IB Chemistry R2.1 R2.1.2
R2.1.2

Limiting & Excess Reagents

Identifying which reagent runs out first and calculating percentage yield.

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 reagent

The reactant that is completely consumed during the reaction, thereby determining the theoretical maximum yield of product.

How to Identify the Limiting Reagent

Three-Step Method

  1. Calculate moles of each reactant from the given quantities.
  2. Divide each by its stoichiometric coefficient from the balanced equation.
  3. The reactant with the smallest value is the limiting reagent.

Limiting Reagent Method

Diagram: Limiting Reagent Method Step 1 Calculate moles Step 2 Divide by coefficient Step 3 Smallest = limiting Example: Fe(s) + S(s) → FeS(s) n(Fe) = 5.0/55.85 = 0.0895 mol n(S) = 5.0/32.07 = 0.156 mol Ratio is 1:1, so divide by 1: Fe = 0.0895, S = 0.156 Fe = 0.0895 (smallest) → Fe is limiting

Think About It

In the example above, how much excess sulphur (in grams) remains after the reaction?

Moles S used = 0.0895 mol (same as Fe, 1:1 ratio). Moles S remaining = 0.156 − 0.0895 = 0.067 mol. Mass remaining = 0.067 × 32.07 = 2.1 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:

\( \% \text{ Yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \)

✏️ Worked Example
A student reacts 5.00 g of iron with excess sulphur and collects 7.20 g of FeS.
Calculate the percentage yield.

Step 1: Theoretical yield (using the limiting reagent, Fe):

n(Fe) = 5.00 / 55.85 = 0.0895 mol → n(FeS) = 0.0895 mol

Theoretical mass = 0.0895 × 87.91 = 7.87 g

Step 2: Percentage yield:

\( \% \text{ Yield} = \frac{7.20}{7.87} \times 100 = \textbf{91.5\%} \)

Why Yields Fall Below 100%

Common Reasons for Reduced Yield

  • Transfer losses: Material lost during filtering, pouring, or scraping
  • Side reactions: Competing reactions that produce unwanted by-products
  • Reversible reactions: Equilibrium prevents complete conversion
  • Impure reagents: Some starting material is not the intended reactant

What About Yields Above 100%?

A yield above 100% is physically impossible for the desired product. It typically results from incomplete drying (trapped solvent adds mass) or co-precipitated impurities that add weight to the product.

← R2.1.1 StoichiometryR2.1.3 Atom Economy →