# Which one is the limiting reactant?

Finds the limiting reactant of a reaction a·A + b·B → p·P from the two masses, molar masses and coefficients, with the theoretical yield of the product and how much of the excess reactant is left.

- Page: https://www.acalculator.org/chemistry/limiting-reactant-calculator
- JSON spec: https://www.acalculator.org/chemistry/limiting-reactant-calculator.json
- Version: 39fda1930c20

## Default answer

Example with the default inputs (Reactant A mass 2 g, Reactant A molar mass, g/mol 28.09, Reactant A coefficient (a) 3, Reactant B mass 1.5 g, Reactant B molar mass, g/mol 28.02, Reactant B coefficient (b) 2, Product coefficient (p) 1, Product molar mass, g/mol 140.31): The limiting reactant is Reactant A: 3.33001 g of product; 0.169989 g of B left.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| mA | Reactant A mass | The mass of reactant a you start with. |
| MA | Reactant A molar mass, g/mol | The molar mass in grams per mole. |
| a | Reactant A coefficient (a) | The coefficient in the balanced equation. |
| mB | Reactant B mass | The mass of reactant b you start with. |
| MB | Reactant B molar mass, g/mol | The molar mass in grams per mole. |
| b | Reactant B coefficient (b) | The coefficient in the balanced equation. |
| p | Product coefficient (p) | The coefficient in the balanced equation. |
| MP | Product molar mass, g/mol | The molar mass in grams per mole. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| limiting | Limiting reactant | The reactant that runs out first. |
| detail | Yield and excess | The product made and the other reactant left, to 6 significant figures. |
| molA | Reactant A (mol) | Moles of A: mass ÷ molar mass. |
| molB | Reactant B (mol) | Moles of B: mass ÷ molar mass. |
| productMol | Theoretical yield (mol) | Moles of product the limiting reactant can make. |
| productG | Theoretical yield (g) | Grams of product: moles × product molar mass. |
| excessG | Excess reactant left (g) | Grams of the other reactant left when the limiting one is used up. |

## Method

n = mass ÷ molar mass for each reactant. The reactant with the smaller n ÷ coefficient is limiting. Product moles = p × (n ÷ coefficient) of the limiting reactant; product grams = moles × product molar mass. Excess left = (n − coefficient × limiting quotient) × molar mass of the other reactant.

## Assumptions

- The reaction goes to completion as written, and A and B react only with each other.
- The theoretical yield is the most product possible; a real yield is usually lower.

## Worked examples

1. mA = 0.002, MA = 28.09, a = 3, mB = 0.0015, MB = 28.02, b = 2, p = 1, MP = 140.31 gives molA = 0.0712, productMol = 0.023733, productG = 3.330011, excessG = 0.169989, limiting = Reactant A, detail = 3.33001 g of product; 0.169989 g of B left. Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields, Example 4.12 Identifying the Limiting Reactant: silicon is limiting, 0.0237 mol of Si₃N₄.
2. mA = 0.004, MA = 2, a = 2, mB = 0.032, MB = 32, b = 1, p = 2, MP = 18 gives productMol = 2, productG = 36, excessG = 0, limiting = Neither (exact ratio), detail = A and B run out together and make 36 g of product. Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields.
3. mA = 0.01, MA = 2, a = 2, mB = 0.032, MB = 32, b = 1, p = 2, MP = 18 gives productG = 36, excessG = 6, limiting = Reactant B, detail = 36 g of product; 6 g of A left. Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields (limiting reactant: compare the reactants’ moles with the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields.

## FAQ

### What is a limiting reactant?

The reactant that is used up first. Once it is gone the reaction stops, so it sets the most product that can form (the theoretical yield). The other reactant is in excess.

### How do I find the limiting reactant?

Turn each mass into moles (mass ÷ molar mass) and divide by that reactant’s coefficient in the balanced equation. The smaller result is the limiting reactant. Comparing grams directly does not work, because molar masses and coefficients differ.

### Which reactant limits 3Si + 2N₂ → Si₃N₄ with 2.00 g of Si and 1.50 g of N₂?

Silicon: 2.00 ÷ 28.09 = 0.0712 mol, ÷ 3 = 0.0237; nitrogen: 1.50 ÷ 28.02 = 0.0535 mol, ÷ 2 = 0.0268. Silicon gives the smaller number, so it runs out first and makes 0.0237 mol of Si₃N₄.

### How is the excess reactant left over worked out?

The limiting reactant’s quotient times the other reactant’s coefficient is how many moles of it react. Subtract that from the moles you had and multiply by its molar mass.

### What if both run out at the same time?

Then the amounts are in the exact ratio of the equation. Neither is limiting and nothing is left over: 4 g of H₂ with 32 g of O₂ makes 36 g of water.

### Why is my real yield lower than the theoretical yield?

Side reactions, incomplete reactions and losses during handling. Divide the actual yield by the theoretical yield for the percent yield.

## Sources

- OpenStax, Chemistry 2e, section 4.4 Reaction Yields: limiting reactant, Example 4.12 (3Si + 2N₂ → Si₃N₄). CC BY 4.0, retrieved 2026-10-03. https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields
- NIST Special Publication 811, Appendix B.8: 1 lb = 0.45359237 kg; 1 oz = 0.028349523125 kg. Retrieved 2026-10-03. https://www.nist.gov/pml/special-publication-811/nist-guide-si-appendix-b-conversion-factors/nist-guide-si-appendix-b8