# Stoichiometry Lab

Questions

A. From your balanced equation, what is the theoretical yield of your product?

Theoretical yield of the CaCO3 is expected to be .69g.

B. According to your data table, what is the actual yield of the product?

The mass of the filter paper was 1.1g, and the total mass of the filter paper when dried with the CaCO3 was 1.8 total. Thus the actual yield of the product was .70g.

C. What is the percent yield?

Percentage yield is actual yield over the theoretical yield x100. Actual yield: .70gTheoretical yield: .69g

percentage yield: 101%

D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error. My percentage yield is actually quite close to a perfect percentage yield, which I am pleased about. The theoretical vs actual in this experiment is so close where the possible sources of error are quite slim in this case. It is always possible that all of the substance from the beaker was not accurately weighed due to some CaCO3 reside being left on the sides of the beaker.

Na2CO3 + CaCl22H2O = CaCO3 + 2NaCl + 2H2O

Convert 1g of CaCl22H2O to moles, 1g CaCl22H2O x 1mol = .00689Moles CaCl22H2O 144.9994

Consider the mole ratio of Na2CO3 to CaCl22H2O

1:1

Thus, 1 mole of CaCl22H2O will equal 1 mole of Na2CO3

Convert moles of Na2CO3 to grams: .00689moles Na2CO3 x 105.99 = .73027g Na2CO3 1 mole Na2CO3

How much CaCO3 can we expect?

.00689 moles of CaCl22H2O x 1mole CaCO3 = .00689Moles CaCO3 1mole CaCl22H2O

Convert .00689Moles CaCO3 to grams, .00689Moles CaCO3 x 100.7 = .69gCaCO3 produced 1MoleCaCl2

Double check by calculating the amount of 2NaCl + 2H2O

.00689molesCaCO3 x 2moles H2O = .01378Moles H2O x 18.02 =.2483g 2H2O 1mole CaCO3 1mole H2O

.00689molesCaCO3 x 2moles NaCl = .01378moles NaCl x 58.44 = .8053g 1mole CaCO3 1 mole NaCl

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