Stoichiometry of a Precipitation Reaction
Name: Brian James|Date:3/10/13|
Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: 73426|
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 (s)= CaCO3 mol *CaCO3 g
=0.0068 mol*100.01 g
=.68 g
Step 4:
Mass of weighing dish _0.6_g
Mass of weighing dish and Na2CO3 .72_g
Net mass of the Na2CO3 .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 __.8__g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate.
.0068 mol of CaCO3 *100.06 g CaCO3/1mol of CaCO3
=.6804 g of CaCO3
Show the calculation of the percent yield.
(Actual yield/theoretical yield)*100
= (.8/.68)*100
=117%
Conclusion:
The primary objectives of this experiment were to predict the amount of product produced in a precipitation reaction using stoichiometry, to accurately measure the reactants and products of the reaction, to determine the actual yield vs theoretical yield, and to accurately calculate percent yield.
The mass of the weighing dish was .6 grams and the mass of the weighing dish plus the sodium carbonate was .72 grams thus the mass of the sodium carbonate alone was .12 grams. After the precipitation reaction the mass of the filter paper and calcium carbonate was 1.8 grams and the calcium carbonate was .8 grams. Through this experiment we used a precipitation reaction by combining 2 aqueous solutions to form an insoluble solid (calcium carbonate). I got 117% in the percent yield which means that I made an error. The error could have been that it was not fully dried throughout even though I left it for a week to dry. The only thing I can think of is that the humidity and temperature of my apartment are not ideal for the
Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: 73426|
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 (s)= CaCO3 mol *CaCO3 g
=0.0068 mol*100.01 g
=.68 g
Step 4:
Mass of weighing dish _0.6_g
Mass of weighing dish and Na2CO3 .72_g
Net mass of the Na2CO3 .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 __.8__g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate.
.0068 mol of CaCO3 *100.06 g CaCO3/1mol of CaCO3
=.6804 g of CaCO3
Show the calculation of the percent yield.
(Actual yield/theoretical yield)*100
= (.8/.68)*100
=117%
Conclusion:
The primary objectives of this experiment were to predict the amount of product produced in a precipitation reaction using stoichiometry, to accurately measure the reactants and products of the reaction, to determine the actual yield vs theoretical yield, and to accurately calculate percent yield.
The mass of the weighing dish was .6 grams and the mass of the weighing dish plus the sodium carbonate was .72 grams thus the mass of the sodium carbonate alone was .12 grams. After the precipitation reaction the mass of the filter paper and calcium carbonate was 1.8 grams and the calcium carbonate was .8 grams. Through this experiment we used a precipitation reaction by combining 2 aqueous solutions to form an insoluble solid (calcium carbonate). I got 117% in the percent yield which means that I made an error. The error could have been that it was not fully dried throughout even though I left it for a week to dry. The only thing I can think of is that the humidity and temperature of my apartment are not ideal for the