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5.1.5Additional lessons

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

5 min read 6 parts Lesson 63 of 98

📚 IB Chemistry SL Hub / Topical Revision
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Part 1 of 6Limiting and Excess

5.1.5 Limiting and Excess Reactants

Reactants are rarely mixed in exactly the ratio of the equation.

Limiting reactant
  • The reactant that is completely used up first, so it determines how much product forms.
Excess reactant
  • The reactant present in more than the amount needed, so some is left over when the reaction stops.
Why it matters
  • Product amounts must always be calculated from the limiting reactant, not the excess.
Everyday example
  • Making sandwiches with 10 slices of bread and 4 slices of cheese: the cheese limits the number of sandwiches.

Part 2 of 6Finding the Limiting Reactant

1
Write the balanced equation and convert the mass or volume of each reactant to moles.
2
Divide each amount by its coefficient in the equation.
3
The reactant with the smaller value is limiting.
4
Use the amount of the limiting reactant and the mole ratio to calculate the amount and then the mass or volume of product.
5
Find the excess by subtracting the amount that reacted from the amount originally present.

Part 3 of 6Checking and Applying

Checking and Applying
Quick check
  • If the moles of A are more than needed to react with all of B, then A is in excess and B is limiting.
Gases
  • For gases at the same temperature and pressure, volumes may be compared directly in the same way as moles.
Industry
  • Cheaper reagents are often used in excess to ensure that the more expensive one reacts completely.
Rates
  • The reaction stops when the limiting reactant is used up, whatever amount of excess reactant is left.
ReactantAmount / molAmount ÷ coefficient
Mg0.09990.0999
HCl0.05000.0250 (limiting)

Part 4 of 6Typical Reaction Outcomes

Typical Reaction Outcomes
Hydrogen volume
  • The volume of gas collected depends only on the limiting reactant.
Mass of precipitate
  • In precipitation reactions, the mass of solid formed is set by the ion present in the smaller proportion.
Left-over reagent
  • Excess metal may be seen as unreacted solid after the end of a reaction.
Percentage yield
  • Yield calculations compare the actual product with the theoretical amount from the limiting reactant.

Part 5 of 6Worked Example

Question

  • 2.43 g of Mg reacts with 50.0 cm³ of 1.00 mol dm⁻³ HCl: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
  • Identify the limiting reactant, the volume of H₂ at STP and the mass of Mg left.
  • (Ar Mg = 24.31)

  1. n(Mg) = 2.43 ÷ 24.31 = 0.0999 mol and n(HCl) = 1.00 × 0.0500 = 0.0500 mol.
  2. Divide by coefficientsMg 0.0999 and HCl 0.0250, so HCl is limiting.
  3. n(H₂) = 0.0500 ÷ 2 = 0.0250 mol, so V = 0.0250 × 22.7 = 0.568 dm³.
  4. Mg reacted = 0.0250 mol, so Mg left = 0.0999 − 0.0250 = 0.0749 mol = 1.82 g.

Answer

  • HCl is limiting;
    • 0.568 dm³ of H₂;
    • 1.82 g of Mg remains.

Part 6 of 6Exam Checklist & Tips

5.1.5 – Key Points for the Exam

  • Limiting reactant The reactant that runs out first and limits the amount of product.
  • Excess reactant A reactant present in more than the amount required.
  • Stoichiometric ratio The ratio of amounts of reactants in the balanced equation.
  • Theoretical yield The maximum amount of product predicted from the limiting reactant.

Common Mistakes

  • Choosing the reactant with the smaller number of moles as limiting without dividing by the coefficient.
  • Calculating the product from the excess reactant.
  • Forgetting to calculate the amount of excess that remains when asked.
  • Mixing up mass and moles when comparing reactants.

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