How to work out limiting reactant problems

When two reactants are mixed, they rarely run out at the same instant. One is used up first and stops the reaction, that is the limiting reactant, and it alone decides how much product forms. The other is left over in excess. Deciding which is which is a standard SPM skill that students often get wrong by comparing the raw amounts instead of the mole ratio. This guide gives a reliable method and two worked examples, using the mole ratio in equations.

Why the raw amounts can fool you

It is tempting to say “there’s more of X, so Y runs out”. That is only safe when the two react in a 1 : 1 ratio. If the balanced equation needs two of one reactant for every one of the other, the amounts must be weighed against the coefficients, not against each other. That is the whole idea of a limiting reactant problem.

The four-step method

1. Write the balanced equation and note the coefficient of each reactant.
2. Find the moles of each reactant given, using moles = mass ÷ molar mass, or moles = concentration × volume.
3. Divide each reactant’s moles by its coefficient. The reactant with the smaller result is the limiting reactant; the other is in excess.
4. Use the limiting reactant’s moles and the mole ratio to find the moles, then the mass or volume, of the product.

The dividing step is the key move. Comparing moles-per-coefficient puts both reactants on the same footing, so the smaller value is genuinely the one that runs out first.

Worked example 1: magnesium and acid

0.10 mol of magnesium is added to 0.15 mol of hydrochloric acid. Which is limiting, and how much hydrogen forms?

Step 1, equation. Mg + 2HCl → MgCl₂ + H₂. Mg has coefficient 1, HCl has coefficient 2.

Step 2, moles given. Mg = 0.10 mol, HCl = 0.15 mol.

Step 3, divide by coefficients. Mg: 0.10 ÷ 1 = 0.10. HCl: 0.15 ÷ 2 = 0.075. The smaller value is HCl’s, so HCl is the limiting reactant and magnesium is in excess.

Step 4, find the product. From the equation, 2 mol HCl gives 1 mol H₂, so moles of H₂ = 0.15 ÷ 2 = 0.075 mol. At room conditions that is 0.075 × 24 = 1.8 dm³ of hydrogen, using the molar volume of 24 dm³ mol⁻¹.

Notice that if you had compared raw amounts, you might wrongly have picked magnesium; dividing by the coefficient is what gives the correct answer. More practice is on our limiting reactant page.

Worked example 2: burning in oxygen

4.8 g of carbon is burned in 12.8 g of oxygen. Which is limiting, and what mass of carbon dioxide forms? (Relative atomic masses: C = 12, O = 16.)

Step 1, equation. C + O₂ → CO₂. Both have coefficient 1.

Step 2, moles given. Carbon: 4.8 ÷ 12 = 0.40 mol. Oxygen (O₂, molar mass 32): 12.8 ÷ 32 = 0.40 mol.

Step 3, divide by coefficients. Both give 0.40 ÷ 1 = 0.40. They are exactly equal, so neither is in excess, the reactants are in the exact stoichiometric ratio.

Step 4, find the product. Moles of CO₂ = 0.40 mol; mass = 0.40 × 44 = 17.6 g. When the divided values tie, the reaction is a “perfect” mix with nothing left over, a valid and examinable outcome.

Checking your answer

Two quick checks catch most errors:

• The excess reactant should have some left. You can confirm the amount by working out how much of it the limiting reactant actually consumed and subtracting. In example 1, the 0.15 mol of HCl consumes 0.075 mol of Mg (half as many, from the 1 : 2 ratio), leaving 0.10 − 0.075 = 0.025 mol of magnesium unreacted.
• Never base the product on the excess reactant. Product always comes from the limiting reactant’s moles, using the excess is the most common error and gives too large an answer.

Turning it into mass or volume

Once you have the limiting reactant’s moles, the rest is ordinary stoichiometry: apply the mole ratio to get the product’s moles, then convert to mass with moles × molar mass, or to gas volume with moles × 24 dm³ mol⁻¹ at room conditions. The mass conversion is walked through on our mass of product or reactant page.

Practise until the order is automatic

The sequence never changes: balance, find moles, divide by coefficients, pick the smaller, then build the product from it. Work through one gas-collection question and one mass question and the method becomes second nature. If choosing the limiting reactant keeps tripping you, that is a quick fix with a teacher, our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial to begin.