Limiting and excess reactants

IB Chemistry HL· 15 min read

1. 1. Key Definitions and Identification★☆☆☆☆⏱ 5 min

📘 Definition

Limiting Reactant

The reactant that is completely consumed when a reaction proceeds to completion. It restricts the maximum amount of product that can form.

📘 Definition

Excess Reactant

A reactant that is present in a greater amount than required to react completely with the limiting reactant. Some remains unreacted after the reaction stops.

The most reliable method to identify the limiting reactant uses the ratio of moles present to stoichiometric coefficient from the balanced equation:

Calculate moles of each reactant present
Divide each reactant's moles by its stoichiometric coefficient from the balanced equation
The reactant with the smaller value is the limiting reactant

📐 Worked Example

2.5 mol of hydrogen gas reacts with 1.8 mol of oxygen gas to form water. Balanced equation: $2 H_{2} + O_{2} \to 2 H_{2} O$ . Identify the limiting reactant.

1
Write down stoichiometric coefficients: H₂ = 2, O₂ = 1. Calculate the ratio for each:
2
$\frac{\text{moles } H_2}{\text{coefficient}} = \frac{2.5}{2} = 1.25$
3
$\frac{\text{moles } O_2}{\text{coefficient}} = \frac{1.8}{1} = 1.8$
4
The smaller value corresponds to the limiting reactant. 1.25 < 1.8, so hydrogen is the limiting reactant.

Exam tip:

Always use moles for this calculation, never mass. Mass ratios do not equal stoichiometric mole ratios, so this will always give the wrong result if molar masses differ.

2. 2. Calculating Mass of Product Formed★★☆☆☆⏱ 5 min

Once the limiting reactant is identified, all product calculations are based on its moles, because the reaction stops when this reactant is fully consumed. The steps for calculation are:

Confirm the limiting reactant using the method above
Use the stoichiometric ratio between limiting reactant and product to find moles of product
Convert moles of product to mass (or volume/concentration, as required)

📐 Worked Example

What mass of water is produced from 2.5 mol H₂ and 1.8 mol O₂ reacting as $2 H_{2} + O_{2} \to 2 H_{2} O$ ? Molar mass of water = 18.02 g mol⁻¹.

1
H₂ is confirmed as the limiting reactant, with 2.5 mol H₂ present.
2
From the balanced equation, the mole ratio H₂:H₂O = 2:2 = 1:1. So moles of H₂O formed = moles of H₂ = 2.5 mol.
3
Calculate mass of water:
4
$\text{mass} = \text{moles} \times M_r = 2.5 \times 18.02 = 45.05 \text{ g}$
5
Final answer (3 significant figures): 45.1 g of water is produced.

3. 3. Calculating Remaining Excess Reactant★★☆☆☆⏱ 5 min

Exam questions frequently ask for the mass of excess reactant left over after the reaction completes. This follows directly from the limiting reactant calculation:

Calculate moles of excess reactant that reacted with the limiting reactant, using the mole ratio
Subtract the reacted moles from the initial moles of excess reactant to get remaining moles
Convert remaining moles to mass (or other required unit)

📐 Worked Example

For the reaction of 2.5 mol H₂ and 1.8 mol O₂, calculate the mass of unreacted oxygen remaining after completion. Molar mass of O₂ = 32.00 g mol⁻¹.

1
Initial moles of O₂ = 1.8 mol. From the balanced equation, 2 mol H₂ reacts with 1 mol O₂.
2
Calculate moles of O₂ that reacted:
3
$\text{Moles reacted} = \frac{\text{moles } H_2}{2} = \frac{2.5}{2} = 1.25 \text{ mol}$
4
Calculate remaining moles of O₂:
5
$\text{Remaining moles} = 1.8 - 1.25 = 0.55 \text{ mol}$
6
Convert to mass:
7
$\text{Mass remaining} = 0.55 \times 32.00 = 17.6 \text{ g}$
8
Final answer (2 significant figures): 18 g of oxygen remains unreacted.

