DAT General Chemistry: Limiting and Excess Reactants

DAT High-Yield Concept

The limiting reactant is completely consumed in a chemical reaction and dictates the maximum product that can form.
To find the limiting reactant quickly, convert initial quantities to moles and divide by their stoichiometric coefficients ( $\text{moles} / \text{coefficient}$ ); the smallest ratio is limiting.
An excess reactant remains unreacted after the limiting reactant is exhausted.
To find leftover excess reactant, calculate consumed moles from the limiting reactant and subtract from initial moles.

Limiting Reactants in Reaction Stoichiometry

In chemical stoichiometry, reactants are rarely present in exact stoichiometric proportions. The limiting reactant (or limiting reagent) is the reactant that is entirely consumed first during a chemical reaction. Because the chemical reaction cannot proceed once this reactant is depleted, it sets the ultimate boundary for total product formation.

Limiting Reactant Definition: The reactant in a chemical reaction that is completely consumed first, thereby limiting the maximum amount of product that can be formed.

Worked Example 1: Determining Reaction Termination
Consider the direct reaction of iron and sulfur to produce iron(II) sulfide: $\ce{Fe + S -> FeS}$ . If $\pu{1.0 mol}$ of $\ce{Fe}$ is reacted with $\pu{2.0 mol}$ of $\ce{S}$ , $\ce{Fe}$ is completely consumed after reacting with $\pu{1.0 mol}$ of $\ce{S}$ to produce $\pu{1.0 mol}$ of $\ce{FeS}$ . The reaction stops, leaving $\pu{1.0 mol}$ of $\ce{S}$ unreacted. Thus, $\ce{Fe}$ is the limiting reactant.

Common DAT Exam Trap: Assuming the reactant with the smaller initial mass or lower initial mole count is automatically the limiting reactant. Stoichiometric coefficients must always be factored in, as a reaction may require multiple moles of one reactant per mole of another.

Identifying the Limiting Reactant

To identify the limiting reactant quantitatively, convert all given starting amounts into moles and evaluate them against the balanced chemical equation. Two reliable approaches are:

Mole-to-Coefficient Ratio Method: Divide the available moles of each reactant by its stoichiometric coefficient in the balanced reaction. The reactant with the lowest quotient runs out first.
Product Comparison Method: Calculate the theoretical moles of product generated by complete consumption of each reactant independently. The reactant yielding the smaller quantity of product is limiting.

This method consists of calculating the expected amount of product based on each reagent. (Source: Juan Carcache · CC BY-SA 4.0 · Wikimedia Commons)

Mole-to-Coefficient Ratio Formula:

\text{Reaction Batches} = \frac{\text{Moles of Reactant}}{\text{Stoichiometric Coefficient}}

The reactant with the smallest value of

\text{Reaction Batches}

is the limiting reactant.

Limiting reactant: whoever runs out of batches first — A balanced equation is a recipe, and the reactant that runs out first is the one that decides how much product you can make. Comparing how much of each reactant you have is not enough, because the recipe consumes them at different rates. Divide the moles of each reactant by its coefficient instead, and you get the number of times the recipe could run on that reactant alone; the smallest of those numbers is the real limit. Here there is more than twice as much hydrogen as nitrogen, yet hydrogen still limits the reaction, because three hydrogen molecules are used for every one of nitrogen. Product is then calculated from the limiting reactant only, and the excess reactant simply sits there once its partner is gone.

Worked Example 2: Identifying the Limiting Reactant
In the synthesis of ammonia, nitrogen reacts with hydrogen: $\ce{N2 + 3H2 -> 2NH3}$ . If a mixture of $\pu{2.00 mol}$ of $\ce{N2}$ and $\pu{5.00 mol}$ of $\ce{H2}$ is allowed to react, identify the limiting reactant.

Solution using Dimensional Analysis:
Calculate the number of reaction batches available from each reactant: $\text{Batches from }\ce{N2} = \frac{\pu{2.00 mol N2}}{1} = 2.00$ $\text{Batches from }\ce{H2} = \frac{\pu{5.00 mol H2}}{3} = 1.67$ Since $1.67 < 2.00$ , $\ce{H2}$ produces fewer reaction batches and is completely consumed first. Therefore, $\ce{H2}$ is the limiting reactant.

