Limiting Reagent
Divide each reactant’s moles by its coefficient — the smaller ratio limits the reaction.
A: 2 mol vs B: 1.6667 mol (moles ÷ coefficient). The smaller ratio is B, so B runs out first and A is in excess.
To find the limiting reagent, divide each reactant’s available moles by its coefficient in the balanced equation; the smaller ratio limits the reaction. With 2 mol of A (coefficient 1) and 5 mol of B (coefficient 3): A gives 2 ÷ 1 = 2 and B gives 5 ÷ 3 ≈ 1.67. Since 1.67 is smaller, B is the limiting reagent and A is in excess.
What the limiting reagent is
In a real reaction the reactants are rarely mixed in exact stoichiometric proportions. The limiting reagent is the one that is fully consumed first — it caps how much product can form. Whatever is left over is the excess reagent. Comparing raw moles is not enough, because each reactant is used up at a rate set by its coefficient in the balanced equation.
Compute moles ÷ coefficient for every reactant; the lowest value is the limiting reagent.
Worked example
Two reactants A and B feed a balanced equation. You have 2 mol of A with coefficient 1, and 5 mol of B with coefficient 3. Which one runs out first?
- 1 Balance the equation first. You need each reactant’s whole-number coefficient before any ratio is meaningful.
- 2 Divide moles by coefficient. A: 2 mol ÷ 1 = 2. B: 5 mol ÷ 3 ≈ 1.67.
- 3 Pick the smallest ratio. 1.67 < 2, so B reaches zero first — B is the limiting reagent.
- 4 Name the excess. A has the larger ratio, so A is left over in excess after the reaction stops.
Reading the ratios
Worked numbers for the example above: the smaller moles ÷ coefficient wins.
| Reactant | Moles | Coefficient | Ratio (moles ÷ coeff) | Role |
|---|---|---|---|---|
| A | 2 | 1 | 2 ÷ 1 = 2 | Excess |
| B | 5 | 3 | 5 ÷ 3 ≈ 1.67 | Limiting |
Why it matters
The limiting reagent sets the ceiling on every product amount — the theoretical yield is computed from it, not from the excess reactant. Add more of the excess and nothing changes; add more of the limiting reagent and you make more product (until something else runs out). Because the whole method rests on coefficients, an unbalanced equation gives the wrong answer, so always balance the reaction before comparing ratios.