Stiochiometry of a Precipitaiton Reaction Essays

Stiochiometry of a Precipitaiton Reaction Essays Stiochiometry of a Precipitaiton Reaction Essay Stiochiometry of a Precipitaiton Reaction Essay Abstract: The objective of this lab is to calculate the theoretical, actual, and percent yield of the product from a precipitation reaction. It is also to learn concepts of solubility and the formation of a precipitate. The Experiment and Observation: Weigh out your 1. 0g of CaCl2-2H20 and put it into the 100mL beaker, add your 25mL of distilled water and stir to form the calcium chloride solution. Next, use stoichiometry to determine how much Na2CO3 and put it into a small paper cup. Then add the 25mL of distilled water to make the sodium carbonate solution. Mix the two solutions in the beaker and a precipitate of calcium carbonate will form instantly. Next, set up your filtration assembly. After the filtration assembly is ready, swirl the contents of the beaker to dislodge any precipitate from the sides. Then, pour the content of the beaker into the filter paper-lined funnel carefully. Afterwards you will need to measure out 2 to 5mL of distilled water into a graduated cylinder. Pour it down the sides of the beaker, swirl, and pour into the filter paper-lined funnel. Once all the liquid has drained from the funnel, lay the filter paper containing the precipitate on folded layers of paper towels and set it somewhere to air-dry. Once the filter paper and the precipitated calcium carbonate are completely dry, weigh them, subtract the original weight of the empty filter paper, and record the net weight of the calcium carbonate. That is your actual yield of calcium carbonate. Then, you can calculate the percent yield, using your theoretical yield and actual yield. Be sure to clean up properly, rinse any remaining chemicals down the sink and throw paper cups and towels in the garbage. Clean and dry all equipment you used. The following results will be observed: 1g CaCl*2H? O*(1moleCaCl? *2H? O/147g CaCl*2H? O)*(1mol Na? CO? /1mol CaCl? *2H? O)*160g Na? CO? /1mo lNa? CO? = 1g of CaCl? *2H? O and . 72 Na? CO? Which will provide a Ca CO? Theoretical yield as : 1 g CaCl22H2O x 1 mole CaCl22H2O x 1 mol CaC03 x 100 g CaC03 =. 68 g CaCO? 147 g CaCl22H2O 1 mol CaCl22H2O 1 mol CaC03 Then double check our work: 0. 72 g Na2C03 x 1 mol Na2C03 x 1 mol CaC03 x 100 g CaC03 =. 68g CaCO? 106 g Na2C03 1 mol Na2C03 1 mol CaC03 Questions A. From your balanced equation what is the theoretical yield of your product? o. 68 g CaCO? B. According to your data table, what is the actual yield of the product? .7G C. What is the percent yield? 97% D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error. The results were off by 3% It could be from the weight to a slight change in the amount of solution. As an example some of the solution could have spilled during measuring and weighing. E. How could these errors be reduced in the future? The errors could be reduced by watching the product more closely and ensuring the scale is accurate prior to use. F. Part A- 1g CaCl? *1mol CaCl? =. 00901 mols 110. 984 .00901*110. 984/1=. 999 g CaCl? First we convert grams into mols. Then we calculate the mols into grams based on the mass of the product. Part B- . 00042 excess reagent in still remaining in solution G. The limiting reactiant is salicylic acid. The theoretical yield is . 113moles. The percent yield is 88. 9 %