Chemistry

How to find the limiting reactant

The limiting reactant is the reactant that runs out first, so it decides how much product can form. To find it, convert each reactant to moles, then work out how much product each one could make on its own. The reactant that makes less product is limiting. The other reactant is in excess, and some of it is left over when the reaction stops.

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Key ideas

Think of making sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 7 slices of cheese, the bread is enough for 5 sandwiches and the cheese for 7. You can only make 5, so bread is limiting and 2 slices of cheese are left over. Chemistry works the same way, except you count in moles.

  1. Balance the equation.
  2. Convert every reactant amount to moles (divide grams by molar mass).
  3. For each reactant, use the mole ratio to find how much product it could make.
  4. The reactant that gives the smaller amount of product is limiting. That smaller amount is the theoretical yield.
  5. To find the leftover, work out how much of the excess reactant was used, then subtract it from what you started with.
nproduct=nreactant×coefficient of productcoefficient of reactantn_{\text{product}} = n_{\text{reactant}} \times \frac{\text{coefficient of product}}{\text{coefficient of reactant}}

Here nn is an amount in moles (mol). Run this once for each reactant and compare the answers.

Worked examples

Example 1: in moles

Problem In 2 H2+O2→2 H2O\mathrm{2\,H_2 + O_2 \rightarrow 2\,H_2O}, you start with 5.0 mol of H2\mathrm{H_2} and 3.0 mol of O2\mathrm{O_2}. Which is limiting, how much water forms, and how much of the excess reactant is left?

  1. Water from hydrogen: the ratio is 2 to 2.
    5.0 mol H2×2 mol H2O2 mol H2=5.0 mol H2O5.0\ \text{mol H}_2 \times \frac{2\ \text{mol H}_2\text{O}}{2\ \text{mol H}_2} = 5.0\ \text{mol H}_2\text{O}
  2. Water from oxygen: the ratio is 2 to 1.
    3.0 mol O2×2 mol H2O1 mol O2=6.0 mol H2O3.0\ \text{mol O}_2 \times \frac{2\ \text{mol H}_2\text{O}}{1\ \text{mol O}_2} = 6.0\ \text{mol H}_2\text{O}
  3. Hydrogen gives less water, so H2\mathrm{H_2} is limiting and 5.0 mol of water forms.
  4. Oxygen used by 5.0 mol of hydrogen, then the leftover.
    5.0 mol H2×1 mol O22 mol H2=2.5 mol O23.0−2.5=0.5 mol O25.0\ \text{mol H}_2 \times \frac{1\ \text{mol O}_2}{2\ \text{mol H}_2} = 2.5\ \text{mol O}_2 \qquad 3.0 - 2.5 = 0.5\ \text{mol O}_2

Answer H2\mathrm{H_2} is limiting, 5.0 mol of water forms, and 0.5 mol of O2\mathrm{O_2} is left over

Example 2: in grams, with leftover

Problem Aluminum reacts with chlorine: 2 Al+3 Cl2→2 AlCl3\mathrm{2\,Al + 3\,Cl_2 \rightarrow 2\,AlCl_3}. You mix 10.0 g of Al with 35.0 g of Cl2\mathrm{Cl_2}. How many grams of AlCl3\mathrm{AlCl_3} form?

  1. Grams to moles for both reactants. MCl2=2(35.45)=70.90M_{\mathrm{Cl_2}} = 2(35.45) = 70.90 g/mol.
    10.0 g26.98 g/mol=0.37064 mol Al35.0 g70.90 g/mol=0.49365 mol Cl2\frac{10.0\ \text{g}}{26.98\ \text{g/mol}} = 0.37064\ \text{mol Al} \qquad \frac{35.0\ \text{g}}{70.90\ \text{g/mol}} = 0.49365\ \text{mol Cl}_2
  2. Product from each. Al: ratio 2 to 2. Cl2\mathrm{Cl_2}: ratio 2 to 3.
    0.37064×22=0.37064 mol0.49365×23=0.32910 mol AlCl30.37064 \times \frac{2}{2} = 0.37064\ \text{mol} \qquad 0.49365 \times \frac{2}{3} = 0.32910\ \text{mol AlCl}_3
  3. Chlorine gives less, so Cl2\mathrm{Cl_2} is limiting, even though there are more grams of it.
  4. Moles of product to grams. MAlCl3=26.98+3(35.45)=133.33M_{\mathrm{AlCl_3}} = 26.98 + 3(35.45) = 133.33 g/mol.
    0.32910 mol×133.33 g/mol=43.88 g0.32910\ \text{mol} \times 133.33\ \text{g/mol} = 43.88\ \text{g}
  5. Leftover aluminum: 0.32910 mol of AlCl3\mathrm{AlCl_3} used 0.32910 mol of Al (ratio 2 to 2).
    0.32910 mol×26.98 g/mol=8.879 g used10.0−8.879=1.1 g left0.32910\ \text{mol} \times 26.98\ \text{g/mol} = 8.879\ \text{g used} \qquad 10.0 - 8.879 = 1.1\ \text{g left}

Answer Cl2\mathrm{Cl_2} is limiting, 43.9 g of AlCl3\mathrm{AlCl_3} forms, and 1.1 g of Al is left over

Example 3: a combustion

Problem CH4+2 O2→CO2+2 H2O\mathrm{CH_4 + 2\,O_2 \rightarrow CO_2 + 2\,H_2O}. You have 16.0 g of methane and 48.0 g of oxygen. How many grams of CO2\mathrm{CO_2} can form?