4. 4. HL Extended Problems for Gases and Solutions★★★☆☆HL only⏱ 7 min

✓ Calculator OK

For HL exams, limiting reactant questions often involve reactants as gases (given by volume, pressure, temperature) or in solution (given by concentration and volume). The method is the same, just convert the given quantities to moles first.

📐 Worked Example

50.0 cm³ of 0.200 mol dm⁻³ AgNO₃ reacts with 25.0 cm³ of 0.150 mol dm⁻³ Na₂CO₃: $2 A g N O_{3} (a q) + N a_{2} C O_{3} (a q) \to A g_{2} C O_{3} (s) + 2 N a N O_{3} (a q)$ . Calculate the mass of Ag₂CO₃ precipitate formed. M(Ag₂CO₃) = 275.75 g mol⁻¹.

1
Calculate initial moles (moles = concentration × volume in dm³):
2
Moles AgNO₃ = $0.200 \times \frac{50.0}{1000} = 0.0100$ mol. Moles Na₂CO₃ = $0.150 \times \frac{25.0}{1000} = 0.00375$ mol.
3
Identify limiting reactant:
4
$\frac{0.0100}{2} = 0.00500 \text{ (AgNO}_3\text{)}, \quad \frac{0.00375}{1} = 0.00375 \text{ (Na}_2\text{CO}_3\text{)}$
5
Na₂CO₃ has the smaller ratio, so it is the limiting reactant.
6
Mole ratio Na₂CO₃:Ag₂CO₃ = 1:1, so moles Ag₂CO₃ = 0.00375 mol.
7
Calculate mass:
8
$\text{mass} = 0.00375 \times 275.75 = 1.03 \text{ g}$
9
Final answer (3 significant figures): 1.03 g of silver carbonate forms.

5. Common Pitfalls

Wrong move:

Using mass instead of moles to identify the limiting reactant

Why:

Mass ratios do not match stoichiometric mole ratios, so this gives an incorrect result when molar masses of reactants differ

Correct move:

Always convert mass of each reactant to moles first, then compare the ratio of moles to stoichiometric coefficient

Wrong move:

Using the excess reactant to calculate the mass of product formed

Why:

The reaction stops when the limiting reactant is consumed, so excess reactant cannot all react, leading to overestimated product mass

Correct move:

Always base all product and excess calculations on the moles of the limiting reactant only

Wrong move:

Reporting the initial amount of excess reactant as the remaining amount

Why:

Most of the excess reactant reacts with the limiting reactant, so this leads to a large overestimation of remaining excess

Correct move:

Calculate how many moles of excess reactant reacted, then subtract this value from the initial moles to get the remaining amount

Wrong move:

Ignoring significant figure requirements for the final answer

Why:

IB exams penalize incorrect significant figures even if the core calculation is numerically correct

Correct move:

Round your final answer to match the number of significant figures of the least precise given data in the question

6. Quick Reference Cheatsheet

Step	Action
1	Calculate moles of all reactants from given data
2	Divide moles by each reactant's stoichiometric coefficient
3	Smallest ratio = limiting reactant, larger = excess
4	Calculate product moles from limiting reactant moles
5	Subtract reacted excess from initial excess to get remaining

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · Paper 1
MCQ limiting reactant identification
2022 · Paper 2
6 mark product mass calculation
2023 · Paper 1
Remaining excess reactant MCQ

Going deeper

syllabusIB Chemistry HL Stoichiometry SyllabusOfficial syllabus reference

What's Next

Limiting and excess reactants are the foundation for all advanced stoichiometry calculations in IB Chemistry HL. Mastering this skill is critical for titration calculations, precipitation problems, gas stoichiometry, and equilibrium problems that appear frequently on both Paper 1 and Paper 2 exams. Next, you will build on this concept to calculate theoretical yield and percentage yield, which connects limiting reactant theory to experimental results. Later, you will apply this same logic to equilibrium calculations to find equilibrium concentrations when one reactant is limiting.