Name: Tori Cohen
Date: 03/15/15
Exp 9: Stoichiometry of a Precipitation Reaction
Lab Section:

Data Tables:

Step 3: Show the calculation of the needed amount of Na2CO3

CaCl2 H2O(aq)=m/M
=1/147
=0.0068 mol
CaCO3(s)=0.0068 1/1
=0.0068 mol
CaCO3(s)=CaCO3 mol CaCO3g
=0.0068 mol 100.01g = .68g

Step 4:

Mass of weighing dish __0.6__g
Mass of weighing dish and Na2CO3 __0.72__g
Net mass of the Na2CO3 __0.12__g Step 6:

Mass of filter paper __1.0__g

Step 10:

Mass of filter paper and dry calcium carbonate __1.8__g
Net mass of the dry calcium carbonate _0.8___g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate.
0.0068 mol of CaCO3 100.06g CaCO3/1mol of CaCO3= 0.6804g of CaCO3

Show the calculation of the percent yield.
(Actual yield/theoretical yield)100=(0.8/0.68)100=117%

Conclusion:
In this experiment, stoichiometry was used to predict the amount of product made in a precipitation reaction. Using stoichiometry accurately measures the reactants and products of the reaction, which determines the actual yield vs the theoretical yield and accurately calculates the percent yield. The mass of the weighing dish was 0.6 grams, the mass of the weighing dish plus the sodium carbonate was 0.72 grams and the sodium carbonate alone was 0.12 grams. The filter paper and calcium carbonate was 1.8 grams and the calcium carbonate was 0.8 grams after the precipitation reaction. In this experiment