Editing Stoichiometry (section)

== Limiting reagent and percent yield ==
{{Main|Limiting reagent|Yield (chemistry)}}
The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed when the reaction is complete. An excess reactant is a reactant that is left over once the reaction has stopped due to the limiting reactant being exhausted.

Consider the equation of roasting [[lead(II) sulfide]] (PbS) in oxygen ({{chem2|O2}}) to produce [[lead(II) oxide]] (PbO) and [[sulfur dioxide]] ({{chem2|SO2}}):
: {{chem2|2 PbS + 3 O2 -> 2 PbO + 2 SO2}}

To determine the theoretical yield of lead(II) oxide if 200.0&nbsp;g of lead(II) sulfide and 200.0&nbsp;g of oxygen are heated in an open container:
: <math>m_\mathrm{PbO} = \left(\frac{200.0 \mbox{ g }\mathrm{PbS}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{PbS}}{239.27 \mbox{ g }\mathrm{PbS}}\right)\left(\frac{2 \mbox{ mol }\mathrm{PbO}}{2 \mbox{ mol }\mathrm{PbS}}\right)\left(\frac{223.2 \mbox{ g }\mathrm{PbO}}{1 \mbox{ mol }\mathrm{PbO}}\right) = 186.6 \mbox{ g}</math>
: <math>m_\mathrm{PbO} = \left(\frac{200.0 \mbox{ g }\mathrm{O_2}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{O_2}}{32.00 \mbox{ g }\mathrm{O_2}}\right)\left(\frac{2 \mbox{ mol }\mathrm{PbO}}{3 \mbox{ mol }\mathrm{O_2}}\right)\left(\frac{223.2 \mbox{ g }\mathrm{PbO}}{1 \mbox{ mol }\mathrm{PbO}}\right) = 930.0 \mbox{ g}</math>

Because a lesser amount of PbO is produced for the 200.0&nbsp;g of PbS, it is clear that PbS is the limiting reagent.

In reality, the actual yield is not the same as the stoichiometrically-calculated theoretical yield. Percent yield, then, is expressed in the following equation:
: <math>\mbox{percent yield} = \frac{\mbox{actual yield}}{\mbox{theoretical yield}}</math>

If 170.0&nbsp;g of lead(II) oxide is obtained, then the percent yield would be calculated as follows:
: <math>\mbox{percent yield} = \frac{\mbox{170.0 g PbO}}{\mbox{186.6 g PbO}} = 91.12\%</math>

=== Example ===
Consider the following reaction, in which [[iron(III) chloride]] reacts with [[hydrogen sulfide]] to produce [[iron(III) sulfide]] and [[hydrogen chloride]]:
: {{chem2|2 FeCl3 + 3 H2S -> Fe2S3 + 6 HCl}}

The stoichiometric masses for this reaction are:
: 324.41&nbsp;g {{chem2|FeCl3}}, 102.25&nbsp;g {{chem2|H2S}}, 207.89&nbsp;g {{chem2|Fe2S3}}, 218.77&nbsp;g HCl

Suppose 90.0&nbsp;g of {{chem2|FeCl3}} reacts with 52.0&nbsp;g of {{chem2|H2S}}. To find the limiting reagent and the mass of HCl produced by the reaction, we change the above amounts by a factor of 90/324.41 and obtain the following amounts:
: 90.00&nbsp;g {{chem2|FeCl3}}, 28.37&nbsp;g {{chem2|H2S}}, 57.67&nbsp;g {{chem2|Fe2S3}}, 60.69&nbsp;g HCl

The limiting reactant (or reagent) is {{chem2|FeCl3}}, since all 90.00 g of it is used up while only 28.37 g {{chem2|H2S}} are consumed. Thus, 52.0 − 28.4 = 23.6&nbsp;g {{chem2|H2S}} left in excess. The mass of HCl produced is 60.7&nbsp;g.

By looking at the stoichiometry of the reaction, one might have guessed {{chem2|FeCl3}} being the limiting reactant; three times more {{chem2|FeCl3}} is used compared to {{chem2|H2S}} (324&nbsp;g vs 102&nbsp;g).