Excess and Limiting Reagents
In a chemical reaction, reactants that are not use up when the reaction is finished are
called excess reagents. The reagent that is completely used up or reacted is called the
limiting reagent, because its quantity limit the amount of products formed.
Let us consider the reaction between sodium and chlorine. The reaction can be
represented by the equation:
2 Na + Cl2 = 2 NaCl,
Thus, if you have 6 Na atoms, 3 Cl2 molecules will be
required. If there is an excess number of Cl2
molecules, they will remain unreacted. We can also
state that 6 moles of sodium will require 3 moles of
Cl2 gas. If there are more than 3 moles of Cl2 gas,
some will remain as an excess reagent, and the
sodium is a limiting reagent. It limits the amount of
the product that can be formed.
Example 1
Calculate the number of moles of CO2 formed in the combustion of ethane
C2H6 in a process when 35.0 mol of O2 is consumed.
Hint...
The reaction is
2 C2H6 + 7 O2 = 4 CO2 + 6 H2O
4 mol CO2
35.0 mol O2 ---------- = 20.0 mol CO2
7 mol O2
Example 2
Two moles of Mg and five moles of O2 are placed in a reaction vessel, and then
the Mg is ignited according to the reaction
Mg + O2 = MgO.
Balance this equation and identify the limiting reagent in this experiment.
Hint...
The balanced reaction is,
2 Mg + O2 = 2 MgO
Thus, two moles of Mg require only ONE mole of O2. Four moles of oxygen will
remain unreacted. Oxygen is the excess reagent, and Mg is the limiting reagent.
Percentage Yield
Using stoichiometry, we can mathematically determine the amount of a product
that should be formed during an experiment, yet we sometimes find that we don't
end up with exactly the right amount of product. Percentage yield allows us to
calculate what percent of the expected product we are able to account for by the end
of our experiment. The formula that we use is;
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
Example 1. A student conducts a single displacement reaction that produces 2.755
grams of copper. Mathematically he determines that 3.150 grams of copper should
have been produced. Calculate the student's percentage yield.
Solve:
actual amount of product: 2.755 g
expected amount of product: 3.150 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
2.755g
percentage yield = --------------- x 100
3.150g
percentage yield = 87.4603174 %
percentage yield = 87.46 %
Example 2. A student completely reacts 5.00g of magnesium with an excess of
oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide.
What is the student's percentage yield?
First, write a balanced chemical equation for the reaction:
2Mg(s) + O2(g) ----> 2MgO(s)
Now, label the given and the unknown and cross out the rest:
given
unknown
2Mg(s) + O2(g) ----> 2MgO(s)
5.00 g
Xg
Change the mass given to moles by dividing by the molar mass of Mg.
mass given: 5.00g
molar mass of Mg: 24.3 g
5.00g
Number of moles of Mg = ----------24.3 g/mole
Number of moles of Mg = 0.206 moles
Compare the molar ratio between the given and the unknown to determine the
number of moles produced.
Coefficient of Mg; 2
Coefficient of MgO: 2
Number of moles of Mg: = 0.206 moles
Number of moles of MgO: = ?
number of moles of given number of moles of unknown
-------------------------------- = -------------------------------------coefficient of given coefficient of unknown
0.206 moles X moles
-------------- = ------------2
2
Number of moles of MgO produced = 0.206 moles
Now, change the number of moles of MgO produced to mass by multiplying by the
molar mass of MgO.
# of moles of MgO = 0.206 moles
Molar mass of MgO = 40.3g/mole
mass = # of moles x molar mass
mass of MgO = 0.206 moles x 40.3 g/mole
mass of MgO = 8.30 g
Now, you are ready to solve the percentage yield problem.
actual mass of MgO produced = 8.10 g
expected mass of MgO = 8.30 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
8.10 g
percentage yield = ------------ x 100
8.30 g
percentage yield = 97.6 %
Percentage purity of a compound
In the reaction between aluminium metal and an excess of copper(II) sulphate solution,
4.0 g of aluminium is reacted and 0.48 g of copper metal is formed. Assuming that all the
aluminium present reacted to give copper, what is the purity of the aluminium?
The equation 2Al(s) + 3CuSO4(aq)
3Cu(s) + Al2(SO4)3(aq)
the amount of copper made = 4.8 g/ 64 g mol-1
= 0.075 mol
From the equation, the maximum amount of aluminium metal that could be present = 2/3 ×
the amount of copper made, as the ratio of Al : Cu is 2 : 3.
the maximum possible amount of 2
= /3 × 0.075 mol
aluminium metal
=0.05 mol
the maximum possible mass of aluminium
= 0.05 mol × 27 g mol-1
metal present
=1.35 g
=(mass present / total mass of
the percentage purity of aluminium used
compound)×100
= (1.35 g / 4.0 g)× 100
= 33.75 %