Limiting Reagent: How to Find It (With Worked Examples)

A sandwich shop has 40 slices of bread and 30 slices of cheese. Each sandwich takes 2 slices of bread and 1 slice of cheese. How many sandwiches can they make?

Not 30, even though there are 30 cheese slices. The bread only covers 20 sandwiches, so after 20 the shop is out of bread with 10 cheese slices left over. Bread is the limiting ingredient. It decides how much product you get, and the cheese is in excess.

Chemical reactions work exactly the same way. The only complications are that chemicals are measured in grams, not counted, and the “recipe” is a balanced equation.

The method

1. Balance the equation. The coefficients are the recipe.
2. Convert every reactant to moles. You can’t compare grams of different substances directly.
3. Divide each reactant’s moles by its coefficient.
4. The smallest result is the limiting reagent. That number tells you how many “batches” of the reaction can run.

Step 3 is the one people skip, and it’s the one that matters. It’s what turns “40 slices of bread” into “enough for 20 sandwiches”.

Worked example 1: hydrogen and oxygen

10.0 g of hydrogen reacts with 32.0 g of oxygen to form water. Which is limiting, and how much water forms?

Balance: 2H₂ + O₂ → 2H₂O

Moles:

• H₂: 10.0 g ÷ 2.016 g/mol = 4.96 mol
• O₂: 32.0 g ÷ 32.00 g/mol = 1.00 mol

Divide by coefficients:

• H₂: 4.96 ÷ 2 = 2.48
• O₂: 1.00 ÷ 1 = 1.00

Oxygen is limiting, even though there’s more than three times as much of it by mass. Hydrogen molecules are so light that 10 g of them is a huge number of molecules.

Theoretical yield: the reaction can run 1.00 “batch”, and each batch makes 2 mol of water. So 2.00 mol × 18.015 g/mol = 36.0 g of water.

Leftover hydrogen: 1.00 batch uses 2 × 1.00 = 2.00 mol of H₂. We had 4.96 mol, so 2.96 mol is left over — 2.96 × 2.016 = 5.97 g of unreacted hydrogen.

Check with conservation of mass: 10.0 + 32.0 = 42.0 g in; 36.0 g water + 6.0 g hydrogen = 42.0 g out.

Worked example 2: making ammonia

28.0 g of nitrogen reacts with 4.00 g of hydrogen. How much ammonia can form?

Balance: N₂ + 3H₂ → 2NH₃

• N₂: 28.0 ÷ 28.014 = 1.00 mol → ÷ 1 = 1.00
• H₂: 4.00 ÷ 2.016 = 1.98 mol → ÷ 3 = 0.661

Hydrogen is limiting. This time the reactant with the smaller mass is limiting — but only because of the coefficient of 3. Without step 3 you’d compare 1.00 to 1.98 and wrongly conclude nitrogen runs out first.

Ammonia formed: 0.661 × 2 = 1.32 mol × 17.03 g/mol = 22.5 g.

Theoretical, actual and percent yield

The amount you calculate from the limiting reagent is the theoretical yield — the maximum possible if everything goes perfectly. In a real lab you’ll usually get less, because of side reactions, reactions that don’t go to completion, and product lost while filtering or transferring. What you actually collect is the actual yield, and:

percent yield = actual yield ÷ theoretical yield × 100

If the ammonia reaction above produced 18.0 g, the percent yield would be 18.0 ÷ 22.5 × 100 = 80%. More on this in percent yield explained.

Why chemists deliberately use an excess

In industry and in the lab, one reactant is often added in excess on purpose. If one reactant is expensive and another is cheap — say, oxygen from the air — using extra of the cheap one makes sure every bit of the expensive one reacts. Combustion engines, for example, are tuned to run with a slight excess of air so all the fuel burns.

Common mistakes

• Comparing masses instead of moles. Example 1 shows how badly that goes.
• Comparing moles without dividing by coefficients. Example 2 shows that one.
• Calculating yield from the excess reactant. Product amounts always come from the limiting reagent.
• Using an unbalanced equation. Every step after that is wrong. The equation balancer takes care of this if you’re unsure.

Quick answers

Is the limiting reagent always the one with fewer moles? No — only after you divide by the coefficients. A reactant with a big coefficient is used up faster.

Can there be no limiting reagent? If the reactants are in exactly the stoichiometric ratio, they run out at the same moment. Both are limiting, and nothing is left over.

What is the excess reagent? Any reactant that isn’t limiting. Some of it remains after the reaction stops.

Try it

The limiting reagent calculator balances the equation for you, finds the limiting reagent from grams or moles, and reports the theoretical yield of every product, the leftover excess and the percent yield. For the general method behind all of this, see stoichiometry step by step.