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
Final (Dish and Salt Product
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:
PROJECT ON “ARVIND MILL (THE PRODUCT MIX AND ITS STRATEGY)” Master of Commerce Semester-I (2013-2014) Submitted In Partial Fulfillment of the requirements For the award of degree of M.Com-I By Suraj Shridhar Tripathi Seat No: _______ Tolani College of Commerce Sher-e-Punjab society, Andheri (East), Mumbai-400 093 PROJECT ON “ARVIND MILL (THE PRODUCT MIX AND ITS STRATEGY)” Master of Commerce ...
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.