na2CO3 1000mL 1L b. Moles of calcium chloride used: 25.0mL CaCl2 x __1L___ x 0.40mol CaCl2 = 0.01mol CaCl2 1000mL 1L c. Mass of calcium carbonate produced: This study source was downloaded by 100000829350039 from CourseHero.com on 11-03-2021 20:43:10 GMT -05:00 https://www.coursehero.com/file/9148310/CALCULATIONS-WITH-A-CHEMICAL-REACTION-lab/ 1.86g – 0.82g = 1.04g CaCO3 d. Moles of calcium carbonate produced: 1.04g CaCO3 x ___1mol____ = 1.03 x 10-2mol CaCO3 100.1g CaCO3 2. Write a balanced chemical equation for the reaction that you observed in this experiment: Na2CO3 + CaCl2 → 2NaCl + CaCO3 3. Determine which of the reactants was in excess in this reaction 0.21mol Na2CO3 x 1mol CaCO3 x 100.1g CaCO3 = 21.0g CaCO3 1mol Na2CO3 1mol is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m 0.01mol CaCl2 x 1mol CaCO3 x 100.1g CaCO3 = 1.00g CaCO3 1mol CaCl2 1mol Na2CO3 is the excess reactant. 4. Calculate the amount of calcium carbonate that should theoretically form from the amount of the limiting reactant 0.01mol CaCl2 x 1mol CaCO3 x 100.1g CaCO3 = 1.00g CaCO3 1mol CaCl2 1mol 5. Calculate the percent yield in your reaction: _1.04g_ x 100 = 104% 1.00g 6. Predict what would happen to the percent yield (greater than, less than, or no change) if the following occurred: a. The solid was not completely dry: greater b. The balance measured all values over by 0.12g: no change c. You mixed up the volumes of the two liquids: no change sh Th Follow-up Questions: 1. Suppose you wanted to add just enough 0.40M calcium chloride solution to 75.0mL of 0.60M sodium carbonate solution for all of the sodium carbonate and calcium chloride to react. a. what volume of 0.40M calcium chloride solution would be required? 75.0mL Na2CO3 x __1L __ x 0.60molNa2CO3 x 1molCaCl2 x _1L NaCl2_ = 0.11L NaCl2 1000mL 1L 1molNa2CO3 0.40MNaCl2 b. what mass of calcium carbonate would be produced, assuming a 100% yield? 0.045mol Na2CO3 x 1molCaCO3 x 100.1gCaCO3 = 4.5g CaCO3 1mol Na2CO3 1mol CaCO3 c. what mass of sodium chloride would be produced, assuming a 100% yield? 0.045molNa2CO3 x 2mol NaCl x 58.5gNaCl = 5.3g NaCl 1mol Na2CO3 1mol NaCl 2. How would you be able to recover the sodium chloride from the solution? You decant the solution (or filter it to remove the precipitate) and then boil it until all the water evaporates This study source was downloaded by 100000829350039 from CourseHero.com on 11-03-2021 20:43:10 GMT -05:00 https://www.coursehero.com/file/9148310/CALCULATIONS-WITH-A-CHEMICAL-REACTION-lab/ is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m 3. A similar reaction occurs when barium chloride solution is mixed with sodium carbonate solution. write a balanced equation for the reaction. Na2CO3 + BaCl2 → BaCO3 + 2NaCl 4. calculate each of the following for the reaction that occurs when 56.0mL of 0.50M barium chloride solution are mixed with 78.0mL of 0.75M sodium carbonate solution a. moles of barium chloride added: 56.0mL BaCl2 x 0.5mol BaCl2 x __1L__ = 0.028 mol BaCl2 1L 1000mL b. moles of sodium carbonate added: 78.0mLNaCl2 x 0.75 molNaCl2 x __1L___ = 0.058molNaCl2, you add excess of Na2CO3 1L 1000mL c. mass of barium carbonate produced, if the yield was 78%: 0.028 mol BaCl2 x 207.3 g BaCl2 = 5.80 g BaCO3 if the reaction was 100% completed 1mol 0.78 x 5.80g = 4.53 g BaCO3 sh Th Conclusion: Na2CO3 + CaCl2 → 2NaCl + CaCO3 Through this experiment, we can find out that when sodium carbonate and calcium chloride react, sodium carbonate is the excess reactant, and calcium chloride is the limiting. The theoretical amount of calcium carbonate that should be produced is 1.00g, whereas the actual amount was 1.04g, therefore the percent yield was 104%. The difference could be due to an imprecise measurement of volume of the solutions or contaminants, either in the actual solutions or in the beakers. The same method of calculations could be applied to other compounds, such as barium chloride solution mixed with sodium carbonate solution. In both cases, you must first find limiting and excess reactions before going on to deduce how much of each reagent is used. This study source was downloaded by 100000829350039 from CourseHero.com on 11-03-2021 20:43:10 GMT -05:00 https://www.coursehero.com/file/9148310/CALCULATIONS-WITH-A-CHEMICAL-REACTION-lab/ Powered by TCPDF (www.tcpdf.org)
