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Essay The Preparation of Calcium Carbonate Lab

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11/17/2011 The Preparation of Calcium Carbonate Purpose: To create chalk (calcium carbonate) and to find the percentage yield in order to see the amounts of anhydrous sodium carbonate and calcium chloride were used up. Also to see if there’s any alterations like mass differentials. Objectives: 1. To introduce the concept of “limiting factor” in a chemical reaction 2. To practice a. Writing a balanced equation b. Determining the number of moles of each reactant and product c. Deciding which chemical is the limiting factor d. Predict theoretical yield e. Determine actual yield f. Use error discussion Materials: * 2 beakers * 2 watch glasses * Stirring rods * Filter …show more content…

6. The precipitate may have not dried up properly making it so there was water adding weight on it. 7. Errors with reading the scales, beaker and etc. Observations: Mass of CaCl2 (g) | Mass of Na2CO3 (g) | Mass of CaCO3 (g) | % yield (%) | 3.98 | 4.00 | 3.20 | 89.14 | 4.00 | 4.00 | 3.02 | 83.72 | 4.00 | 4.00 | 3.01 | 82.70 | 4.00 | 4.00 | 3.15 | 86.54 | 3.94 | 4.00 | 3.34 | 93.87 | 4.00 | 4.00 | 3.20 | 88.88 | 4.00 | 4.00 | 3.33 | 92.50 | 4.00 | 4.00 | 3.24 | 90.00 | Mean of % Yield = (89.14 + 83.72 + 82.7 + 86.54 + 93.87 + 88.88 + 92.5 + 90 + 97.5) ÷ 9 = 89.41 % Error Range 97.5-82.7 = (89.41% - 7.4%)(89.41% + 7.4%) 2 = (82.01% – 96.81%) =7.4 Questions and Calculations: 1. Step six in the procedure is necessary because there maybe some left over solution left over in the filter and so we use the water to get rid of that excess solution which is leftover. Step six is also important because this 5 mL of water could make the left over precipitate release excess water in it making it so that the drying process is quicker. 2. Na CO + CaCl CaCO + NaCl m=4.0g m= 4.0g 4g of Na CO x 1 mol of Na CO 4g of CaCl x 1mol of CaCl 106 g of Na CO 110g of CaCl = 0.0377 mol = 0.0364 mol Excess

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