The limiting reactant is the one that runs out first, and it sets the maximum amount of product. To find it, compare the moles of each reactant against what the equation requires.
This lesson is part of conservation reasoning across a reaction. It builds on connecting particle ratios to mole ratios.
How do you spot the limiting reactant?
Write the equation, convert both quantities to moles, then ask how many moles of one reactant the other needs. If one reactant is short, it is the limiting one.
For Mg + 2HCl → MgCl₂ + H₂, each mole of magnesium needs 2 mol of hydrochloric acid. The acid is used twice as fast as the metal.
Worked example: one row of data
A student adds 1.20 g of magnesium to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid. (Mg = 24; 1 mol of gas = 24 dm³ at room conditions.)
- Moles of Mg = 1.20 ÷ 24 = 0.0500 mol.
- Moles of HCl = 1.00 × 50.0 ÷ 1000 = 0.0500 mol.
- 0.0500 mol of magnesium needs 0.100 mol of HCl, but only 0.0500 mol is present.
- So HCl is the limiting reactant. Moles of H₂ = 0.0500 ÷ 2 = 0.0250 mol.
- Volume of H₂ = 0.0250 × 24 = 0.600 dm³.
Notice that both mole values are equal, 0.0500 mol, yet the acid still limits the reaction. That is why comparing raw numbers fails.
Reading a complete dataset
A student adds different masses of magnesium to 40.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid and measures the hydrogen each time.
| Mass of Mg (g) | Moles of Mg | Volume of H₂ (cm³) |
|---|---|---|
| 0.06 | 0.0025 | 60 |
| 0.12 | 0.0050 | 120 |
| 0.24 | 0.0100 | 240 |
| 0.36 | 0.0150 | 240 |
| 0.48 | 0.0200 | 240 |
The acid contains 0.500 × 40.0 ÷ 1000 = 0.0200 mol, enough for 0.0100 mol of Mg. Up to 0.24 g, magnesium is limiting and the gas volume rises with it. From 0.24 g onward the acid is limiting, so the volume stays at 240 cm³.
The mistake that costs marks
The common slip is to compare the moles directly and skip the ratio, or to compare the masses. Both lead to the wrong reactant.
| Method | First row of the worked example | Verdict |
|---|---|---|
| Compare masses 1.20 g and 1.83 g | Seems to say magnesium is limiting | Wrong, masses are not comparable |
| Compare moles 0.0500 and 0.0500 | Seems to say neither is limiting | Wrong, ignores the 1 : 2 ratio |
| Divide by coefficient: 0.0500 ÷ 1 and 0.0500 ÷ 2 | The smaller value, 0.0250, belongs to HCl | Correct |
If you want to see how a plateau appears on a graph, the graph evidence comparison lab plots datasets like the one above.
Check yourself
0.65 g of zinc is added to 20.0 cm³ of 0.400 mol dm⁻³ sulfuric acid: Zn + H₂SO₄ → ZnSO₄ + H₂. Decide which reactant is limiting and find the volume of hydrogen at room conditions. (Zn = 65.)
Answer
Moles of Zn = 0.65 ÷ 65 = 0.0100 mol.
Moles of H₂SO₄ = 0.400 × 20.0 ÷ 1000 = 0.00800 mol.
The ratio is 1 : 1, so the smaller mole value is limiting: the acid, at 0.00800 mol.
Moles of H₂ = 0.00800 mol, so volume = 0.00800 × 24 = 0.192 dm³ = 192 cm³.
Zinc is in excess, with 0.00200 mol left over.
What to study next
Move on to explaining a smaller measured yield, then try the cluster practice set.
To have a teacher check your dataset reasoning, see online one-to-one Chemistry tuition.