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Revised 12/2015
Chem 1104 Lab
EXPERIMENT: LIMITING REAGENT
NOTE: Students should have moles of reactants in DATASHEET converted into masses in
grams prior to the lab period.
INTRODUCTION
Limiting reactant
You have learned in your lectures about stoichiometry and understand the relative
amounts of chemicals needed to carry out a reaction. Let’s take a very simple example in the
reaction of hydrogen gas with chlorine gas to make hydrogen chloride gas:
H2(g) + Cl2(g) → 2 HCl(g)
(1)
In this reaction one mole of H2 reacts with one mole of Cl2 to give two moles of HCl.
However, it is almost impossible and sometimes not practical to use exactly stoichiometric
amounts of reagents (1:1 in this case) and either by chance or on purpose one reactant will be in
excess. The other starting material is the limiting reactant. Let’s assume that a reaction is
carried out loading 1.0 mole of H2 and 1.5 moles of Cl2 into a suitable reactor. How many moles
of HCl should you get?
Balanced equation:
H2
Initial quantities:
Change (reaction):
Final quantities:
1.0 mol
-1.0 mol
0.0 mol
+
Cl2
→
1.5 mol
-1.0 mol
0.5 mol
2 HCl
(1)
0 mol
+2.0 mol
2.0 mol
The first lesson to learn is that changing the quantities of reactants does not mean you
change the product, i.e. you will not make HCl2 because that compound does not exist. What
will happen instead is that the limiting reactant, H2, will be used up and some of the reactant in
excess, Cl2, will be left over after the reaction is done. The same would happen if we were to
react 1.5 moles of H2 with 1.0 mole of Cl2:
Balanced equation:
H2
Initial quantities:
Change (reaction):
Final quantities:
1.5 mol
-1.0 mol
0.5 mol
+
Cl2
1.0 mol
-1.0 mol
0.0 mol
→
2 HCl
(1)
0 mol
+2.0 mol
2.0 mol
In this case Cl2 is the limiting reactant and will be used up, while some H2 will be left
over because it is in excess. The important thing to note is that when you write the quantities on
the Change (reaction) line, that you use the stoichiometric amounts dictated by the balanced
chemical reaction, in this case, 1:1:2. A great representation of limiting reactants can be found in
your textbook.
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Percent Yield
Very few reactions go to completion. Remember, one way or another the reactants have
to come into contact with one another to react. Imagine reacting one mole H2 with one mole of
Cl2 according to Eq 1. How long would it take for the last molecule of unreacted H2 to find the
last molecule of unreacted Cl2 in a flask full of HCl? At some point you have to stop waiting and
realize that the reaction has proceeded as far as possible. Obviously you will not have exactly
2 moles of HCl but slightly less, because some H2 and some Cl2 did not get a chance to react. In
this case 2 moles would have been the theoretical yield of HCl. Chemists determine the success
of their reactions by calculating the percentage yield which is defined as:
% yYield 
actual yield
 100% %
theoretica l yield
(2)
So, if we reacted one mole of H2 with one mole of Cl2 and prepared 1.8 moles of HCl, the %
yield would be:
1.8 moles
% yield 
 100%
(3)
2.0 moles
And therefore the yield would be 90.%. Obviously we would also have 10.% of the original
amount of H2 and 10.% of the original amount of Cl2 left over. In every day experiments we will
have to combine both concepts, that of the limiting reactant and that of percentage yield. As a
final example, let’s assume we reacted 0.8 moles of H2 with 1.0 mole of Cl2 and prepared 1.2
moles of HCl.
