General Chemistry I · Stoichiometry
Limiting and Excess Reactants
On this page 7 sections
In 30 seconds
In a real reaction the reactants are rarely present in exactly the stoichiometric ratio. The limiting reactant is the one that is completely consumed first and therefore determines the maximum amount of product that can form. The excess reactant is any reactant left over after the reaction stops. Because the reaction cannot continue once the limiting reactant runs out, every yield calculation must be based on the limiting reactant, not on whichever reactant has the larger mass.
Why this matters
Industrial reactions are run with one reactant in excess (usually the cheaper one) so that the valuable reactant is fully consumed. Understanding which reactant limits lets chemists calculate true maximum yield, minimize waste, and cut cost — whether synthesizing fertilizer (the Haber process above), a drug, or a polymer. Misidentifying the limiting reactant overestimates yield and wastes money.
The college version
Key Ideas
Limiting vs. excess
- Limiting reactant: used up first; sets the ceiling on product.
- Excess reactant: present in more than the required amount; some remains unreacted.
- Which one is limiting depends on the amounts and the mole ratio, not on which has more grams.
Two equivalent methods
- Compare product yield: calculate how much product each reactant could make; the reactant that makes the least product is limiting.
- Mole-ratio (required vs. available): convert each reactant to moles, then compare the available ratio to the coefficient ratio to see which runs out first.
A key misconception to avoid
- The reactant with the smaller mass is not automatically limiting — molar masses differ, so convert to moles first.
Equations and Variables
- Moles: n = m / M
- Required amount of B for A: n(B needed) = n(A) × (coeff B / coeff A)
- Maximum product from reactant X: n(product) = n(X) × (coeff product / coeff X)
- Excess remaining: n(excess left) = n(excess initial) − n(excess consumed)
How It Works (Problem-Solving Method)
- Balance the equation.
- Convert all given reactant amounts to moles.
- Identify the limiting reactant — either compute product from each reactant (smaller = limiting), or compare needed vs. available using the mole ratio.
- Calculate maximum product using the limiting reactant's moles and the mole ratio.
- If asked, find excess remaining by subtracting consumed from initial for the excess reactant.
Worked Example
For N₂ + 3 H₂ → 2 NH₃, if you have 3.0 mol N₂ and 6.0 mol H₂, which is limiting and how much NH₃ forms?
Mole-ratio method:
- H₂ required to react with 3.0 mol N₂ = 3.0 mol N₂ × (3 mol H₂ / 1 mol N₂) = 9.0 mol H₂ needed.
- Only 6.0 mol H₂ is available, so H₂ is limiting (N₂ is in excess).
Maximum NH₃ from the limiting H₂:
- 6.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 4.0 mol NH₃.
Excess N₂ remaining:
- N₂ consumed = 6.0 mol H₂ × (1 mol N₂ / 3 mol H₂) = 2.0 mol N₂.
- N₂ left = 3.0 − 2.0 = 1.0 mol N₂.
Mass-based version: 10.0 g N₂ and 3.00 g H₂ → NH₃.
- n(N₂) = 10.0 ÷ 28.02 = 0.357 mol; n(H₂) = 3.00 ÷ 2.016 = 1.49 mol.
- H₂ needed for 0.357 mol N₂ = 0.357 × 3 = 1.07 mol H₂; 1.49 mol available → N₂ is limiting.
- NH₃ = 0.357 mol N₂ × (2/1) = 0.714 mol → 0.714 × 17.03 = 12.2 g NH₃.
Common Confusions
- "The reactant with less mass is limiting." — Wrong; a heavy reactant can still be limiting if the ratio demands a lot of it. Always compare in moles.
- "Use the excess reactant to compute product." — Product is set by the limiting reactant only; using excess overestimates the answer.
- "Excess reactant means leftover is automatically known." — You must subtract the amount consumed (found via the mole ratio) from the initial amount.
- "There is always exactly one limiting reactant." — Usually yes, but if reactants are in the exact stoichiometric ratio, both are fully consumed (neither is in excess).
Eli explains
The same idea, in plain words
Explain it like I’m 10
Think of making grilled-cheese sandwiches: each needs 2 slices of bread and 1 slice of cheese. You have 6 slices of bread and 4 slices of cheese. Bread can make 3 sandwiches, cheese can make 4 — so bread runs out first and limits you to 3 sandwiches, with 1 cheese slice left over. Bread is the "limiting reactant." The catch: in chemistry you can't just look at the pile sizes (grams) and guess — you have to convert to "sandwich counts" (moles) first, because a heavy pile of one chemical might still be the one that runs out first.
Key takeaways
- Limiting reactant is used up first and caps the product.
- Excess reactant is left over after reaction.
- Convert to moles before deciding which is limiting (mass alone misleads).
- Product is always calculated from the limiting reactant.
- The smaller mass reactant is not necessarily limiting.
- Excess remaining = initial − consumed.
- Limiting reactant runs out first and sets maximum product.
- Convert to moles, then compare needed vs. available (or compare product yields).
- Compute product from the limiting reactant's moles.
- Excess left = initial − consumed.
Study tools & related lessonsYou’ll learn to · Related
You’ll learn to
- Define limiting reactant and excess reactant.
- Identify the limiting reactant using both the compare-product-yield and the mole-ratio methods.
- Calculate the maximum amount of product and the amount of excess reactant remaining.
- Explain why the limiting reactant controls the reaction's yield.
Sources & references
- OpenStax, "4.4 Reaction Yields." *Chemistry 2e*.
- OpenStax, "4.3 Reaction Stoichiometry." *Chemistry 2e*.
- Chemistry LibreTexts, "7.4: How to Write Balanced Chemical Equations."
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.