5.1.5Additional lessons
Limiting and Excess Reactants — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 63 of 98
Section 1 of 6Limiting and Excess
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.
Section 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.
Section 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.
| Reactant | Amount / mol | Amount ÷ coefficient |
|---|---|---|
| Mg | 0.0999 | 0.0999 |
| HCl | 0.0500 | 0.0250 (limiting) |
Section 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.
Section 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)
- n(Mg) = 2.43 ÷ 24.31 = 0.0999 mol and n(HCl) = 1.00 × 0.0500 = 0.0500 mol.
- Divide by coefficientsMg 0.0999 and HCl 0.0250, so HCl is limiting.
- n(H₂) = 0.0500 ÷ 2 = 0.0250 mol, so V = 0.0250 × 22.7 = 0.568 dm³.
- 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.
Section 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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