Excess Reactants and Calculating Leftover Reactant

An excess reactant is any reactant present in a quantity greater than what is required to react completely with the limiting reactant. Once the limiting reactant is exhausted, the excess reactant stops reacting and remains in the reaction mixture.

To calculate the amount of excess reactant remaining after complete reaction:

Identify the limiting reactant.
Use mole ratios from the balanced chemical equation to determine how many moles of the excess reactant are consumed by the limiting reactant.
Subtract the consumed moles from the initial moles supplied: $\text{Moles Remaining} = \text{Initial Moles} - \text{Consumed Moles}$
Convert remaining moles to mass ( $\pu{g}$ ) using molar mass if requested.

Worked Example 3: Calculating Excess Reactant Remaining
For the reaction $\ce{N2 + 3H2 -> 2NH3}$ , suppose you start with $\pu{28.0 g}$ of $\ce{N2}$ and $\pu{10.0 g}$ of $\ce{H2}$ . Calculate the mass of excess reactant remaining after the reaction completes.

Step 1: Convert starting masses to moles. $\pu{28.0 g N2} \times \frac{\pu{1 mol N2}}{\pu{28.02 g N2}} = \pu{0.999 mol N2}$ $\pu{10.0 g H2} \times \frac{\pu{1 mol H2}}{\pu{2.016 g H2}} = \pu{4.96 mol H2}$

Step 2: Identify the limiting reactant. $\text{Batches from }\ce{N2} = \frac{\pu{0.999 mol}}{1} = 0.999, \quad \text{Batches from }\ce{H2} = \frac{\pu{4.96 mol}}{3} = 1.65$ $\ce{N2}$ is the limiting reactant because $0.999 < 1.65$ .

Step 3: Calculate consumed excess reactant ( $\ce{H2}$ ). $\pu{0.999 mol N2} \times \frac{\pu{3 mol H2}}{\pu{1 mol N2}} = \pu{3.00 mol H2 consumed}$

Step 4: Calculate remaining $\ce{H2}$ and convert to grams. $\pu{4.96 mol H2 initial} - \pu{3.00 mol H2 consumed} = \pu{1.96 mol H2 remaining}$ $\pu{1.96 mol H2} \times \frac{\pu{2.016 g H2}}{\pu{1 mol H2}} = \pu{3.95 g H2 remaining}$

Common DAT Exam Trap: Subtracting the starting mass of limiting reactant directly from the starting mass of excess reactant ( $\pu{10.0 g} - \pu{28.0 g}$ ). Mass subtraction is invalid; calculations must always go through stoichiometric mole conversions.

Summary & Mnemonic

Feature	Limiting Reactant	Excess Reactant
Consumption Level	$100\%$ consumed at reaction completion	Partially consumed; leftover portion remains
Role in Reaction	Dictates maximum product yield	Present in abundance; does not limit product yield
Mole / Coefficient Ratio	Smallest numerical ratio	Larger numerical ratio

LIMIT Mnemonic: Lowest Input Moles Indicate Termination! The reactant that yields the lowest number of stoichiometric reaction batches terminates the reaction first.

Critical Thinking Questions

If equal masses of $\ce{H2}$ and $\ce{O2}$ are reacted according to $\ce{2H2 + O2 -> 2H2O}$ , why is $\ce{O2}$ always the limiting reactant despite starting with the same mass as $\ce{H2}$ ?
How does a non-100% reaction yield (incomplete conversion) impact the actual amount of excess reactant remaining compared to the theoretical leftover amount?
In industrial chemical processes, why is the more expensive reagent intentionally designated as the limiting reactant?

References

Sources returned by Google Search grounding during lesson verification.

7.2: Theoretical Yield, Limiting and Excess Reagents - Chemistry LibreTexts — chem.libretexts.org
Learn: Limiting reactant and reaction yields - Khan Academy — khanacademy.org

Limiting and Excess Reactants

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