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General chemistry · Stoichiometry

Limiting reagent practice problems

The limiting reagent is the reactant that runs out first. Convert every reactant to moles, compare with the balanced equation, and the smallest supply sets how much product forms.

The four-step method

  1. Balance the equation. The coefficients are mole ratios, not mass ratios.
  2. Convert each reactant to moles using its molar mass.
  3. Divide each mole amount by its coefficient. The smallest result is the limiting reagent.
  4. Use only the limiting reagent to calculate the theoretical yield and how much excess is left.

The reactant with fewer grams is not automatically limiting. Molar mass and coefficients both matter.

Worked example: water from hydrogen and oxygen

2 H2 + O2 → 2 H2O. You start with 4.0 g H2 and 16.0 g O2.

  • H2: 4.0 g ÷ 2.016 g/mol = 1.98 mol. Divided by its coefficient 2: 0.99.
  • O2: 16.0 g ÷ 32.00 g/mol = 0.500 mol. Divided by its coefficient 1: 0.500.

0.500 is smaller, so O2 is limiting, even though it has more grams.

Water formed: 0.500 mol O2 × (2 mol H2O / 1 mol O2) = 1.00 mol = 18.0 g H2O.

H2 used: 1.00 mol. H2 left: 1.98 − 1.00 = 0.98 mol, about 2.0 g.

Practice problems

1. N2 + 3 H2 → 2 NH3. You have 28.0 g N2 and 9.0 g H2. Which is limiting, and how many grams of NH3 can form?

Show answer

N2: 1.00 mol (÷1 = 1.00). H2: 9.0 ÷ 2.016 = 4.46 mol (÷3 = 1.49). N2 is limiting. NH3 = 2.00 mol × 17.03 g/mol = 34.1 g. About 1.46 mol (2.9 g) of H2 is left over.

2. 2 Al + 3 Cl2 → 2 AlCl3. You have 5.40 g Al and 10.0 g Cl2. What is the theoretical yield of AlCl3?

Show answer

Al: 5.40 ÷ 26.98 = 0.200 mol (÷2 = 0.100). Cl2: 10.0 ÷ 70.90 = 0.141 mol (÷3 = 0.0470). Cl2 is limiting. AlCl3 = 0.141 × 2/3 = 0.0940 mol × 133.33 g/mol = 12.5 g.

3. A reaction gives 10.2 g of product when the theoretical yield is 12.5 g. What is the percent yield?

Show answer

10.2 ÷ 12.5 × 100 = 81.6%.

Frequently asked questions

Is the limiting reagent always the one with the smallest mass?

No. Compare moles divided by each coefficient. A light reactant with a small molar mass can be in large excess.

Do I use the excess reagent to find theoretical yield?

No. Theoretical yield always comes from the limiting reagent, because it runs out first.

Reference: OpenStax Chemistry 2e, section 4.4 (Reaction Yields). Molar masses from standard atomic weights.

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