The Limiting Reagent
When reagents are not mixed in exact stoichiometric proportions, one will be used up first. This is the limiting reagent, and it determines the maximum amount of product that can form. The other reagent is in excess.
Limiting and Excess Reactants
Limiting reagent: The reactant that is completely consumed first during a chemical reaction, thereby dictating the maximum theoretical amount of product that can form. Any remaining reactants are said to be in excess.
How to Identify the Limiting Reagent
Three-Step Method for Identifying the Limiting Reagent
- Calculate initial moles of each reactant: \( n = \dfrac{m}{M} \text{ or } n = cV \).
- Divide each mole amount by its stoichiometric coefficient from the balanced equation.
- The smallest resulting value identifies the limiting reagent. Use its initial moles for all subsequent theoretical product calculations.
Limiting Reagent Method
Calculating Unreacted Excess Reactant Mass
In the \(\text{Fe} + \text{S}\) reaction: \(0.0895\text{ mol}\text{ Fe}\) reacts with exactly \(0.0895\text{ mol}\text{ S}\).
Initial \(\text{S} = 0.125\text{ mol}\). Unreacted \(\text{S} = 0.125 - 0.0895 = 0.0355\text{ mol}\).
Mass of remaining excess \(\text{S} = 0.0355\text{ mol} \times 32.06\text{ g mol}^{-1} = \mathbf{1.14\text{ g}}\).
Percentage Yield
The theoretical yield is the maximum mass of product calculated from stoichiometry (using the limiting reagent). The actual yield is what you physically collect in the lab. The percentage yield compares these two values:
Percentage Yield Equation
\( \text{Percentage Yield} = \dfrac{\text{Experimental (Actual) Yield}}{\text{Theoretical Yield}} \times 100\% \)
Both yields must be in identical units (both in grams or both in moles).
Calculating Percentage Yield from Experimental Data
Problem: A student reacts \(5.00\text{ g}\) of iron with excess sulphur and collects \(7.20\text{ g}\) of \(\text{FeS}\). Calculate the percentage yield.
\(\text{Fe}(\text{s}) + \text{S}(\text{s}) \rightarrow \text{FeS}(\text{s})\)
Step 1: Calculate theoretical moles of \(\text{FeS}\)
\(n(\text{Fe}) = \dfrac{5.00}{55.85} = 0.0895\text{ mol} \implies n(\text{FeS})_{\text{theoretical}} = 0.0895\text{ mol}\)
Step 2: Calculate theoretical mass of \(\text{FeS}\)
\(M(\text{FeS}) = 55.85 + 32.06 = 87.91\text{ g mol}^{-1}\)
\(m(\text{FeS})_{\text{theoretical}} = 0.0895 \times 87.91 = 7.87\text{ g}\)
Step 3: Calculate percentage yield
\(\text{Percentage Yield} = \dfrac{7.20\text{ g}}{7.87\text{ g}} \times 100 = \mathbf{91.5\%}\)
Why Yields Fall Below 100%
Key Factors Reducing Experimental Yield
- Transfer losses: Mechanical loss of substance sticking to glassware, filter paper, or evaporating dishes.
- Side reactions: Competing secondary reactions forming alternative by-products.
- Reversible reactions: Reaction reaching dynamic equilibrium before complete reactant conversion.
- Impure starting materials: Starting mass includes non-reactive contaminants.
- Incomplete separation: Loss during washing, purification, or recrystallisation.
Apparent Yields Above 100%: Experimental Errors
A true chemical yield above 100% violates the Law of Conservation of Mass and is physically impossible. If experimental data yields \(>100\%\), it indicates practical errors such as incomplete drying (wet product containing solvent/water) or unseparated impurities/by-products.
Study this topic on the go
Get active recall flashcards, notes, and topic quizzes in ChemEasy, or build your revision schedule with ChemPlan IB.