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Limiting Reagent Calculator

Enter two or more reactants with their amounts and balanced equation coefficients to identify the limiting reagent.

Reagent 1
Reagent 2

Limiting reagents and theoretical yield

Real reactions almost never start with reactants in exact stoichiometric proportions. One reactant runs out first, and the moment it does, the reaction stops — even if other reactants are sitting around. That bottleneck reactant is the limiting reagent, and it determines the theoretical yield of every product. The others are “in excess” and have leftover after the reaction completes.

The non-obvious step is that you can’t just compare mole counts directly: a reactant with coefficient 3 and another with coefficient 1 are on different scales. The standard method divides each reactant’s moles by its stoichiometric coefficient and picks the smallest ratio. That ratio is the equivalent of “how many units of reaction can this reactant support,” and the smallest one is what bounds the reaction.

The method

  1. Balance the equation.
  2. Convert each reactant to moles (mass divided by molar mass).
  3. Divide each mole value by the reactant’s coefficient.
  4. The smallest quotient identifies the limiting reagent.
  5. Use the mole ratio to compute theoretical yield in moles, then convert to grams.
  6. Compute how much excess reagent reacted (limiting moles × ratio), subtract from starting amount → excess remaining.

What the calculator does

  1. Enter the balanced equation.
  2. Enter each reactant’s amount in grams or moles.
  3. The calculator identifies the limiting reagent, computes theoretical yield in both moles and grams, and reports the leftover of every excess reagent.
  4. All steps shown — molar mass conversions, the divide-by-coefficient comparison, and the mole-ratio arithmetic.

Worked examples

Synthesis of water. 2H2 + O2 → 2H2O. Start with 10.0 g H2 and 80.0 g O2.

  • Moles: H2 = 10.0/2.016 = 4.960; O2 = 80.0/32.00 = 2.500
  • Divide by coefficient: H2 = 4.960/2 = 2.480; O2 = 2.500/1 = 2.500
  • Limiting: H2 (smallest quotient)
  • H2O produced = 4.960 × (2/2) = 4.960 mol = 89.4 g
  • O2 consumed = 4.960 × (1/2) = 2.480 mol = 79.4 g; 0.6 g O2 excess

Iron oxidation. 4Fe + 3O2 → 2Fe2O3. 100 g Fe and 50.0 g O2.

  • Moles: Fe = 1.791; O2 = 1.563
  • Divide by coefficient: Fe = 0.4477; O2 = 0.5208
  • Limiting: Fe
  • Fe2O3 = 1.791 × (2/4) = 0.8953 mol = 143.0 g
  • O2 consumed = 1.791 × (3/4) = 1.343 mol = 43.0 g; 7.0 g O2 excess

Precipitation. BaCl2 + Na2SO4 → BaSO4 + 2NaCl. 25.0 g BaCl2 + 20.0 g Na2SO4.

  • Moles: BaCl2 = 0.1201; Na2SO4 = 0.1408
  • Coefficients are 1:1, so compare directly: BaCl2 < Na2SO4
  • Limiting: BaCl2
  • BaSO4 = 0.1201 mol = 28.0 g

The 1:1 case shows why the divide-by-coefficient step matters: it just happens to reduce to a direct comparison when coefficients match, but the moment they differ you need it.

Frequently Asked Questions

What is a limiting reagent?
The limiting reagent is the reactant that runs out first, capping how much product the reaction can make. Once it's gone, the reaction stops regardless of how much of the other reactants is left. The leftover reactants are 'in excess.' The limiting reagent sets the theoretical yield — every product calculation is anchored to it, not to the most abundant reactant.
How do you find the limiting reagent?
Convert each reactant amount to moles, then divide each mole count by that reactant's coefficient in the balanced equation. The smallest quotient identifies the limiting reagent — that ratio is what would be consumed per 'unit' of reaction. Comparing raw mole counts works only when coefficients are equal; the divide-by-coefficient step is what makes the comparison fair across stoichiometry.
What is theoretical yield?
Theoretical yield is the maximum product mass or moles you could obtain if the reaction ran to completion with no losses, computed from the limiting reagent via the mole ratio in the balanced equation. It is the ceiling against which actual yields are measured. Real yields are almost always lower because of incomplete reaction, side products, or transfer losses.
What is percent yield?
Percent yield = (actual yield / theoretical yield) × 100. It quantifies how much of the maximum possible product you actually obtained. Below 100% means losses somewhere in the process; above 100% almost always means the product is impure (water of crystallization, residual solvent, unreacted starting material). Percent yield can't be calculated without first finding the theoretical yield, which requires the limiting reagent.
How much excess reagent remains after the reaction?
Use the limiting reagent's moles and the mole ratio from the balanced equation to compute how much of the excess reagent actually reacted. Subtract that from the starting amount of the excess reagent to get the leftover. Convert back to grams via molar mass if needed. The same arithmetic generalizes to any number of excess reactants.