Finding the Limiting and Excess Reactants

The limiting reactant is the reactant that is completely used up first in a reaction and so determines the maximum amount of product, while the excess reactant is the one that remains after the reaction is over. In brief: if 28.0 g of N2 (1.00 mol) reacts with 8.00 g of H2 (3.97 mol) in N2 + 3H2 → 2NH3, only 3.00 mol of H2 is needed for the nitrogen, so nitrogen is limiting, hydrogen is in excess, and the maximum yield is 2.00 mol (34.0 g) of ammonia.

In practice, reactants are rarely mixed in exactly the ratio of the equation. One of them runs out first, and the reaction stops. Identifying the limiting reactant is the key to calculating the theoretical yield correctly. This article explains two methods to find it, shows how to calculate the amount of excess reactant left, uses a table method for organization, works through examples with masses and solutions, explains why chemists deliberately use an excess, and gives some history.

What is a limiting reactant?

A bicycle shop has 40 wheels and 25 frames. Each bicycle needs 2 wheels and 1 frame. The 40 wheels can make 20 bicycles, and the 25 frames could make 25 bicycles, so the wheels are limiting, and 5 frames are left over. Chemical reactions work in the same way. The amount of product is determined by the limiting reactant, and any increase in the other reactant would not give more product.

40 / 2 = 20 25 / 1 = 25

Method 1: compare the moles needed

  1. Write the balanced equation and convert the amounts of all reactants to moles.
  2. Choose one reactant and calculate how many moles of the other reactant it needs, using the mole ratio.
  3. Compare the amount needed with the amount available. If the available amount is smaller than needed, that second reactant is limiting. If it is larger, the first reactant is limiting.

Method 2: compare the products

  1. For each reactant, calculate how much product it could form if it were completely consumed, using the mole ratio.
  2. The reactant that gives the smaller amount of product is the limiting reactant, and the smaller amount is the theoretical yield.

The reaction table

An organized table (sometimes called a “BCA” table: before, change, after) helps to keep track. For N2 + 3H2 → 2NH3 with 1.00 mol of N2 and 3.97 mol of H2:

Before-change-after table for N2 + 3H2 → 2NH3 (moles)
N2 H2 NH3
Before1.003.970
Change−1.00−3.00+2.00
After00.972.00

The nitrogen is used up completely (limiting), and 0.97 mol of hydrogen remains (excess).

Bar chart of moles of nitrogen and hydrogen available and needed 0 1 2 3 4 Amount (mol) 1.00 1.32 3.97 3.00 N2 (limiting) H2 (excess) Solid: available Dashed: needed
For N2 + 3H2 → 2NH3 with 1.00 mol N2 and 3.97 mol H2. Solid bars show the amount available. Dashed bars show the amount of each reactant needed to react completely with all of the other one. Nitrogen is limiting because 1.32 mol would be needed but only 1.00 mol is available. Hydrogen needs only 3.00 mol, so 0.97 mol is left over.

Worked examples

  1. Ammonia synthesis with masses. 28.0 g of N2 and 8.00 g of H2. n(N2) = 28.0 ÷ 28.014 = 0.9995 mol. n(H2) = 8.00 ÷ 2.016 = 3.968 mol. H2 needed for the nitrogen: 3 × 0.9995 = 2.9985 mol. Available: 3.968 mol, which is more, so N2 is limiting. NH3: 2 × 0.9995 = 1.999 mol = 34.0 g. H2 left: 3.968 − 2.9985 = 0.970 mol = 1.96 g.
  2. Aluminum and chlorine. 2Al + 3Cl2 → 2AlCl3. 5.00 g of Al and 10.0 g of Cl2. n(Al) = 5.00 ÷ 26.982 = 0.1853 mol. n(Cl2) = 10.0 ÷ 70.906 = 0.1410 mol. Cl2 needed for the Al: 0.1853 × 3 ÷ 2 = 0.2780 mol, more than the 0.1410 mol available, so Cl2 is limiting. AlCl3: 0.1410 × 2 ÷ 3 = 0.09402 mol, with a mass of 0.09402 × 133.34 = 12.5 g. Al used: 0.09402 × 26.982 = 2.54 g, so 5.00 − 2.54 = 2.46 g of Al remains.
    0.2780 0.1410 mol 0.1853 × 3/2 = 0.2780 > 0.1410
  3. Method 2 (products). From Al: 0.1853 mol of AlCl3 (24.7 g). From Cl2: 0.09402 mol (12.5 g). The smaller amount comes from Cl2, so Cl2 is limiting, and the theoretical yield is 12.5 g. This agrees with the first method.
  4. Solutions. 50.0 mL of 0.100 mol/L AgNO3 and 50.0 mL of 0.200 mol/L NaCl: AgNO3 + NaCl → AgCl + NaNO3. n(Ag+) = 5.00 × 10−3 mol. n(Cl−) = 1.00 × 10−2 mol. The ratio is 1 : 1, so silver nitrate is limiting. AgCl: 5.00 × 10−3 mol = 0.717 g. Excess Cl−: 5.00 × 10−3 mol.
  5. Combustion. A mixture of 1.00 g of CH4 and 4.00 g of O2 reacts, CH4 + 2O2 → CO2 + 2H2O. n(CH4) = 1.00 ÷ 16.043 = 0.0623 mol. n(O2) = 4.00 ÷ 31.998 = 0.1250 mol. O2 needed: 2 × 0.0623 = 0.1247 mol, which is slightly less than 0.1250 mol, so CH4 is limiting (the mixture is almost exactly stoichiometric). CO2: 0.0623 mol = 2.74 g.
  6. Which is in excess? A student mixes 10.0 g of Zn with 20.0 mL of 1.00 mol/L HCl, Zn + 2HCl → ZnCl2 + H2. n(Zn) = 10.0 ÷ 65.38 = 0.1530 mol. n(HCl) = 0.0200 mol. Zn needed: 0.0100 mol. Zinc is in great excess, and HCl is limiting. H2: 0.0100 mol = 0.224 L at STP.

Why use an excess of one reactant?

Chemists often use an excess of an inexpensive or easily removed reactant to ensure that the valuable reactant reacts completely, or to push an equilibrium toward the products. In combustion engines, an excess of air makes sure that the fuel burns completely: a lean mixture has excess air, and a rich mixture has excess fuel. The ideal ratio of air to fuel for gasoline is about 14.7 : 1 by mass (the stoichiometric ratio). In titration, the reagent added in slight excess changes the color of an indicator. Excess reactants must usually be separated from the product afterwards.

Where do limiting reactants appear in everyday life?

How did the concept develop?

Jeremias Richter’s work on proportions of acids and bases in the 1790s implied that a reaction stops when one reagent is used up. Justus von Liebig applied a similar idea to agriculture in the 1840s, in his law of the minimum, which says that growth is controlled by the scarcest resource. The term “limiting reagent” became part of general chemistry teaching in the 20th century.

Which mistakes are common?

Mistake 1: Choosing the reactant with the smaller mass

The limiting reactant depends on the moles and the coefficients, not on the mass.

Mistake 2: Choosing the reactant with the smaller number of moles

Divide each amount by its coefficient. The smaller result identifies the limiting reactant: for 1.00 mol N2 (coefficient 1) and 3.97 mol H2 (coefficient 3), 1.00 and 1.32 show N2 is limiting.

Mistake 3: Calculating the product from the excess reactant

Use the limiting reactant for the theoretical yield.

Mistake 4: Forgetting to calculate the excess left over

Subtract the amount used from the amount you started with.