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Lab Report: Determining Percent Yield in a Chemical Reaction
Date Submitted: 09/10/2006 03:19:41
Purpose:
To find out the percent yield of copper in the reaction between copper sulfate (CuSO4) and Iron (Fe).
Materials:
Balance
100-mL beaker
250-mL beaker
Bunsen burner
Copper sulfate crystals
Glass stirring rod
100-mL graduated cylinder
Iron filings
Ring stand and ring
Wire gauze
Procedure:
1. Record mass of clean 100-mL beaker.
2. Add 8.0 grams of copper sulfate crystals to beaker.
3. Add 50.0 milliliters of distilled water to the crystals.
4. Put wire gauze on ring on ring stand,
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copper<Tab/>2.54 g
Percent yield of copper<Tab/>-17.32%
Calculations:
<Tab/>Actual yield of copper = 69.43 g - 67.33 g = 2.1 g
Percent yield of copper = 2.1 g Cu - 2.54 g Cu
<Tab/><Tab/><Tab/><Tab/><Tab/> 2.54 g Cu
Conclusion:
The percent yield of copper is -17.32%.
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