Limiting Reactant Calculator
This limiting reactant calculator compares two reactants of a balanced equation. For 3 Si + 2 N₂ → Si₃N₄ with 2.00 g of silicon and 1.50 g of nitrogen, silicon runs out first: it is the limiting reactant, and at most 0.0237 mol of Si₃N₄ can form. The table shows every step, the leftover of the excess reactant and the theoretical yield.
How it is calculated
The limiting reactant is the one that is used up first and so limits the amount of product; the other reactant is in excess. To find it, calculate for each reactant how much product it could form on its own – the smaller amount wins. This is the method in the textbook OpenStax Chemistry 2e. Its sandwich analogy: with 28 slices of bread and 11 slices of cheese and a recipe of 2 bread + 1 cheese, only 11 sandwiches can be made – the cheese limits, six slices of bread are left.
Limiting reactant examples
| Balanced equation | Amounts | Limiting reactant | Theoretical product |
|---|---|---|---|
| H₂ + Cl₂ → 2 HCl | 3 mol H₂ · 2 mol Cl₂ | Cl₂ | 4.00 mol HCl |
| 3 Si + 2 N₂ → Si₃N₄ | 2.00 g Si · 1.50 g N₂ | Si | 0.0237 mol Si₃N₄ |
| 2 H₂ + O₂ → 2 H₂O | 5.00 g H₂ · 10.0 g O₂ | O₂ | 0.625 mol H₂O |
| N₂ + 3 H₂ → 2 NH₃ | 14.0 g N₂ · 6.00 g H₂ | N₂ | 1.00 mol NH₃ |
| CH₄ + 2 O₂ → CO₂ + 2 H₂O | 16.0 g CH₄ · 48.0 g O₂ | O₂ | 0.750 mol CO₂ |
| 4 Fe + 3 O₂ → 2 Fe₂O₃ | 10.0 g Fe · 5.00 g O₂ | Fe | 0.0895 mol Fe₂O₃ |
| 2 Na + Cl₂ → 2 NaCl | 2 mol Na · 2 mol Cl₂ | Na | 2.00 mol NaCl |
The first three rows are from OpenStax; the others are calculated with the same method. Example 4.12 step by step: 2.00 g Si ÷ 28.09 g/mol = 0.0712 mol Si and 1.50 g N₂ ÷ 28.02 g/mol = 0.0535 mol N₂. Silicon alone would give 0.0712 × 1/3 = 0.0237 mol Si₃N₄, nitrogen alone 0.0535 × 1/2 = 0.0268 mol. Silicon gives less, so it is the limiting reactant. The calculator uses the current IUPAC atomic weights (Si 28.085, N₂ 28.014), which changes only the last digits.
Excess reactant and leftover
Once the limiting reactant is known, the excess reactant that reacts follows from the coefficients. With 3 mol H₂ and 2 mol Cl₂ (ratio 1 : 1), all 2 mol Cl₂ react with 2 mol H₂, and 1 mol H₂ is left over – exactly the case OpenStax describes. The smaller mass is not automatically the limiting reactant: in 2 H₂ + O₂ the 5.00 g of hydrogen are the smaller mass, but oxygen limits, because hydrogen has a much lower molar mass.
- The equation must be balanced first; the coefficients go into their own fields.
- The theoretical yield is the maximum; the amount actually obtained is usually lower.
- Write formulas with correct capitals: CO is carbon monoxide, Co is cobalt.
How much of that maximum you really got is shown by the percent yield calculator; the molar mass of any formula comes from the molar mass calculator.
Frequently asked questions
How do you find the limiting reactant?
Convert both amounts to moles, then calculate how much product each reactant would give on its own using the coefficients. The reactant that gives less product is limiting.
Is the limiting reactant the one with the smaller mass?
Not necessarily. Masses must be converted to moles and compared with the coefficients. 5.00 g H₂ and 10.0 g O₂ react to water, and oxygen – the larger mass – is limiting.
How much of the excess reactant is left?
Amount of excess reactant minus the amount that reacts with the limiting reactant. For H₂ + Cl₂ with 3 mol and 2 mol, 1 mol H₂ is left.
What is the theoretical yield?
The amount of product formed when the limiting reactant reacts completely. For 2.00 g Si and 1.50 g N₂ it is 0.0237 mol Si₃N₄.
Can I enter moles instead of grams?
Yes. Switch “Amounts in” to moles; the formulas are then only labels, unless you want the yield and leftover in grams too.
Sources and legal basis
- OpenStax Chemistry 2e, 4.4 Reaction Yields – limiting and excess reactant, comparing the product amount from each reactant; H₂ + Cl₂ → 2 HCl (3 mol/2 mol), Example 4.12 (3 Si + 2 N₂ → Si₃N₄: 2.00 g Si, 1.50 g N₂ → 0.0237/0.0268 mol, Si limiting), Check Your Learning (5.00 g H₂, 10.0 g O₂ → O₂) (read 10.10.2026)
- IUPAC CIAAW – Abridged Standard Atomic Weights (Atomic Weights 2021, revised 2024), used for the molar masses (read 10.10.2026)
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