Experiment 9 okiemute
Name:
Exp 9: Stoichiometry of a Precipitation Reaction
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 *CaCO3 g
=0.0068 mol*100.01 g
=0.68 g
Convert moles of Na-2CO3 to grams of Na2CO3
= 0.00680 moles Na-2CO3 x 105.99g Na-2CO3
1 mole Na-2CO3
= 0.72g
0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O
Step 4:
Mass of weighing dish _0.5_g
Mass of weighing dish and Na2CO3 _1.2_g
Net mass of the Na2CO3 _0.7__g Step 6:
Mass of filter paper _1.0_g
Step 10:
Mass of filter paper and dry calcium carbonate __1.5_g
Net mass of the dry calcium carbonate __0.5__g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate.
0.00680 moles CaCO3 x 100 g CaCO3
1 mole CaCO3
= 0.68g CaCO3
Show the calculation of the percent yield.
Actual yield/Theoretical yield x 100
(0.5/0.68) x 100
= 73.5%
Conclusion:
The objectives of this experiment are 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. the theoretical yield and to calculate the percent yield. A Precipitation reaction begins by combining two aqueous solutions to form a precipitate, an insoluble product that is also a solid. With my knowledge of the measuring scale, I was able to use the scale with no problem throughout the whole experiment. The mass of the weighing dish was 0.5 grams.
Initially it was calculated that 0.68 g of CaCO3 was needed for a full reaction. The net mass of Na2CO3 (reactant) was 0.72 g and the net mass of CaCO3 was 0.5 g. I was able to use the graduated cylinder to measure the 25ml of distilled water, before pouring it into the beak and inside the beaker was a 1.0g of
Exp 9: Stoichiometry of a Precipitation Reaction
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 *CaCO3 g
=0.0068 mol*100.01 g
=0.68 g
Convert moles of Na-2CO3 to grams of Na2CO3
= 0.00680 moles Na-2CO3 x 105.99g Na-2CO3
1 mole Na-2CO3
= 0.72g
0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O
Step 4:
Mass of weighing dish _0.5_g
Mass of weighing dish and Na2CO3 _1.2_g
Net mass of the Na2CO3 _0.7__g Step 6:
Mass of filter paper _1.0_g
Step 10:
Mass of filter paper and dry calcium carbonate __1.5_g
Net mass of the dry calcium carbonate __0.5__g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate.
0.00680 moles CaCO3 x 100 g CaCO3
1 mole CaCO3
= 0.68g CaCO3
Show the calculation of the percent yield.
Actual yield/Theoretical yield x 100
(0.5/0.68) x 100
= 73.5%
Conclusion:
The objectives of this experiment are 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. the theoretical yield and to calculate the percent yield. A Precipitation reaction begins by combining two aqueous solutions to form a precipitate, an insoluble product that is also a solid. With my knowledge of the measuring scale, I was able to use the scale with no problem throughout the whole experiment. The mass of the weighing dish was 0.5 grams.
Initially it was calculated that 0.68 g of CaCO3 was needed for a full reaction. The net mass of Na2CO3 (reactant) was 0.72 g and the net mass of CaCO3 was 0.5 g. I was able to use the graduated cylinder to measure the 25ml of distilled water, before pouring it into the beak and inside the beaker was a 1.0g of