The purpose of this experiment is to use stoichiometry to predict how much of a product will be made in a precipitation reaction, to measure the reactants and products of the reaction correctly, to figure out the actual yield vs. the theoretical yield and to calculate the percent yield.
First, 1.0 g of CaCl2·2H2O was put into a 100-mL beaker and 25 mL of distilled water was added. The two ingredients were stirred to create a calcium chloride solution. Then, stoichiometry was used to determine how much Na2CO3 was needed for a full reaction: First, 1 g of CaCl22H2O was converted to moles: 0.
00680 moles. The mole ratios of CaCl22H2O and Na2CO3 was seen to be 1:1. Then, moles of Na2CO3 were converted to grams: 0 .72 g. The measure of CaCo3 was predicted to be 0.00680 moles. 0.00680 moles converted to grams is 0.68 grams.
Then, 0 .72 grams of Na2CO3 was measured into a paper cup because that was the measure calculated for Na2CO3 using stoichiometry in the step before.
25 mL of distilled water was added and stirred. Then, that solution was poured into the 100 mL beaker and it formed a precipitate (calcium carbonate) instantly with the calcium chloride solution.
Next, a filteration system was set up: A small cup was placed inside a larger cup for support and a funnel placed in to the small cup. Then, a 1.1 gram circle of filter paper was folded in half twice and one section of the folds in the filter paper was opened to fit into the funnel.
Then, the solution was poured slowly into the funnel. After all the liquid strained through the filter system, the filter paper with its contents which did not strain through was put aside on a few paper towels to dry. Once it was dry, the filter paper was weighed again and the weight was 1.9 grams. The initial weight of the filter paper was subtracted from 1.9 grams, leaving 0.8 grams of precipitate.
Then, using the theoretical yield and actual yield, the percent yield was figured out: .80/ .68= 1. 176 117.6%.
1 g of CaCl22H2O was converted to moles| 0.00680 moles|
mole ratios of CaCl22H2O and Na2CO3| 1:1|
moles of Na2CO3 were converted to grams| .72 g|
. 0.00680 moles converted to grams| 0.68 grams|
initial weight of the filter paper subtracted from final weight with precipitate| 1.9 – 1.1= .8 g| using the theoretical yield and actual yield, the percent yield was calculated| .80/ .68= 1. 176 117.6%.|
No additional questions.
This experiment was successful. Stoichiometry was used to predict how much of a product will be made in a precipitation reaction, the reactants and products of the reaction were measured, the actual yield vs. the theoretical yield was figured out and the percent yield was calculated.
Remember! This is just a sample.
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