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Solution Stoichiometry

The document details a laboratory experiment conducted at Dawson College on solution stoichiometry, specifically focusing on the reaction between Na2CO3 and CaCl2. Key data includes the mass of the dry product, moles of reactants, limiting reagent identification, and calculated yields. The report also addresses pre-laboratory and post-laboratory questions regarding the experiment's procedures and outcomes.

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202 views3 pages

Solution Stoichiometry

The document details a laboratory experiment conducted at Dawson College on solution stoichiometry, specifically focusing on the reaction between Na2CO3 and CaCl2. Key data includes the mass of the dry product, moles of reactants, limiting reagent identification, and calculated yields. The report also addresses pre-laboratory and post-laboratory questions regarding the experiment's procedures and outcomes.

Uploaded by

boutannoura4
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Dawson College

Solution Stoichiometry Laboratory #2

Fatima Ezzahra Boutannoura

2344428

Partner : Marwa Bouladjine

Angélique Fortier

202-NYA-05 section 21

Date performed: September 26th, 2023.

Date submitted: October 03rd, 2023.


Calculated Data:

1- Mass of dry product: 38.5560g – 38.3086g = 0,2484g

2- Moles of CaCl2 used: 0.2728mol/L x 0.0100L = 2.73x10^-3 mol

3- Moles of Na2CO3 used: 0.3330mol/L x 0.0103L = 3.43x10^-3 mol

4- Limiting reagent: CaCl2

5- Calculated mass of excess reagent remaining in the mixture after reaction:

−3 1 mol N a 2 CO 3 −3
a) 2.73 x 10 mol CaCl2 × =2.73 x 10 m ol Na 2 CO 3
1mol CaCl 2
b) 3. 43 x 10−3 mol−2.73 x 10−3 mol = 7.0x10−4 mol Na2 CO 3

−4 1 05. 99 g
c) 7 .0 x 10 mol Na2 CO 3 × =0.0 7420 g Na2 CO 3
1 mol

=0.27 3 2 g CaCO3
−3 1 molCaCO 3 100 ,09 gCaCO 3
6- Theoretical yield: 2.73 x 10 mol × ×
1 molCaCl 2 1 molCaCO 3

0.2484 g
7- % Yield: ×100=90.9 %
0.27 32 g
Pre-Laboratory Questions:

1- M.E.: Na2CO3(aq) + CaCl2(aq) ⟶ CaCO3(s) + 2NaCl(aq)


C.I.E.: 2Na+(aq) + (CO3) -(aq) + Ca2+(aq) + 2Cl-(aq) ⟶ CaCO3(s) + 2Na+(aq) + 2Cl-(aq)

N.I.E.: Ca2+(aq) + (CO3) –(aq) ⟶ CaCO3(s) or CaCO3(aq) ⟶ CaCO3(s)


Spectators: 2Na+, 2Cl-

2- M=n/V a) n = 0.1437M x 0.01L b) V = 0.001437mol/0.0725M


n=MxV n = 0.001437 mol V = 1.04 x 10−4
V=n/M
10.00mL
=0.01L

Post-Laboratory Questions:

1- If it wasn’t completely dissolved, we would not have had the concentration of the solution we
wanted to work with.
2- a) We want to make sure that there is no water left in the precipitate because we only want the
mass of CaCO3
b) If we omit this step, the experimental yield would not have been a 100% CaCO3, but CaCO3
plus water, which will give us a wrong % yield.
3- a) Adding a little bit of water helps us get rid of the solvent and only leaves the solute which is
CaCO3.
b) If we don’t do this step, when the precipitate will dry it wouldn’t only be CaCO3, but the
solvent too. So, by “rinsing” it we remove what we don’t want to dry with the solute.
4- a) ethanol is a substance that evaporates more easily than water, so by adding it we want to
make sure that when we put the precipitate in the oven it will dry quickly and completely.
b) the precipitate will take too much time to dry, and it might not even dry completely.
5- a) it should have the same number of moles as Na2CO3: 3.43 x 10−3 mol

−3 110.98 g
3.43 x 10 mol × =0.3807 g of CaCl 2
1 mol
b) No it will not influence the theoretical yield, because changing the mass influence the
experimental yield, instead we could say that it will influence the percentage yield.

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