5.11: Limiting Reactant and Percent Yield

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Mixing of exact amounts of reactants such that all are consumed and none left over in a chemical reaction almost never occurs. Instead, one of the reactants is usually a limiting reactant. Suppose, for example that 100 g of elemental zinc (atomic mass 65.4) and 80 g of elemental sulfur (atomic mass 32.0) are mixed and heated undergoing the following reaction:

\[\ce{Zn + S \rightarrow ZnS}\]

What mass of ZnS, formula mass 97.4 g/mol, is produced? If 100 g of zinc react completely, the mass of S reacting and the mass of ZnS produced would be given by the following calculations:

\(\textrm{Mass s = 100.0 G Zn} \times \frac{\textrm{1 mol Zn}}{\textrm{65.4 g Zn}} \times \frac{\textrm{1 mol S}}{\textrm{1 mol Zn}} \times \frac{\textrm{32.0 g S}}{\textrm{1 mol S}} = \textrm{48.9 g S}\)

\(\textrm{Mass ZnS = 100.0g Zn} \times \frac{\textrm{1 mol Zn}}{\textrm{65.4 g Zn}} \times \frac{\textrm{1 mol S}}{\textrm{1 mol Zn}} \times \frac{\textrm{97.4 g ZnS}}{\textrm{1 mol ZnS}} = \textrm{149 g S}\)

Only 48.9 g of the available S react, so sulfur is in excess and zinc is the limiting reactant. A similar calculation for the amount of Zn required to react with 80 g of sulfur would show that 164g of Zn would be required, but only 100 g is available.

Exercise

A solution containing 10.0 g of HCl dissolved in water (a solution of hydrochloric acid) was mixed with 8.0 g of Al metal undergoing the reaction

\(\ce{2Al + 6HCl \rightarrow 2AlCl3 + 3H2}\)

Given atomic masses H 1.0, Al 27.0, and Cl 35.5, which reactant was left over? How much? What mass of AlCl₃ was produced?

Answer: HCl was the limiting reactant. Only 2.47 g of Al were consumed leaving 5.53 g of Al unreacted. The mass of AlCl₃ produced was 12.2 g

Percent Yield

The mass of product calculated from the mass of limiting reactant in a chemical reaction is called the stoichiometric yield of a chemical reaction. By measuring the actual mass of a product produced in a chemical reaction and comparing it to the mass predicted from the stoichiometric yield it is possible to calculate the percent yield. This concept is illustrated by the following example.

Suppose that a water solution containing 25.0 g of CaCl₂ was mixed with a solution of excess sodium sulfate,

\[\ce{CaCl2 (aq) + Na2SO4 (aq) \rightarrow CaSO4 (s) + 2NaCl(aq)}\]

to produce a solid precipitate of CaSO₄, the desired product of the reaction. (Recall that a precipitate is a solid formed by the reaction of species in solution; such a solid is said to precipitate from the solution.) Removed by filtration and dried, the precipitate was found to have a mass of 28.3 g, the measured yield. What was the percent yield?

Using atomic masses Ca 40.0, Cl 35.5, Na 23.0, and O, 16.0 gives molar masses of 111 g/mol for CaCl₂ and 136 g/mol for CaSO₄. Furthermore, 1 mole of CaCl₂ yields 1 mol of CaSO₄. The stoichiometric yield of CaSO₄ is given by the following calculation

\[\textrm{Mass CaSO}_{4} = \textrm{25.0 g CaCl}_{2} \times \frac{\textrm{1 mol CaCl}_{2}}{\textrm{111 g CaCl}_{2}} \times \frac{\textrm{1 mol CaSO}_{4}}{\textrm{1 mol CaCl}_{2}} \times \frac{\textrm{136 g CaSO}_{4}}{\textrm{1 mol CaSO}_{4}} = \textrm{30.6 g CaSO}_{4}\]

The percent yield is calculated by the following:

\[\textrm{Percent yield} = \frac{\textrm{Measured yield}}{\textrm{Stoichiometric yield}} \times 100\]

\[\textrm{Percent yield} = \frac{\textrm{28.3 g}}{\textrm{30.6 g}} \times 100 = 92.5 \%\]