Percent Yield

1. 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. (3 points) Table of Masses

Empty Dish
NaHCO
HCl+NaCO3
Final (Dish and Salt Product
24.35 (grams)
10.06 (grams)
40.06 (grams)
31.52 (grams)

2. 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. (3 points)

NaHCO3 + HCl → CO2 + H2O + NaCl

3. 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. (5 points) NaHCO3 – The Empty Dish

37.06 (grams) – 24.35 (grams) = 12.71 (grams) NaHCO3
12.71 (grams) NaHCO3 ÷ 84.01 (grams/mole) = 0.1513 moles of NaHCO3 0.135 Moles of NaHCO3 × 58.4428 (Molar mass of NaCl)
Giving me 8.8240 (grams) NaCl as my theoretical yield.

4. What is the actual yield of NaCl in your experiment? Show your work below. (4 points) The actual yield is:
31.52 (grams) – 24.35 (grams) = 7.14 (grams) NaCl
(After the evaporation of water)
5. Determine the percent yield of NaCl in your experiment, showing all work
neatly in the space below. (5 points) Actual yield ÷ Theoretical yield
7.17 (grams) ÷ 8.8420 (grams) = 81% Yield

6. 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. (5 points) If the product was not heated long enough (or not heated at all), the actual yield would increase. The product would be slightly heavier do to the fact that it would still be partially damp to the lack of evaporation of water. The actual yield would be increased, therefore the percent yield would increase since the increased yield would be in its numerator.