Limiting Reactant & Percent Yield

Which ingredient runs out first decides how much you can make — and reality rarely gives you the full amount.

High schoolIntro Gen ChemUni Year 1
⏱️ About 22 min

You want to make sandwiches, each needing 2 slices of bread and 1 slice of cheese. With 10 slices of bread but only 3 of cheese, you can make 3 sandwiches — then you're out of cheese, with bread to spare. Reactions work exactly like this: whichever ingredient runs out first caps the product.

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The big idea: The limiting reactant is the one that runs out first and sets the maximum product — but you find it with the reaction's mole ratio, not by eyeballing which amount is smallest. The most you could make is the theoretical yield; what you actually collect, as a fraction of that, is the percent yield.
🎯 By the end, you'll be able to
  • Use the balanced equation's mole ratio to identify the limiting reactant
  • See why the limiting reactant isn't simply 'the smaller amount'
  • Calculate the theoretical yield from the limiting reactant
  • Compute percent yield from actual and theoretical yields
📎 Helpful to know first

The mole ratio is the recipe

A balanced equation is a recipe written in moles. For N₂ + 3H₂ → 2NH₃, the coefficients say: every 1 mole of N₂ needs 3 moles of H₂ and yields 2 moles of NH₃. Those ratios — 1 : 3 : 2 — are the mole ratios, and every reaction calculation runs on them.

When you mix reactants, they'll almost never be in the exact ratio the recipe wants. One will run out while some of the other is still left over. The one that runs out is the limiting reactant; it decides how much product you can make. The leftover one is in excess.

⚠️ The trap: 'limiting' ≠ 'smallest amount'
It's tempting to say whichever reactant you have fewer moles of is limiting. That's wrong. What matters is how each amount compares to what the recipe demands. A reactant needed in a 3 : 1 ratio gets used up three times as fast — so you can have more of it and still run out first. Always check against the mole ratio, never by raw size.
\[ \ce{N2 + 3H2 -> 2NH3} \]
Mole ratio 1 : 3 : 2. One mole of nitrogen reacts with three moles of hydrogen to make two moles of ammonia.
📝 Worked example: React 1.0 mol N₂ with 6.0 mol H₂ (N₂ + 3H₂ → 2NH₃). Which is limiting, and how much NH₃ forms?
  1. The recipe needs H₂ and N₂ in a 3 : 1 ratio. For 1.0 mol N₂ you need 3 × 1.0 = 3.0 mol H₂.
  2. You have 6.0 mol H₂ — more than the 3.0 mol required. So H₂ is in excess and N₂ is the limiting reactant.
  3. Base the product on the limiting reactant: 1 mol N₂ → 2 mol NH₃, so 1.0 mol N₂ → 2.0 mol NH₃.
  4. The extra 3.0 mol H₂ (6.0 supplied − 3.0 used) is left over, unreacted.
✓ N₂ is limiting; 2.0 mol NH₃ forms, with 3.0 mol H₂ left over.

Theoretical yield vs what you actually get

The amount of product the limiting reactant could make, if everything went perfectly, is the theoretical yield. In the real world you always get less — some product is lost in transfers, side reactions eat reactant, or the reaction doesn't fully finish. What you actually collect is the actual yield.

The percent yield tells you how efficient the reaction was: actual divided by theoretical, times 100.

\[ \%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \]
A percent yield of 100% means you got every bit the recipe promised; real reactions usually land below that.
✏️ Practice: For 2H₂ + O₂ → 2H₂O, you mix 3.0 mol H₂ with 2.0 mol O₂. What is the maximum number of moles of H₂O that can form?
mol
Solution
  1. Check each reactant against the 2 : 1 ratio. The 3.0 mol H₂ needs 3.0 ÷ 2 = 1.5 mol O₂; you have 2.0 mol O₂, so O₂ is in excess.
  2. That makes H₂ the limiting reactant — even though 3.0 mol H₂ is a larger number than 2.0 mol O₂. (This is the 'smallest amount' trap in action.)
  3. H₂ and H₂O are in a 2 : 2 (= 1 : 1) ratio, so 3.0 mol H₂ → 3.0 mol H₂O. (Using O₂ instead would wrongly predict 4.0 mol, but there isn't enough H₂ for that.)
✏️ Practice: A reaction has a theoretical yield of 25.0 g, but you actually collect 20.0 g of product. What is the percent yield?
%
Solution
  1. Use % yield = (actual ÷ theoretical) × 100.
  2. = (20.0 g ÷ 25.0 g) × 100.
  3. = 0.800 × 100 = 80.0%.

Check your understanding

1. The limiting reactant is best described as the one that…
It's the reactant that runs out first once you compare each amount to what the balanced equation's mole ratio demands — not simply the smallest amount.
2. For 2H₂ + O₂ → 2H₂O with 3.0 mol H₂ and 2.0 mol O₂, which is limiting?
3.0 mol H₂ needs only 1.5 mol O₂, so O₂ (2.0 mol) is in excess and H₂ is limiting — despite H₂ being the larger number of moles.
3. Theoretical yield is 40.0 g and you collect 30.0 g. The percent yield is:
(30.0 ÷ 40.0) × 100 = 75.0%. Percent yield is a ratio of actual to theoretical, so it's unitless and normally at or below 100%.
🔎 Related lessons
Previous: Balancing Chemical Equations Next: Solution Stoichiometry & Molarity
✅ Key takeaways
  • A balanced equation gives mole ratios — the recipe every reaction calculation runs on.
  • The limiting reactant runs out first and sets the maximum product.
  • Find it with the mole ratio, not by picking the smallest amount — a larger amount can still run out first.
  • Theoretical yield = the most the limiting reactant could make; actual yield = what you collect.
  • Percent yield = (actual ÷ theoretical) × 100, usually below 100% in practice.
➡️ So far reactants have been dry moles and grams. But most chemistry happens in solution, where amount is set by concentration and volume. Next: molarity, dilution, and titration — stoichiometry in a beaker of liquid.
Want to test yourself on this? Try the Chemistry practice test →