Experiment 8: Limiting Reagents
Taieba Khan
Anna Thony
Course: Chemistry 1300
Section: D09
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Instructor: Professor Carnevale
Laboratory Assistant: L.A. Nilli Modi
Date of Experiment: February 13, 2020
Abstract:
In this experiment the limiting reactant in a mixture of two soluble salts and the percent
composition of each salt in the mixture was determined. After adding the salt mixture to water,
the supernatant from the precipitate was gathered and tested for the limiting reactant. Through
adding drops of each test reagent into the supernatant, the limiting reactant was found to be
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CaCl2. This also meant that the excess reactant was C2O4. The percent composition of CaCl2 was
calculated to be 20.96%, making the percent composition of C2O4 to be 79.04% in the salt
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mixture.
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Introduction:
The purpose of this experiment was to find the limiting reactant in a mixture of two salts
and then determine the percent of each salt in the mixture. In a chemical reaction, the two main
factors that affect the amount of products that result from the reaction are the moles of the
starting reactants and the overall percent yield of the reaction. However, chemicals react by
following specific mole ratiosThis was a two-part experiment that relied on the use of a filtering
apparatus made with Whatman No. 42 filter paper.
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Materials and Methods:
Please refer to Experiment 8 on pages 123-128 of Laboratory Manual for Principles of
General Chemistry by J.A. Beran. In this experiment, the unknown “Phineas” was used in which
the amount of each reactant in the mixture was what needed to be found. One deviation from the
procedure is that instead of using 1 g of the salt mixture, only .5 g was used. In addition, only
100 mL of deionized water was used to dissolve the salt mixture instead of 150 mL. Also, instead
of heating up the solutions and water on a hot plate up to 75°C, the temperature was kept at
around 55°C. Lastly, the filter paper with the product was only oven-dried for 20 minutes, rather
than the 60 minutes recommended in the lab manual.
Data:
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Results:
Table 1: Precipitation of Ca2C2O4ꞏH2O from the Salt Mixture
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Trial 1
Trial 2
1. Mass of salt mixture (g)
.6355
.7147
2. Mass of filter paper (g)
.2953
.4528
3. Mass of filter paper and product after drying (g)
.4547
.6392
4. Mass of dried product (g)
.1594
.1864
5. Formula of dried product
Ca2C2O4
Ca2C2O4
Table 1 displays the various measured values of the materials used initially and calculated values
needed to determine the mass of the dried product from precipitation of the salt mixture.
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Table 2: Determination of Limiting Reactant Data Analysis
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Trial 1
Trial 2
1. Moles of Ca2C2O4ꞏH2O precipitated (mol)
.0012
.0015
2. Moles of limiting reactant (CaCl2) in salt mixture (mol)
.0012
.0015
3. Mass of limiting reactant in salt mixture (g)
.1332
.1615
4. Mass of excess reactant (K2C2O4) in salt mixture (g)
.5023
.5532
5. Percent limiting reactant in salt mix (%)
20.96%
22.60%
6. Percent excess reactant in salt mix (%)
79.04%
77.40%
7. Mass of excess reactant that reacted (g)
.1995
.2419
8. Mass of excess reactant, unreacted (g)
.3028
.3113
Table 2 displays the calculated values needed to find the amount of limiting reactant and excess
reactant in the unknown salt mixture “Phineas”.
Calculations:
1. Mass of filter paper and product after drying – Mass of filter paper = Mass of dried
product
.4547g - .2953g = .1594 g CaC2O4
2. Mass of dried product X Molecular mass of product = Moles of CaC2O4 precipitated
.1594g CaC2O4 X
1 mol
128.097 g /mol
= .0012 mol CaC2O4
3. Moles of dried product X Mole ratio of limiting reactant to dried product* Molecular =
Moles of limiting reactant
.0012 mol CaC2O4 X
1mol CaCl 2
1 mol CaC 2 O 4
= .0012 mol CaCl2
4. Moles of limiting reactant X Molecular mass of limiting reactant / 1 mol of limiting
reactant = Mass of limiting reactant
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.0012 mol CaCl2 X
110.98 g CaCl 2
1 mol
= .1332 g CaCl2
5. Mass of salt mixture – Mass of limiting reactant = Mass of excess reactant
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.6355 g salt mixture - .1332 g CaCl2 = .5023 g K2C2O4
6. Mass of limiting reactant / Mass of salt mixture X 100% = Percent mass of limiting
reactant
(.1332 g CaCl2 / .6355 g salt mixture) X 100% = 20.96% CaCl2
7. Mass of excess reactant / Mass of salt mixture X 100% = Percent mass of excess
reactant
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(.5023 g K2C2O4 / .6355 g salt mixture) X 100% = 79.04% K2C2O4
8. Moles of excess reactant X (Molecular weight of excess reactant / 1 mol) = Mass of
excess reactant that reacted
.0012 mol K2C2O4 X
166.22 g K 2 C 2 O 4
1 mol
= .1995 g K2C2O4
9. Mass of excess reactant – Mass of excess reactant that reacted = Mass of excess
reactant, unreacted
.5023 g K2C2O4 - .1995 g K2C2O4 = .3028 g K2C2O4
Discussion:
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Conclusion:
Post Lab Questions:
2. Due to the couple of drops of water placed on the properly folded filter paper before its mass
was measured, the reported mass of the limiting reactant will be too low. This is because the loss
of water from the filter paper after heating will become a part of the calculated mass loss of the
Ca2C2O4ꞏH2O precipitate.
3. Due to the porosity of the filter paper allowing for some of the Ca2C2O4ꞏH2O precipitate to
pass through the filter paper, the percent of the limiting reactant in the original salt mixture will
be reported too low because some of the precipitate was lost in the filtering process.
5. If the Ca2C2O4ꞏH2O precipitate is not completely air-dried when mass is determined, the
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reported mass of the limiting reactant in the original salt mixture will be reported too high
because there is a larger mass of Ca2C2O4ꞏH2O, leading to the mass of the limiting reactant being
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calculated to a much higher number.
8. As a result of using this wrong reagent in the test, when adding NaCl to the test tube
containing the supernatant, no visible precipitate will form because the Na+ will not be able to
react with C2O42-.
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