In a simple reaction model, the reactant that would run out first is the limiting quantity, and it decides how much product forms. The method is to compare each amount, divided by its coefficient, from a balanced equation. It builds on the equation ratio lesson and appears whenever a question gives the amounts of two reactants.
How do I find the limiting reactant?
The equation tells you what is needed for each other, so ask: how much of reactant B is needed to use up all of reactant A? Then compare with what is given.
A quick method is to divide each amount in mol by its coefficient. The smaller value belongs to the limiting reactant.
Steps
- Balance the equation.
- Convert each given mass to moles.
- Divide each amount by its coefficient, or calculate how much of one reactant the other needs.
- The smaller result is limiting.
- Use the limiting amount with the ratio to find the product.
- Find what is left over in the excess reactant, if asked.
Worked example
This is a model calculation. In Mg + 2HCl → MgCl₂ + H₂, 2.4 g of magnesium is combined with 0.15 mol of hydrochloric acid, treated as just an amount.
Find the limiting reactant and the mass of MgCl₂ formed. Use Mg = 24, Cl = 35.5.
Step 1, moles of Mg: 2.4 ÷ 24 = 0.10 mol.
Step 2, compare: Mg ÷ 1 = 0.10. HCl ÷ 2 = 0.075. The smaller is HCl, so HCl is limiting.
Step 3, product: ratio HCl : MgCl₂ = 2 : 1, so MgCl₂ = 0.15 ÷ 2 = 0.075 mol. Mr of MgCl₂ = 24 + 71 = 95, so mass = 0.075 × 95 = 7.125 g, about 7.1 g.
Step 4, excess: Mg used = 0.075 mol, so 0.10 − 0.075 = 0.025 mol remains. That is 0.025 × 24 = 0.60 g of Mg.
Second check: H₂ formed is 0.075 mol, and mass conservation holds: Mg used 1.8 g + HCl 0.15 × 36.5 = 5.475 g gives 7.275 g. Products: 7.125 g MgCl₂ + 0.075 × 2 = 0.15 g H₂ = 7.275 g. They agree.
A plausible mistake
A student compares 0.10 mol of Mg with 0.15 mol of HCl and says Mg is limiting because 0.10 is smaller.
Mistaken answer: Mg is limiting, MgCl₂ = 0.10 mol
This compares raw moles. The equation needs 2 mol of HCl for every 1 mol of Mg, so 0.10 mol of Mg would need 0.20 mol of HCl, and only 0.15 mol is there.
The correction is to divide by the coefficients, or to work out how much the other reactant needs.
Check yourself
1. In N₂ + 3H₂ → 2NH₃, 1.0 mol of N₂ and 2.4 mol of H₂ are mixed in the model. Which is limiting, and how many moles of NH₃ form?
Show answer
N₂ ÷ 1 = 1.0. H₂ ÷ 3 = 0.80. H₂ is limiting. NH₃ = 2.4 × (2 ÷ 3) = 1.6 mol. (N₂ used is 0.80 mol, so 0.20 mol remains.)
2. In Fe + S → FeS, 5.6 g of Fe and 4.0 g of S are used in the model. Which is limiting, and what mass of FeS forms? (Fe = 56, S = 32)
Show answer
Fe = 5.6 ÷ 56 = 0.10 mol. S = 4.0 ÷ 32 = 0.125 mol. Ratio 1 : 1, so Fe is limiting. FeS = 0.10 mol. Mr = 88, so mass = 8.8 g. S left: 0.025 mol = 0.80 g.
3. In CH₄ + 2O₂ → CO₂ + 2H₂O, 0.50 mol of CH₄ meets 0.80 mol of O₂ in the model. Which is limiting, and how many moles of CO₂ form?
Show answer
CH₄ ÷ 1 = 0.50. O₂ ÷ 2 = 0.40. O₂ is limiting. CO₂ = 0.80 ÷ 2 = 0.40 mol.
Where this leads next
The last lesson in this module is a habit that protects every step: checking an amount calculation by units. The mole and equation-ratio tutor explains the limiting-quantity reasoning for an equation you enter, and the equation balance reasoning trainer checks that equation first.
Limiting-quantity questions often go wrong because of a small setup choice. A teacher who sees your working can spot it quickly in online one-to-one Chemistry tuition.