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5.1.5Additional lessons

Limiting and Excess Reactants — IB Chemistry HL Revision Notes & Topical Questions

5 min read 4 parts Lesson 93 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5Key Ideas

5.1.5 Limiting and Excess Reactants
Limiting reactant
  • The reactant that is completely used up first and therefore determines the maximum amount of product.
Excess reactant
  • A reactant present in a larger amount than needed, so some remains after the reaction finishes.
Why it matters
  • Calculations of product amount must always be based on the limiting reactant, not on the excess.
Comparison
  • Compare the available moles, divided by their coefficients, to see which gives the smallest value.

Part 2 of 5Method

1
Write a balanced equation and convert every given quantity to moles.
2
Divide each amount in moles by the coefficient of that reactant in the equation.
3
The smallest result identifies the limiting reactant.
4
Use the moles of the limiting reactant and the mole ratio to calculate the moles of product.
5
Subtract the moles used from the moles supplied to find the amount of excess left.

Part 3 of 5Practical Meaning

Practical Meaning
Industry
  • An inexpensive reactant is often used in excess to ensure the costly reactant is fully converted.
Experiments
  • Adding excess acid ensures a metal reacts completely so that gas volume depends only on the metal.
Yield
  • Using the wrong reactant as limiting leads to an overestimate of the theoretical yield.
Checking
  • The mass of products plus unused excess must equal the total mass of reactants supplied.

Part 4 of 5Worked Example

Question

  • 2.00 g of magnesium is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
  • Identify the limiting reactant and calculate the volume of hydrogen at STP (22.7 dm³ mol⁻¹).

  1. n(Mg) = 2.00 ÷ 24.31 = 0.0823 mol; n(HCl) = 1.00 × 0.0500 = 0.0500 mol.
  2. Divide by coefficients
    • Mg = 0.0823 ÷ 1 = 0.0823;
      • HCl = 0.0500 ÷ 2 = 0.0250.
    • HCl is smaller, so it is limiting.
  3. n(H₂) = 0.0500 ÷ 2 = 0.0250 mol.
  4. V = 0.0250 × 22.7 = 0.568 dm³.

Answer HCl is limiting; 0.568 dm³ of H₂ forms

Part 5 of 5Exam Checklist & Tips

5.1.5 – Key Points for the Exam

  • Limiting reactant The reactant that runs out first and sets the amount of product.
  • Excess reactant The reactant left over when the reaction is complete.
  • Theoretical yield The maximum amount of product predicted from the limiting reactant.
  • Stoichiometric ratio The exact ratio of reactants shown by the balanced equation.

Common Mistakes

  • Choosing the reactant with the smaller mass as limiting without converting to moles.
  • Forgetting to divide by the coefficient when comparing amounts.
  • Calculating the product amount from the excess reactant.
  • Giving the volume of hydrogen using the mass of acid rather than the moles.

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