Chemical reactions require specific proportions of substances to mix and react completely. A reactant is any starting material in a chemical reaction. When two or more reactants are mixed, one will be consumed first, causing the reaction to stop. This substance is the limiting reactant because it determines the maximum amount of product that can be formed. Any other substance present beyond what is needed to react with the limiting reactant is called the excess reactant. The total amount of excess reactant remaining unreacted must be determined through a systematic calculation.
Laying the Groundwork for Calculation
Before any calculation can begin, two preliminary steps establish the correct quantitative relationships between the reaction components. The first involves ensuring the chemical equation is properly balanced. A balanced equation uses whole-number coefficients, which are the source of the mole ratios. The law of conservation of mass dictates that the number of atoms for each element must be equal on both the reactant and product sides.
The second step involves converting the known masses of the starting materials into moles. Moles represent the fundamental counting unit in chemistry. The conversion from mass (typically measured in grams) to moles is accomplished by dividing the mass of the substance by its molar mass. Molar mass is calculated from the atomic masses of its constituent elements and is expressed in units of grams per mole. Calculations comparing reacting quantities must always use the mole unit to accurately apply the mole ratios derived from the balanced equation.
Identifying the Limiting Reactant
Identifying the limiting reactant is the single most important step because it sets the scale for the entire reaction and dictates how much of the excess reactant is consumed. The identification process compares the available moles of each reactant to the ratio required by the balanced chemical equation. Consider the general reaction \(A + 2B \rightarrow C\), which indicates that one mole of A requires two moles of B to react completely.
Imagine starting with \(150.0 \text{ g}\) of A (Molar Mass: \(50.0 \text{ g/mol}\)) and \(100.0 \text{ g}\) of B (Molar Mass: \(25.0 \text{ g/mol}\)). The initial moles are \(3.0 \text{ moles}\) of A and \(4.0 \text{ moles}\) of B. To determine the limiting reactant, calculate the amount of the other reactant needed for complete consumption.
If all \(3.0 \text{ moles}\) of A reacted, the stoichiometry requires \(3.0 \text{ moles of A} \times (2 \text{ moles B} / 1 \text{ mole A})\), which equals \(6.0 \text{ moles}\) of B. Since only \(4.0 \text{ moles}\) of B are available (less than the \(6.0 \text{ moles}\) required), substance B is the limiting reactant.
Conversely, if all \(4.0 \text{ moles}\) of B reacted, the amount of A needed is \(4.0 \text{ moles of B} \times (1 \text{ mole A} / 2 \text{ moles B})\), which equals \(2.0 \text{ moles}\) of A. Because \(3.0 \text{ moles}\) of A are present (more than the \(2.0 \text{ moles}\) needed), substance A is the excess reactant.
Finding the Amount of Reactant Used
With the limiting reactant (B) identified, the next step is to calculate the precise quantity of the excess reactant (A) consumed. The key principle is that the entire amount of the limiting reactant is used up, and this total amount determines the extent to which the excess reactant is depleted. Stoichiometric calculations must be performed using the moles of the limiting reactant, as this value represents the maximum possible reaction extent.
Using the example where \(4.0 \text{ moles}\) of B are consumed, the amount of A that reacted is found by applying the \(1:2\) mole ratio from the balanced equation \(A + 2B \rightarrow C\). This ratio converts the consumed moles of B to the consumed moles of A.
The moles of A used are calculated as \(4.0 \text{ moles of B consumed} \times (1 \text{ mole A} / 2 \text{ moles B})\), resulting in \(2.0 \text{ moles}\) of A that reacted. This value is then converted back to a mass measurement using the molar mass of A (\(50.0 \text{ g/mol}\)).
The mass of A consumed is \(2.0 \text{ moles of A used} \times 50.0 \text{ g/mol}\), which equals \(100.0 \text{ grams}\). This consumed mass links the initial reactant amounts to the final amount remaining.
Calculating the Final Excess Remaining
The final step determines the exact quantity of the excess reactant that did not react and is left over. This unreacted amount is found by subtracting the mass of the excess reactant used from the initial mass provided. While the calculation can use moles, the final answer is generally presented in units of mass for practical laboratory use.
In the running example, the initial amount of the excess reactant A was \(150.0 \text{ grams}\). Since the reaction consumed \(100.0 \text{ grams}\) of A, the amount remaining is the difference between the starting mass and the used mass: \(150.0 \text{ g} – 100.0 \text{ g}\).
This subtraction reveals that \(50.0 \text{ grams}\) of the excess reactant A remain unreacted in the reaction vessel. The excess mass calculated represents the quantity of substance that could have reacted if more of the limiting reactant had been present.