Limiting Reagent Calculator
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The limiting reagent is the reactant that runs out first. Divide the moles of each reactant by its coefficient in the balanced equation; the smaller result is limiting. For 2H2 + O2 → 2H2O with 10 g H2 and 64 g O2, oxygen runs out first and 72.06 g of water can form.
| Reactants | |
| H2 moles | 4.9603 mol |
| O2 moles | 2 mol |
| Product | |
| Theoretical yield | 72.06 g |
| Theoretical yield (moles) | 4 mol |
| Excess reactant | |
| Excess reactant | H2 |
| Left over | 1.936 g |
Show steps
- Moles of H2 = 10 g ÷ 2.016 g/mol = 4.9603 mol.
- Moles of O2 = 64 g ÷ 32 g/mol = 2 mol.
- Divide by the coefficients: H2 4.9603 ÷ 2 = 2.4802; O2 2 ÷ 1 = 2.
- The smaller number belongs to O2, so O2 is the limiting reagent.
- Moles of H2O = 2 × 2 = 4 mol; mass = 4 × 18.015 = 72.06 g.
- Excess H2 used = 8.064 g, so 1.936 g is left over.
How to use the calculator
- Balance the equation first, for example 2H2 + O2 → 2H2O.
- For each reactant enter its mass, its molar mass and its coefficient. Then enter the product’s molar mass and coefficient.
- Press Calculate to see which reactant runs out, the theoretical yield and how much of the other reactant is left.
Need a molar mass? Use the molar mass calculator linked below.
How to find the limiting reagent
- Convert each reactant’s mass to moles: moles = mass ÷ molar mass.
- Divide each by its coefficient in the balanced equation.
- The reactant with the smaller result is the limiting reagent.
- Use the limiting reagent and the mole ratio to find moles of product, then multiply by the product’s molar mass.
Comparing raw masses or raw moles is a common mistake. In the water example there are more grams of oxygen, yet oxygen still runs out first.
Worked example
| Step | H2 | O2 |
|---|---|---|
| Mass | 10.0 g | 64.0 g |
| Moles | 10.0 ÷ 2.016 = 4.96 | 64.0 ÷ 32.00 = 2.00 |
| ÷ coefficient | 4.96 ÷ 2 = 2.48 | 2.00 ÷ 1 = 2.00 |
| Result | Excess | Limiting |
2.00 mol O2 makes 4.00 mol H2O, which is 4.00 × 18.015 = 72.06 g. It uses 4.00 mol H2 (8.06 g), leaving 1.94 g of hydrogen.
Limiting reagent practice questions
Test yourself with 6 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
Masses and raw moles can mislead; moles ÷ coefficient compares how many “batches” of the reaction each reactant can supply.
C: 24 ÷ 12 = 2 mol. O2: 32 ÷ 32 = 1 mol. The ratio is 1 : 1, so oxygen runs out first.
Quiz complete
Frequently asked questions
What is a limiting reagent?
It is the reactant that is used up first in a reaction. Once it runs out no more product can form, so it sets the theoretical yield.
How do you find the limiting reactant from grams?
Change each mass to moles, divide by the coefficient from the balanced equation, and choose the smaller value.
How do you calculate the excess reactant left over?
Use the limiting reagent and the mole ratio to find how many moles of the other reactant react, change that to grams, and subtract from the starting mass.
Does the limiting reagent always have the smaller mass?
No. Molar masses and coefficients differ, so a reactant with more grams can still run out first, as oxygen does when it reacts with hydrogen.
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