The limiting quantity is the amount that runs out first. The method is the same every time: convert to moles, divide by the coefficient, and pick the smaller value.
This lesson sits in moles, formulas and equations. For a dataset-based view with a plateau, read identifying the limiting reactant from a dataset.
What is the rule for choosing?
For each reactant, calculate moles ÷ coefficient. The reactant with the smaller value is limiting.
Dividing by the coefficient shows how many “reaction units” each reactant can supply. The smallest supply stops the reaction.
Worked example: a 1 : 1 precipitation
20.0 cm³ of 0.100 mol dm⁻³ silver nitrate is mixed with 30.0 cm³ of 0.0500 mol dm⁻³ sodium chloride: AgNO₃ + NaCl → AgCl + NaNO₃. Find the mass of silver chloride. (Ag = 108, Cl = 35.5.)
- Moles of AgNO₃ = 0.100 × 20.0 ÷ 1000 = 0.00200 mol.
- Moles of NaCl = 0.0500 × 30.0 ÷ 1000 = 0.00150 mol.
- Both coefficients are 1, so NaCl (0.00150) is smaller and is limiting.
- Moles of AgCl = 0.00150 mol. Mass = 0.00150 × 143.5 = 0.215 g.
Silver nitrate is in excess by 0.00050 mol.
When the smaller mole value is not the limit
Consider N₂ + 3H₂ → 2NH₃ with 0.50 mol of nitrogen and 1.2 mol of hydrogen.
| Reactant | Moles | ÷ coefficient |
|---|---|---|
| N₂ | 0.50 | 0.50 ÷ 1 = 0.50 |
| H₂ | 1.2 | 1.2 ÷ 3 = 0.40 |
Hydrogen is limiting even though 1.2 is the larger mole value, because it is needed three times as fast. Moles of NH₃ = 1.2 × 2 ÷ 3 = 0.80 mol.
The mistake that costs marks
The shortcut of choosing the smaller mole value, or the smaller mass, works only by luck. In the ammonia example, comparing moles picks the wrong reactant.
| Shortcut | Applied to the ammonia example | Result |
|---|---|---|
| Smaller mass | 14 g N₂ against 2.4 g H₂ picks H₂ | Right by luck |
| Smaller moles | 0.50 against 1.2 picks N₂ | Wrong |
| Moles ÷ coefficient | 0.50 against 0.40 picks H₂ | Right every time |
Try the same steps on your own numbers with the mole and stoichiometry steps tool.
Check yourself
4.0 g of copper(II) oxide reacts with 25.0 cm³ of 1.00 mol dm⁻³ sulfuric acid: CuO + H₂SO₄ → CuSO₄ + H₂O. Find the limiting reactant and the moles of copper(II) sulfate formed. (Cu = 64, O = 16.)
Answer
Moles of CuO = 4.0 ÷ 80 = 0.050 mol. Moles of H₂SO₄ = 1.00 × 25.0 ÷ 1000 = 0.0250 mol.
The ratio is 1 : 1, so the acid is limiting with 0.0250 mol.
Moles of CuSO₄ = 0.0250 mol. Copper(II) oxide is in excess, with 0.025 mol left.
What to study next
Go back to converting between mass, moles and particles if the first step felt slow. Then test the chapter with the moles practice set.
For a teacher to check your reasoning on new data sets, see online one-to-one Chemistry tuition.