  1. Moles of each reactant. MCH4=16.04M_{\mathrm{CH_4}} = 16.04 g/mol and MO2=32.00M_{\mathrm{O_2}} = 32.00 g/mol.
    16.016.04=0.9975 mol CH448.032.00=1.500 mol O2\frac{16.0}{16.04} = 0.9975\ \text{mol CH}_4 \qquad \frac{48.0}{32.00} = 1.500\ \text{mol O}_2
  2. CO2\mathrm{CO_2} from each: methane 1 to 1, oxygen 1 to 2.
    0.9975 mol CO21.500×12=0.7500 mol CO20.9975\ \text{mol CO}_2 \qquad 1.500 \times \frac{1}{2} = 0.7500\ \text{mol CO}_2
  3. Oxygen gives less, so O2\mathrm{O_2} is limiting. Convert to grams with MCO2=44.01M_{\mathrm{CO_2}} = 44.01 g/mol.
    0.7500 mol×44.01 g/mol=33.01 g0.7500\ \text{mol} \times 44.01\ \text{g/mol} = 33.01\ \text{g}

Answer O2\mathrm{O_2} is limiting and 33.0 g of CO2\mathrm{CO_2} can form

Common mistakes and how to fix them

  • Picking the reactant with less mass or fewer moles. Fix: compare the product each reactant can make, which builds in the coefficients.
  • Using the excess reactant to find the yield. Fix: once you know which reactant is limiting, every later calculation starts from it.
  • Subtracting the wrong things for the leftover. Fix: leftover = amount of excess reactant at the start minus the amount used, both in the same unit.
  • Forgetting to balance first. Fix: wrong coefficients give the wrong limiting reactant. Balance, then start.

Practice problems

  1. In N2+3 H2→2 NH3\mathrm{N_2 + 3\,H_2 \rightarrow 2\,NH_3}, you start with 2.0 mol of N2\mathrm{N_2} and 4.5 mol of H2\mathrm{H_2}. Which reactant is limiting?

    1. N2
    2. H2
    3. Neither, they run out at the same time
    Show answer

    Answer: H2

    N2\mathrm{N_2} could make 2.0×2=4.02.0 \times 2 = 4.0 mol of NH3\mathrm{NH_3}. H2\mathrm{H_2} could make 4.5×23=3.04.5 \times \tfrac{2}{3} = 3.0 mol. Hydrogen makes less, so it runs out first.

  2. In the same reaction (2.0 mol N2\mathrm{N_2}, 4.5 mol H2\mathrm{H_2}), how many moles of ammonia form?

    1. 3.0 mol
    2. 4.0 mol
    3. 6.5 mol
    4. 9.0 mol
    Show answer

    Answer: 3.0 mol

    The limiting reactant sets the yield: 4.5 mol H2×23=3.04.5\ \text{mol H}_2 \times \tfrac{2}{3} = 3.0 mol NH3\mathrm{NH_3}.

  3. 2 Na+Cl2→2 NaCl\mathrm{2\,Na + Cl_2 \rightarrow 2\,NaCl}. You react 4.60 g of Na with 10.0 g of Cl2\mathrm{Cl_2}. How many grams of NaCl can form?

    Show answer

    Answer: 11.7 g

    Na: 4.60÷22.99=0.20014.60 \div 22.99 = 0.2001 mol, which could make 0.2001 mol NaCl. Cl2\mathrm{Cl_2}: 10.0÷70.90=0.141010.0 \div 70.90 = 0.1410 mol, which could make 0.2821 mol NaCl. Sodium is limiting: 0.2001×58.44=11.690.2001 \times 58.44 = 11.69, so 11.7 g.

  4. Zn+2 HCl→ZnCl2+H2\mathrm{Zn + 2\,HCl \rightarrow ZnCl_2 + H_2}. You have 0.50 mol of Zn and 0.80 mol of HCl. Which reactant is limiting?

    1. Zn
    2. HCl
    3. Neither, they run out at the same time
    Show answer

    Answer: HCl

    Zn could make 0.50 mol of H2\mathrm{H_2}. HCl could make 0.80×12=0.400.80 \times \tfrac{1}{2} = 0.40 mol. HCl makes less, so it is limiting.

  5. In the zinc reaction above, how many moles of zinc are left over?

    Show answer

    Answer: 0.10 mol of Zn

    0.80 mol of HCl uses 0.80×12=0.400.80 \times \tfrac{1}{2} = 0.40 mol of Zn. Leftover: 0.50−0.40=0.100.50 - 0.40 = 0.10 mol.

Frequently asked questions

What is the difference between the limiting reactant and the excess reactant?

The limiting reactant is used up completely and stops the reaction. The excess reactant is the one you have more than enough of, so some of it is still there at the end. A reaction with two reactants has one of each, unless they are mixed in exactly the right ratio.

Is there a faster method?

Yes. Divide the moles of each reactant by its coefficient. The smallest result is limiting. In Example 2, Al gives 0.37064÷2=0.1850.37064 \div 2 = 0.185 and Cl2\mathrm{Cl_2} gives 0.49365÷3=0.1650.49365 \div 3 = 0.165, so Cl2\mathrm{Cl_2} is limiting. You still need the product calculation for the yield.

Why does the limiting reactant matter in real life?

Factories choose which reactant to put in excess. They usually let the cheaper or easier one be in excess so the expensive one gets used up completely. In a car engine, the amount of air limits how much fuel can burn.

Sources

  1. OpenStax Chemistry 2e, 4.4 Reaction Yields, accessed October 1, 2026
  2. CIAAW, Standard atomic weights, accessed October 1, 2026

Try asking Ducky

  • “Which one is limiting in problem 4? Check my mole math before I go on.”
  • “I picked the reactant with fewer grams and got it wrong. Can you show me why on my own numbers?”
  • “Quiz me on finding the leftover amount until I get three right in a row.”

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