Stoichiometry of a Precipitation Reaction

Topics: Solubility, Precipitation, Sodium Pages: 2 (555 words) Published: November 15, 2011
Stoichiometry of a Precipitation Reaction

The purpose of this lab is to calculate the theoretical, actual, and percent yield of the product from a precipitation reaction. Also, to learn concepts of solubility and the formation of a precipitate.

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. After that 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. After that 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.

Theoretical yieldActual yieldPercent yield
0.82g0.74g0.74g/0.82g x 100= 90.2%


A.From your balanced equation what is the theoretical yield of your product? 1g CaCl2 2H2O x 1 mol...
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