Stoichiometry is the branch of chemistry concerned with the relative quantities of reactants and products in chemical reactions. In a perfect scenario, all reactants would be consumed completely, but in the laboratory or industry, reactants are often mixed in unequal amounts. This mixing imbalance means that one reactant will be entirely used up before the others, putting a natural limit on how much product can be created. Finding this limiting component, which determines the maximum possible output of a reaction, is a fundamental step in chemical analysis.
Understanding Limiting Reactant Concepts
The available quantity of each starting material dictates the overall result of a chemical reaction. The Limiting Reactant (LR) is the substance that is completely consumed first, causing the entire process to stop. Once the LR is gone, no more product can be formed, regardless of how much of the other materials are present.
Consider the analogy of making grilled cheese sandwiches: one sandwich requires two slices of bread and one slice of cheese. If you have 28 slices of bread and only 11 slices of cheese, you can only make 11 sandwiches before running out of cheese. In this scenario, the cheese is the limiting reactant because it restricts the total number of sandwiches produced.
The other starting material, the bread, is called the Excess Reactant (ER) because a portion of it remains unreacted after the limiting reactant is fully depleted. Identifying the limiting reactant is the first step in calculating the maximum possible yield of a chemical process.
Preparation: Balancing the Equation and Calculating Molar Mass
Two foundational steps must be completed to establish the mathematical relationship between the reactants and products. First, the chemical equation for the reaction must be correctly written and balanced. Balancing ensures that the number of atoms for each element is equal on both sides of the equation, following the law of conservation of mass.
The whole-number coefficients in the balanced equation represent the mole ratio of the substances. This ratio is the heart of stoichiometry, showing the exact proportional relationship in which the reactants combine. These coefficients allow conversion from moles of one substance to moles of another.
The second preparatory step involves calculating the molar mass for each reactant given in grams. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (\(\text{g/mol}\)). This value is calculated by summing the atomic masses of all atoms present in the compound’s chemical formula.
Molar mass serves as the bridge between the mass of the reactants (grams) and the mole quantity needed for stoichiometric calculations. Since the mole ratio is expressed in moles, converting the initial grams of each reactant into moles is necessary for comparing the materials on a common basis.
Step-by-Step Procedure for Identifying the Limiting Reactant
Identifying the limiting reactant when starting with a measured mass of each reactant requires converting mass into a comparable measure of chemical quantity. The most reliable method is to calculate the amount of product that could be formed from the given mass of each starting material. The reactant that produces the smallest amount of product is the limiting reactant.
Imagine the reaction between sodium hydroxide (\(\text{NaOH}\)) and phosphoric acid (\(\text{H}_3\text{PO}_4\)): \(3\text{NaOH} + \text{H}_3\text{PO}_4 \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}\). Suppose you start with 35.60 grams of \(\text{NaOH}\) and 30.80 grams of \(\text{H}_3\text{PO}_4\).
Convert Mass to Moles
The first step is to convert the initial grams of each reactant into moles using their respective molar masses: 40.00 \(\text{g/mol}\) for \(\text{NaOH}\) and 98.00 \(\text{g/mol}\) for \(\text{H}_3\text{PO}_4\). For \(\text{NaOH}\), dividing 35.60 grams by 40.00 \(\text{g/mol}\) reveals the number of moles available. Similarly, dividing 30.80 grams of \(\text{H}_3\text{PO}_4\) by its molar mass yields the moles of \(\text{H}_3\text{PO}_4\). This conversion translates the physical mass into the chemical unit of quantity needed for comparison.
Calculate Potential Product Yield
The next step uses the mole ratio from the balanced equation to convert the moles of each reactant into the moles of a common product, such as \(\text{Na}_3\text{PO}_4\). For \(\text{NaOH}\), the stoichiometric ratio is 3 moles of \(\text{NaOH}\) to 1 mole of \(\text{Na}_3\text{PO}_4\). For \(\text{H}_3\text{PO}_4\), the ratio is 1 mole of \(\text{H}_3\text{PO}_4\) to 1 mole of \(\text{Na}_3\text{PO}_4\).
Determine the Limiting Reactant
After performing these two calculations, you will have two different amounts of product, expressed in moles. The reactant that results in the smaller number of moles of product is the limiting reactant because it will be completely consumed first. In this example, the calculation shows that \(\text{NaOH}\) would produce a smaller quantity of \(\text{Na}_3\text{PO}_4\), making \(\text{NaOH}\) the limiting reactant.
Calculating the Theoretical Yield
Once the limiting reactant has been identified, the maximum possible amount of product that can be formed is determined. This amount is known as the Theoretical Yield. All subsequent calculations must start with the quantity of the limiting reactant, as the excess reactant is irrelevant to the final product amount.
Continuing with the example, the theoretical yield calculation begins with the moles of the limiting reactant, \(\text{NaOH}\). This mole quantity of \(\text{NaOH}\) is then converted to moles of the desired product, \(\text{Na}_3\text{PO}_4\), using the 3:1 mole ratio from the balanced equation.
The final step is to convert the calculated moles of product back into a measurable mass, typically grams. This is achieved by multiplying the moles of \(\text{Na}_3\text{PO}_4\) by its molar mass, which is 163.94 \(\text{g/mol}\). The resulting mass is the maximum amount of sodium phosphate that could be produced under ideal conditions.
The theoretical yield provides a benchmark against which the mass of product actually collected in a real-world experiment can be compared. In this specific case, the moles of \(\text{NaOH}\) ultimately allow for the formation of 48.64 grams of \(\text{Na}_3\text{PO}_4\), which is the theoretical yield.