Limiting Reactants, Theoretical Yield and Percent Yield
Most of the time when a reaction is carried out the reactants will not be present in the proportions indicated by the balanced equation. Consequently some reactant will be consumed while others will be left over when the reaction is finished. The reaction used up first is called the limiting reactant. The amount of limiting reactant present determines the amount of product which can be produced. This amount of product produced is called the theoretical yield, it is the amount of product that would form if all of the limiting reactant is consumed in the reaction. The theoretical yield is the maximum amount of a product that can be produced from a given amount of limiting reactant. The amount of product actually produced by the reaction is the actual yield and will usually be less than the theoretical yield. To express the efficiency of a reaction, chemists often calculate the percent yield, which is defined by:
actual yield |
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%yield |
= |
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x |
100% |
theoretical yield |
Lets work through an example problem that will illustrate all of these concepts.
Example problem
Calculate the theoretical yield of C2H5Cl if 112 g of C2H5OH is reacted with 34.7 g of PCl3 based on the reaction below. If 23.7 g of C2H5Cl is produced, what is the percent yield?
| 3 C2H5OH + PCl3 ==> 3 C2H5Cl + H3PO3 |
Solution
First we will assume that C2H5OH is the limiting reactant and calculate how much C2H5Cl would be formed. Then we will assume that PCl3 is the limiting reactant and calculate how much C2H5Cl would be formed. The reactant producing the smaller amount of C2H5Cl is the limiting reactant and the corresponding quantity of C2H5Cl is the theoretical yield.
1 mol C2H5OH |
3 mol C2H5Cl |
64.51 g C2H5Cl |
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112 g C2H5OH |
x |
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x |
------------------ |
x |
------------------ |
= |
157 g C2H5Cl |
46.07 g C2H5OH |
3 mol C2H5OH |
1 mol C2H5Cl |
1 mol PCl3 |
3 mol C2H5Cl |
64.51 g C2H5Cl |
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34.7 g PCl3 |
x |
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x |
------------------ |
x |
------------------ |
= |
48.9 g C2H5Cl |
137.3 g PCl3 |
1 mol PCl3 |
1 mol C2H5Cl |
Since the calculation with PCl3 gives the smaller amount of C2H5Cl, PCl3 is the limiting reactant and 48.9 g is the theoretical yield of C2H5Cl. The percent yield is
actual yield |
23.7 g |
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%yield |
= |
--------------- |
x |
100% |
= |
-------------- |
x |
100% |
= |
48.5% |
theoretical yield |
48.9 g |