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Limiting Reactants and Percent Yield

How do you find the limiting reactant?

IntermediateReactions & StoichiometryLast reviewed 3 October 2026
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What is it?

In most real reactions the reactants are not mixed in exactly the ratio of the balanced equation. One of them runs out first.

  • The limiting reactant is the reactant that is used up first. It limits how much product can form.
  • Any other reactant is in excess: some of it is left over when the reaction stops.
  • The theoretical yield is the maximum amount of product, calculated from the limiting reactant.
  • The actual yield is the amount of product really obtained in an experiment.
  • The percent yield compares the two.

Key idea

The amount of product depends only on the reactant that runs out first, the limiting reactant, not on how much of the other reactants you add.

Why does it matter?

  • Chemists choose the limiting reactant on purpose. The most expensive or hardest-to-make reactant is often the limiting one, while a cheap one (such as oxygen from air) is used in excess so that none of the valuable reactant is wasted.
  • Percent yield measures efficiency. In the lab and in industry, a low yield means wasted materials, money and energy.
  • Excess reactant must be dealt with. Leftover reactant often has to be separated from the product, recycled or disposed of safely.

How does it work?

Finding the limiting reactant

  1. Convert the amount of each reactant into moles.
  2. For each reactant, calculate how much product it could make on its own, using the mole ratio from the balanced equation.
  3. The reactant that gives the smaller amount of product is the limiting reactant. That smaller amount is the theoretical yield.

An equivalent method: take one reactant, calculate how much of the other it needs, and compare that with how much you actually have.

Excess reactant left over

Use the limiting reactant to calculate how much of the excess reactant is used up, then subtract that from the amount you started with.

Percent yield

% yield=actualtheoretical×100%\%\ \text{yield} = \frac{\text{actual}}{\text{theoretical}} \times 100\%

Both yields must be in the same units (usually grams).

Think of it like this

Making sandwiches needs 2 slices of bread + 1 slice of cheese → 1 sandwich. You have 10 slices of bread but only 3 slices of cheese. The cheese runs out after 3 sandwiches, with 4 slices of bread left over. The cheese is the limiting “reactant”, even though you have fewer slices of it. If you drop one sandwich on the floor and only 2 are eaten, your “percent yield” is 2 ÷ 3 × 100% = 67%.

More precisely

Percent yields are usually below 100% because reactions may not go to completion, side reactions can make unwanted products, and some product is lost while it is separated and purified. A yield above 100% doesn’t mean extra product was created. It signals a problem, for example a product that is still wet or contains impurities.

Visualise it

Particle diagram for 2 H2 plus O2 gives 2 H2O. Before the reaction there are 5 hydrogen molecules and 2 oxygen molecules. After the reaction there are 4 water molecules and 1 hydrogen molecule left over. Oxygen runs out first, so oxygen is the limiting reactant and hydrogen is in excess.
Two O₂ molecules can react with only four H₂ molecules, so one H₂ is left over.

Worked example

Worked example: Limiting reactant, theoretical yield and excess

Question: 28.0 g of nitrogen and 9.00 g of hydrogen react to make ammonia. Which is the limiting reactant? What mass of ammonia can form, and what mass of the excess reactant is left over? (N = 14.01, H = 1.008)

NX2+3 HX2→2 NHX3\ce{N2 + 3H2 -> 2NH3}
  1. Moles: NX2\ce{N2}: 28.028.02=0.9993\dfrac{28.0}{28.02} = 0.9993 mol; HX2\ce{H2}: 9.002.016=4.464\dfrac{9.00}{2.016} = 4.464 mol
  2. Hydrogen needed by all the nitrogen: 0.9993×31=2.9980.9993 \times \dfrac{3}{1} = 2.998 mol. We have 4.464 mol, which is more than enough, so nitrogen is the limiting reactant.
  3. Theoretical yield: 0.9993×21=1.9990.9993 \times \dfrac{2}{1} = 1.999 mol NHX3\ce{NH3}, and 1.999×17.03=34.041.999 \times 17.03 = 34.04 g, which is 34.0 g
  4. Hydrogen left over: 4.464−2.998=1.4664.464 - 2.998 = 1.466 mol, and 1.466×2.016=1.466 \times 2.016 = 2.96 g

Check: 28.0 g + 9.00 g = 37.0 g of reactants, and 34.0 g + 2.96 g = 37.0 g of products and leftovers.

Worked example: Percent yield

Question: In the reaction above, the chemist actually collects 27.2 g of ammonia. What is the percent yield?

  1. Actual yield = 27.2 g; theoretical yield = 34.04 g (use the unrounded value from the previous calculation)
  2. Percent yield = 27.234.04×100%=79.9%\dfrac{27.2}{34.04} \times 100\% = 79.9\%

Answer: the percent yield is 79.9%.

Common mistake

Common mistake: Picking the reactant with the smaller mass

In the example, there are 28.0 g of nitrogen but only 9.00 g of hydrogen, yet nitrogen is the limiting reactant. Mass alone tells you nothing; you must compare moles, using the mole ratio.

Common mistake: Picking the reactant with fewer moles

Even fewer moles doesn’t automatically mean limiting. In 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O} with 1.0 mol of HX2\ce{H2} and 0.6 mol of OX2\ce{O2}, the hydrogen needs only 0.50 mol of oxygen, so hydrogen is limiting, even though there are fewer moles of oxygen. Always use the ratio from the balanced equation.

