Stoichiometry: Limiting & Excess Reactant

Identify which reactant runs out first and quantify how much of the others are wasted.

Material & Energy BalancesChemical Engineering Year 1Free preview

⏱️ About 14 min

You feed 100 mol of N₂ and 250 mol of H₂ into a reactor making ammonia. Which reactant runs out first — and what happens to the rest?

💡

The big idea: The limiting reactant determines the maximum possible product; everything else is in excess.

🎯 By the end, you'll be able to

Write and verify balanced chemical equations
Identify the limiting reactant from given feed amounts
Calculate percent excess of the excess reactant
Explain why the limiting reactant controls product yield

📎 Helpful to know first

Unsteady-State Material Balances

Balanced Equations & Stoichiometric Coefficients

A balanced chemical equation shows the molar ratios in which reactants are consumed and products are formed. For ammonia synthesis:

N₂ + 3H₂ → 2NH₃

The numbers in front of each species are the stoichiometric coefficients ν_i: ν_N₂ = 1, ν_H₂ = 3, ν_NH₃ = 2. They tell us that 1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃. A balanced equation must have the same number of each atom on both sides — check: 2 N atoms and 6 H atoms on each side. ✓

\[ \text{Percent excess} = \frac{n_{\text{fed}} - n_{\text{stoichiometric}}}{n_{\text{stoichiometric}}} \times 100\% \]

Percent excess compares how much of a reactant is fed beyond what the limiting reactant requires.

Limiting vs Excess Reactant

The limiting reactant is the one that runs out first — it determines the maximum amount of product that can form. The other reactants are in excess: more is fed than can possibly react.

To find the limiting reactant, divide each feed amount by its stoichiometric coefficient. The species with the smallest ratio is limiting. For N₂ + 3H₂ → 2NH₃ with 100 mol N₂ and 250 mol H₂: N₂ ratio = 100/1 = 100; H₂ ratio = 250/3 = 83.3. Since 83.3 < 100, H₂ is limiting.

⚠️ Compare ratios, not raw amounts

A common mistake is to say 'we have less N₂ than H₂, so N₂ is limiting.' That logic fails because stoichiometry demands 3 mol H₂ per 1 mol N₂. Always divide feed amount by stoichiometric coefficient before comparing.

📝 Worked example: Ammonia is synthesized by N₂ + 3H₂ → 2NH₃. The feed contains 100 mol N₂ and 250 mol H₂. Identify the limiting reactant and calculate the percent excess of the other reactant.

Stoichiometric ratio: 1 mol N₂ : 3 mol H₂
Divide feed by stoichiometric coefficient: N₂ → 100/1 = 100; H₂ → 250/3 = 83.3
Smaller ratio is H₂ (83.3 < 100), so H₂ is the limiting reactant
N₂ required to react with all H₂: 250/3 = 83.33 mol N₂
N₂ fed = 100 mol; excess = 100 - 83.33 = 16.67 mol
Percent excess N₂ = (16.67 / 83.33) × 100 = 20.0%

✓ H₂ is the limiting reactant; N₂ is 20.0% in excess

✏️ Practice: For the reaction N₂ + 3H₂ → 2NH₃, a feed contains 50 mol N₂ and 200 mol H₂. What is the percent excess of the excess reactant?

Solution

Ratios: N₂ → 50/1 = 50; H₂ → 200/3 = 66.7. N₂ is limiting (smaller ratio)
H₂ required for 50 mol N₂: 3 × 50 = 150 mol
Excess H₂ = 200 - 150 = 50 mol
Percent excess H₂ = (50 / 150) × 100 = 33.3%

Check your understanding

1. In the reaction N₂ + 3H₂ → 2NH₃, if you feed equal moles of N₂ and H₂, which is the limiting reactant?

With equal moles, H₂ has the smaller feed/coefficient ratio (n/3 vs n/1), so H₂ is limiting.

2. Percent excess of a reactant is calculated as:

Percent excess = (amount fed − amount stoichiometrically required) / amount required × 100.

✅ Key takeaways

Stoichiometric coefficients give the molar ratios in a balanced reaction
The limiting reactant has the smallest feed/coefficient ratio and controls maximum product
Percent excess = (fed − required) / required × 100
Always divide by stoichiometric coefficients before comparing feed amounts

➡️ Now that we can identify the limiting reactant and percent excess, we need a systematic way to track how far a reaction has proceeded. That tool is the extent of reaction.

Want to test yourself on this? Try the Chemical Aptitude test →