Please help! Initial: CaCl2•2H2O (g) 1.0 Initial: CaCl2•2H2O (moles) 0.068 Initi
ID: 493880 • Letter: P
Question
Please help!
Initial: CaCl2•2H2O (g)
1.0
Initial: CaCl2•2H2O (moles)
0.068
Initial: CaCl2 (moles)
0.068
Initial: Na2CO3 (moles)
0.068
Initial: Na2CO3 (g)
.7209
Theoretical: CaCO3 (g)
0.680
Mass of Filter paper (g)
1
Mass of Filter Paper + CaCO3 (g)
1.7
Actual: CaCO3 (g)
.72
% Yield:
103%
1. Let’s say we decided to run this experiment again. This time we used 1.0 gram of CaCl2·2H2O and 1.0 gram of Na2CO3. How many grams of CaCO3 would we produce? Please show/explain how you found your answer.
2. Of the two reactants in the experiment, one was the limiting reagent and the other was the excess reagent. Calculate the grams of the excess reagent still remaining in solution (using the amounts from the lab procedures).
3. What if we ran the experiment and used 1.5 grams CaCl2·2H2O and 1.0 gram Na2CO3. Show how many grams of CaCO3 would be produced.
4. Before the advent of Advil and Tylenol, did people simply have to “grin and bear it” when it came to pain? One of the most common ancient medicines for pain, fever, and inflammation came as a byproduct of the willow tree. While the first uses date back to 400 BCE, American historians cite the use of willow bark tea by the Lewis and Clark exploration party in the early 1800’s. Salicylic acid derived from the willow tree’s bark was the key chemical involved with the relief of pain and the reaction to make aspirin is a fairly simple one performed in numerous chemistry classrooms nationwide.
Aspirin can be made by reacting acetic anhydride (C4H6O3) with salicylic acid (C7H6O3) to form aspirin (C9H8O4).
C4H6O3 + C7H6O3 --> C2H4O2 + C9H8O4
When synthesizing aspirin, a student began with 3.05 mL of acetic anhydride (density = 1.08 g/mL) and 1.85 g of salicylic acid. The reaction was allowed to run its course and 1.84 grams of aspirin was collected by the student. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction.
5. In this lab, calcium carbonate was a precipitate. Use the solubility rules in Table 4.1 of your textbook to predict if the following reactions will form precipitates. If the reaction does not form a precipitate write “no reaction”. For the reactions that form a precipitate, write the balanced chemical equation and the net ionic equation for the reaction.
i. cobalt (II) nitrate and sodium iodide
ii. cobalt (II) nitrate and sodium phosphate
iii. barium nitrate and sodium sulfate
iv. nickel nitrate and sodium chloride
Also,
show work for all calculations required to generate the data in all the data tables from the lab. Show work for percent error calculation.
Conclusion: Summarize the main experimental finding of the lab. Calculate percent error for this value.
Initial: CaCl2•2H2O (g)
1.0
Initial: CaCl2•2H2O (moles)
0.068
Initial: CaCl2 (moles)
0.068
Initial: Na2CO3 (moles)
0.068
Initial: Na2CO3 (g)
.7209
Theoretical: CaCO3 (g)
0.680
Mass of Filter paper (g)
1
Mass of Filter Paper + CaCO3 (g)
1.7
Actual: CaCO3 (g)
.72
% Yield:
103%
Explanation / Answer
CaCl2 + Na2CO3 -- > CaCO3(S)
ratio is 1:1 so
if we use:
mass = 1 g of NA2CO3 --> mol = mass/MW= 1/105.9888 = 0.009434
previously, we used --> 0.068
since we only have 0.068 mol of CaCl2, this will be the limiting reactant
So actually, increasing amount of Na2CO3 will NOT increase our yield, expect 0.680 mol of CaCO3 or
mass = mol*MW = 0.0068*100 = 0.68 g of CaCO3
2. Of the two reactants in the experiment, one was the limiting reagent and the other was the excess reagent. Calculate the grams of the excess reagent still remaining in solution (using the amounts from the lab procedures).
Actually, we did not had excess/leftover, since this is 1:1 ratio
0.0068016 mol of CaCl2*H2O + 0.0068016 mol of NA2CO3 = 0.0068016 mol of CaCO3
3. What if we ran the experiment and used 1.5 grams CaCl2·2H2O and 1.0 gram Na2CO3. Show how many grams of CaCO3 would be produced.
now, we are increasing both species so:
mol of CaCl2*2H2O = mass/MW = 1.5/(146.98) = 0.0102 mol of CaCl2*2H2O
mass = 1 g of NA2CO3 -->
mol = mass/MW= 1/105.9888 = 0.009434 of Na2CO3
the Na2CO3 limits now, so 0.009434 mol of Na2CO3 --> 0.009434 mol of CaCO3 will form
mass = mol*MW = 0.009434 *100 = 0.94 g of CaCO3 will be produced
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