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04.08 Percent Yield: Percent Yield Lab Report

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cheyenne leksell

on 13 July 2014

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Transcript of 04.08 Percent Yield: Percent Yield Lab Report

04.08 Percent Yield: Percent Yield Lab Report
By: Cheyenne Leksell
Create a data table to record your data from the procedure. Be sure that the data table is neat and organized and that all measurements have the correct significant figures and units.
Mass of the Empty Dish
Mass of the Dish with NaHCO3
Mass of the Dish after Adding Drops of HCI
Mass of the Dish and the Resulting Salt
24.35 Grams
37.0 Grams
40.6 Grams
31.52 Grams
Write the complete balanced equation for the reaction that occurred in this lab. Hint: H2CO3 is not a final product of the double-replacement reaction; it breaks down (decomposes) immediately into two products.
NaHCO3 + HCI CO2 + H2O + NaCI
The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product, showing all of your work in the space below.
1. NaHCO3 minus the Empty Dish
37.06 Grams - 24.35 Grams = 12.71 Grams NaHCO3

2. NaHCO3 / 84.01 Grams Mol = 0.1513 Mol of NaHCO3

3. .1513 mols x Mass of NaCI = 8.8420 Grams NaCI

Theoretical Yield is 8.8420 Grams NaCI
What is the actual yield of NaCl in your experiment? Show your work below.
Mass of Dish and Salt - Mass of Empty Dish = Yield of NaCI
31.52 grams - 24.35 grams = 7.14 g NaCI
Actual Yield of NaCI is 7.14 grams of NaCI
Determine the percent yield of NaCl in your experiment, showing all work neatly in the space below.
Actual Yield Theoretical Yield = % Yield of NaCI
7.17 grams
8.8420 grams
Percent Yield of NaCI is 81%
If you had not heated the product long enough to remove all of the water, explain in detail how that would have specifically affected your calculated actual yield and percent yield.
If I had not heated the product long enough to remove all of the water, the theoretical yield would have been higher. This is because the water would cause the salt product to be heavier since it is not dried out. When calculating, the calculation would increase because I would have had to use higher numbers due to its heavier mass.
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