When using 1.0 g of CaCl2·2H2O, stoichiometric calculations should give the following results: 1 g CaCl2·2H2O x 1 mole CaCl2·2H2O x 1 mol Na2C03 x 106 g Na2C03 = 147 g CaCl2·2H2O 1 mol CaCl2·2H2O 1 mol Na2C03
Using 1 g of CaCl2·2H2O and .72 or .8 g (slight excess) Na2CO3 Should give a CaCO3 theoretical yield as follows:
1 g CaCl2·2H2O x 1 mole CaCl2·2H2O x 1 mol CaC03 x 100 g CaC03 = 147 g CaCl2·2H2O 1 mol CaCl2·2H2O 1 mol CaC03
To double-check, we can calculate CaCO3 theoretical yield by using Na2CO3 0.72 g Na2C03 x 1 mol Na2C03 x 1 mol CaC03 x 100 g CaC03 =
106 g Na2C03 1 mol Na2C03 1 mol CaC03
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Experiment Stoichiometry of a Precipitation Reaction
A. From your balanced equation what is the theoretical yield of your product? B. According to your data table, what is the actual yield of the product? C. What is the percent yield?
D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error.
E. How could these errors be reduced in the future?