Common mistake: Calculating the yield from the excess reactant

The theoretical yield must come from the limiting reactant. Using the excess reactant overestimates it (in the example, hydrogen alone could make 50.7 g of ammonia, but only 34.0 g can form).

Notation note

  • The limiting reactant is also called the limiting reagent; an excess reactant is also called an excess reagent.
  • “Yield” can be given as a mass, an amount in moles, or a percentage. Check which one a question asks for.

Remember this

Remember this

  • The limiting reactant runs out first and decides how much product forms.
  • Compare reactants in moles using the mole ratio, never by mass alone.
  • Theoretical yield comes from the limiting reactant.
  • Percent yield = actual ÷ theoretical × 100%.

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What is the limiting reactant?
    Answer

    The reactant that is used up first. It limits how much product can form.

  2. Question
    What is an excess reactant?
    Answer

    A reactant that is not used up completely; some is left over when the reaction stops.

  3. Question
    How do you find the limiting reactant?
    Answer

    Convert each reactant to moles, calculate how much product each could make, and pick the one that gives less product.

  4. Question
    What is the theoretical yield?
    Answer

    The maximum amount of product that can form, calculated from the limiting reactant.

  5. Question
    What is the formula for percent yield?
    Answer

    Percent yield = (actual yield ÷ theoretical yield) × 100%

  6. Question
    Is the reactant with the smaller mass always the limiting reactant?
    Answer

    No. Compare moles, using the mole ratio from the balanced equation. Mass alone tells you nothing.

  7. Question
    In 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, 1.0 mol of HX2\ce{H2} mixes with 0.6 mol of OX2\ce{O2}. Which is limiting?
    Answer

    HX2\ce{H2}. It needs only 0.50 mol of OX2\ce{O2}, and 0.6 mol is available, so oxygen is in excess.

  8. Question
    Give three reasons why a percent yield is usually less than 100%.
    Answer

    The reaction may not go to completion; side reactions make other products; some product is lost during separation and purification.

  9. Question
    What does a percent yield above 100% suggest?
    Answer

    An error, for example the product is still wet or contains impurities. Extra product cannot be created.

Quiz

7 questions

  1. Question 1EasyWhat is the limiting reactant in a reaction?
    Show answer

    Answer: The reactant that is used up first

    The limiting reactant runs out first, so it decides how much product can form. The reactant left over at the end is the excess reactant.

  2. Question 2MediumIn 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, 1.0 mol of HX2\ce{H2} is mixed with 0.6 mol of OX2\ce{O2}. Which is the limiting reactant?
    Show answer

    Answer: HX2\ce{H2}

    1.0 mol of HX2\ce{H2} needs 1.0×12=0.501.0 \times \dfrac{1}{2} = 0.50 mol of OX2\ce{O2}. There is 0.6 mol available, so oxygen is in excess and hydrogen runs out first, even though there are fewer moles of oxygen.

  3. Question 3MediumIn NX2+3 HX2→2 NHX3\ce{N2 + 3H2 -> 2NH3}, 2.0 mol of NX2\ce{N2} is mixed with 3.0 mol of HX2\ce{H2}. How many moles of NHX3\ce{NH3} can form?
    Show answer

    Answer: 2.0 mol

    All the NX2\ce{N2} would need 6.0 mol of HX2\ce{H2}, but only 3.0 mol is available, so HX2\ce{H2} is limiting: 3.0×23=2.03.0 \times \dfrac{2}{3} = 2.0 mol of NHX3\ce{NH3}. The answer 4.0 mol uses the excess reactant.

  4. Question 4EasyThe theoretical yield of a reaction is 50.0 g, and 42.5 g of product is collected. What is the percent yield?
    Show answer

    Answer: 85.0%

    Percent yield = 42.550.0×100%=85.0%\dfrac{42.5}{50.0} \times 100\% = 85.0\%. The answer 118% divides the wrong way round.

  5. Question 5MediumIn 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, 4.0 mol of HX2\ce{H2} reacts with 3.0 mol of OX2\ce{O2}. How much of the excess reactant is left over?
    Show answer

    Answer: 1.0 mol of OX2\ce{O2}

    4.0 mol of HX2\ce{H2} needs only 2.0 mol of OX2\ce{O2}, so hydrogen is limiting. Oxygen left over = 3.0 − 2.0 = 1.0 mol.

  6. Question 6Hard16.0 g of methane burns with 48.0 g of oxygen: CHX4+2 OX2→COX2+2 HX2O\ce{CH4 + 2O2 -> CO2 + 2H2O}. What mass of COX2\ce{CO2} can form? (C = 12.01, H = 1.008, O = 16.00)
    Show answer

    Answer: 33.0 g

    Moles: CHX4\ce{CH4} = 16.0 ÷ 16.04 = 0.9974 mol; OX2\ce{O2} = 48.0 ÷ 32.00 = 1.50 mol. The methane would need 1.995 mol of OX2\ce{O2}, so oxygen is limiting. COX2\ce{CO2} = 1.50 ÷ 2 = 0.750 mol, and 0.750 × 44.01 = 33.0 g. The answer 43.9 g uses the excess reactant (methane).

  7. Question 7EasyA student reports a percent yield of 112%. What is the most likely explanation?
    Show answer

    Answer: The product was still wet or contained impurities

    Mass cannot be created, so a yield above 100% signals an error, most often a product that still contains water (solvent) or other substances.

Notes and downloads

  • Worksheet

    Limiting Reactants Worksheet

    8 questions on limiting and excess reactants, theoretical yield and percent yield, in moles and grams. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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