Stoichiometry of a Precipitation Reaction

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  • Topic: Stoichiometry, Solubility, Calcium carbonate
  • Pages : 3 (567 words )
  • Download(s) : 3867
  • Published : March 24, 2013
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Data Tables:

Step 3: Show the calculation of the needed amount of Na2CO3 Convert 1.0g of CaCl2-.2H2O to moles of CaCl2-.2H2O

1.0g x 1 mole CaCl2-.2H2O
147.0 g CaCl2-.2H2O

= 0.00680 moles CaCl2-.2H2O

The mole ratio is 1:1

Hence if we have 0.00680 moles of CaCl2-.2H2O we will as well need 0.00680 moles of


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

This means that we need 0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O

Step 4:

Mass of weighing dish_0.7___g
Mass of weighing dish and Na2CO3__1.4__g
Net mass of the Na2CO3 __0.7__g

Step 6:

Mass of filter paper __0.7__g

Step 10:

Mass of filter paper and dry calcium carbonate__1.2__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. The mole ration between CaCl2-.2H2O and CaCO3 is 1:1 that means that if we have

0.00680 moles of CaCl2-.2H2O we will get 0.00680 moles CaCO3

Convert the moles of CaCO3 to grams of CaCO3

= 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 objective of the experiment is to predict the amount of product produced in a precipitation reaction using stoichiometry. Secondly, the experiment accurately measures the reactants and products of a reaction. Also, the experiment is to determine actual yield vs. theoretical yield and to calculate the percent yield. For example in this experiment, we were able to predict that we need 0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O. Another example is that, we calculate the amount of...
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