Let’s calculate the limiting reactant and the % yield. The calculations below are the
proper way to calculate both and should be done each time you do an experiment:
Overall balanced chemical equation: H2 + Cl2 → 2 HCl
0.8 moles of H2 can generate 1.6 moles of HCl, according to the balanced equation and the 1 H2:
2 HCl ratio.
1.0 mole of Cl2 can generate 2.0 moles of HCl, according to the balanced equation and the 1 Cl2:
2 HCl ratio.
Since 0.8 moles of H2 can generate less HCl than 1.0 mole of Cl2 can:
The limiting reactant is: H2
Reactant in excess is: Cl2
The theoretical yield is: 1.6 moles of HCl
The actual yield is: 1.2 moles of HCl
%
yield

1.2 moles
 100%  75%
1.6 moles
So why would we use an excess of one reactant? Once again remember, one way or another the
starting materials have to come into contact with one another to react. When one mole of H2 is
reacted with one mole of Cl2 according to Eq 1 it would take a long time for the last molecule of
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unreacted H2 to find the last molecule of unreacted Cl2 in a flask full of HCl. However, if we
reacted one mole of H2 with two moles of Cl2, the last molecule of unreacted H2 would be
surrounded by 1 mole of unreacted Cl2 in the flask. In most cases you will get a much better %
yield if an excess of one reactant is used. In addition, reactions tend to proceed much more
quickly when an excess of one reactant is used. You have to keep in mind however that you
have to be able to separate your product from the excess reactant. In the case of H2, Cl2 and HCl
this would be extremely difficult since all three are gases.
In this experiment we will examine the effects of varying stoichiometries in the reaction
of CaCl2·2H2O (calcium chloride dihydrate) with K2C2O4·H 2O (potassium oxalate
monohydrate) to form CaC2O4·H2O (calcium oxalate monohydrate; major constituent of kidney
stones) according to the following overall balanced equation:
CaCl2·2H2O(aq) + K2C2O4·H 2O(aq) → CaC2O4·H2O(s) + 2KCl(aq) + 2H2O(l)
(4)
Note: CaCl2·2H2O and K2C2O4·H2O are hydrates. Hydrates, which can simplistically be
described as a clustering of x number of water molecules around the salt, where x varies
depending on the salt (i.e., for calcium chloride, x is 2, while for potassium oxalate, x is 1). The
waters of hydration do count in determining the molar mass of a hydrate compound.
This reaction is well suited for our purpose because only one component of the reaction
mixture is insoluble while all reactants and side products (i.e., all products other than the desired
product) are water soluble. You will weigh both reactants and after completion of the reaction
you will weigh the insoluble product. Using the conversions to number of moles, you will
determine the limiting reactant and the % yield. You will also check for excess reactant using
some simple tests, confirming your limiting reactant calculations.
Vacuum filtration
The vacuum filtration apparatus is more rapid than simple gravity filtration. In order to
carry out vacuum filtration a special funnel, known as a Buchner funnel is used. A piece of filter
paper is placed flat in the funnel, and the vacuum is applied by turning on the aspirator. The
filter paper is then made wet with the solvent that you are using. (Methanol is used in this
experiment).
Check to see that the paper is free of creases and completely seals the holes in the funnel to make
sure no solution will leak around the edges of the filter paper.
With the vacuum turned on the solution to be filtered is poured into the Buchner funnel. Scrape
out the container with a spatula and use a small amount of cold methanol to rinse out the
container and wash the crystals. Be careful not to use too much solvent since it may dissolve
your product which will then be carried into the filter flask. Finally, before turning off the
aspirator, break the vacuum by disconnecting the hose from the aspirator so that water from the
aspirator does not get sucked back into the filter flask.
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Figure 1
Vacuum filtration apparatus.
PROCEDURE:
1.
Precisely weigh CaCl2·2H2O and K2C2O4·H2O according to Trial 1 in Table 1 into separate
weighing dishes and immediately transfer the solids into a 250 mL Erlenmeyer flask.
Record the actual weights in Table 2. Add 150 mL of distilled water to the flask and stir the
mixture with a stir rod for a few minutes.
2.
Cover the flask with a watch glass and heat the mixture to a maximum of 75°C and
maintain that temperature for 15 minutes. Once the mixture has reached 70°C, remove from
the heat if necessary (if the temperature drops below 70°C, place it back on the hotplate).
