Limiting and Excess Reactants

  • Limiting and Excess Reactants: The reactant that is completely consumed when a reaction is run to completion is known as the limiting reactant. The other reactants are termed as excess reactants.

    • To find the limiting reactant: First, balance the stoichiometric equation. Then, take the ratio of the reactant feed rate to their stoichiometric coefficients. The limiting reactant is the reactant that has the lowest ratio.

    • Fractional excess: The fractional excess of a reactant is the ratio of excess to the stoichiometric requirement. \[\text{Fractional excess of $A$} = \frac{n_A \text{ in feed} - n_A \text{ stoichiometric}}{n_A \text{ stoichiometric}} \tag*{(3)}\] ‘\(n_A\)’ is the number of moles of \(A\)—the reactant. ‘\(n_A\) stoichiometric’ is the amount of \(A\) needed, to react completely with the limiting reactant.

    Example: Two moles of nitrogen (\(\ce{N2}\)) react with four moles of hydrogen (\(\ce{H2}\)) to form ammonia (\(\ce{NH3}\)). The reaction is: \[\ce{N2} + 3\ce{H2} \rightarrow 2\ce{NH3}\] The ratio of reactant amount to their stoichiometric coefficients: \[\ce{N2} = \frac{2}{1} = 2 \qquad \ce{H2} = \frac{4}{3} = 1.33\] The ratio for \(\ce{H2}\) is the least; therefore, \(\ce{H2}\) is the limiting reactant.

    Stoichiometric \(\ce{N2}\) requirement: 3 mole of \(\ce{H2}\) demands 1 mole of \(\ce{N2}\). Therefore, for 4 mole of \(\ce{H2}\), the requirement of \(\ce{N2}\) = 4/3 mole. \[\% \text{excess of } \ce{N2} = \frac{2-\frac{4}{3}}{\frac{4}{3}} \times 100 = 50\%\]

  • Incomplete Reactions: Chemical reactions are generally not complete; that is, even the limiting reactant is not completely used up. This idea is expressed as percent completion, and it is important to note that this completion must be based on the limiting reactant.

  • Percent Excess: Percent excess of excess reactant is based on the 100% completion of the limiting reactant.