Stir the mixture every few minutes to speed up dissolution. After 15 minutes, remove the
flask from the heat and let it cool by placing your flask inside an 800-mL beaker filled with
250 mL of tap water and a few ice cubes. Once the mixture is below 25°C, remove from the
ice/water bath. While the precipitate is settling at the bottom of the flask prepare hot wash
water (for step 4) by heating about 30 mL of distilled water in a 50 mL beaker to a
maximum of 75 °C.
3.
After the precipitate has settled, take two centrifuge tubes and fill them half-way with the
supernatant liquid (the liquid resting on top of the precipitate) from Trial 1 using a dropping
pipette, making sure the liquid level is identical in both of the tubes so they can be properly
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balanced in a centrifuge. Centrifuge both mixtures for 1 minute (ask your instructor for
operating instructions; centrifuges can be very dangerous if not used properly). Be sure to
record your observations in Table 3.
a) To one of the centrifuge tubes add 2 drops of 0.5 M CaCl2. If a precipitate forms you
know that K2C2O4 was in excess and CaCl2 was the limiting reactant. See note below.
b) To the other centrifuge tube add two drops of 0.5 M K2C2O4. If a precipitate forms you
know that CaCl2 was in excess and K2C2O4 was the limiting reactant. See note below.
NOTE: Remember, your initial reaction may not go to completion and small amounts of
each of the reactants may be present and give a small amount of precipitate in the tests.
However, if one centrifuge tube shows a lot more precipitate than the other, you will know
which reactant was in excess and which one was the limiting reactant. If both centrifuge
tubes show very little or no precipitate you know that the reaction was done using
stoichiometric amounts. In that case enter “none” for the appropriate entries in Table 3.
4.
Set up a Vacuum Filtration Apparatus as shown in Figure 1 below. Label a sample vial
with masking tape and your names and lab day/time, and trial number. Place one piece of
filter paper into the vial. Weigh the filter paper and vial together (after the vial has been
labeled with tape!) and record the mass. Then remove the filter paper and place it into the
center of the filter and using a few mL of water from the wash bottle get the entire surface
of the filter paper wet so that it sticks to the funnel. At this point, ask your instructor if the
vacuum is on. Get them to show you how to attach your flask to the vacuum. While swirling
(to help transfer of more crystals), add the contents of your reaction flask to the filter. Once
all of the mixture has been filtered, add approximately 5 mL of the hot wash water to the
flask so you can rinse remaining crystals into the filter. Do this 3 or 4 times until the
reaction flask contains only a miniscule amount of crystals. Let the vacuum work for about
5 minutes. Then disconnect your flask from the vacuum hose and turn off the water to the
aspirator. Please do not waste time at the filtration apparatus, as some sharing may be
necessary.
5.
Place the product with the filter paper into the labeled vial (be sure you have the same vial
that you weighed in step 4) and set it aside to hand to your instructor at the end of the lab
session.
6.
Repeat steps 1 – 5 with the remaining 2 Trials as indicated in Table 1. This lab is timeconsuming and you must be able to do several tasks simultaneously. The next reaction
MUST BE started as soon as the hot plate is available, otherwise you will run out of time
and not be able to complete the experiment.
7.
After one week of drying, place the filter paper and product into a weighing dish and weigh
them. Determine the mass of your product from all 3 reactions and calculate the limiting
reactant and % yield for each trial.
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Quiz Hints:
1. Define the term “limiting reagent.”
2. Know the technique of vacuum filtration and the differences between vacuum and gravity
filtration.
3. Know how to determine the limiting reagent and the percent yield.
REFERENCES:
1. Reeves V. Chemistry 1006 Laboratory Handout, University of New Brunswick, Fredericton,
N.B. 2015.
2. This experiment is based on the experiment “Limiting Reactant” in J.A. Beran’s Laboratory
Manual for Principles of General Chemistry, 2009, John Wiley & Sons, Inc