Limiting Reactant Calculator
Give the amounts you start with. Find which reactant runs out first, the theoretical yield of every product and exactly how much excess is left in the flask.
Separate substances with +, sides with -> or =. Coefficients are optional; the equation is balanced for you.
N₂ + 3H₂ → 2NH₃
LR
Limiting reactant
Runs out after 0.9995 mol of reaction
TY
Theoretical yield of NH₃
g
1.999 mol
XS
H₂ left over
g
0.9698 mol unused of 3.968
| Reactant | Moles available | Coefficient | Moles ÷ coefficient |
|---|---|---|---|
| N₂ ← limiting | 0.9995 | 1 | 0.9995 |
| H₂ | 3.968 | 3 | 1.323 |
See it with particles: sandwiches
A sandwich needs 2 slices of bread and 1 slice of cheese: 2 Br + 1 Ch → 1 Sw. You have 10 bread and 4 cheese. Step through to see which runs out.
Made 0. Bread ÷ 2 = 5, cheese ÷ 1 = 4, so cheese will run out first.
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Limiting reactant in four steps
- Balance the equation (done above automatically).
- Convert every reactant amount to moles using its molar mass.
- Divide each by its coefficient. The smallest value is the limiting reactant and equals the “moles of reaction”.
- Multiply that value by each product’s coefficient for the theoretical yield, and by each excess reactant’s coefficient to find how much of it is used.
Frequently asked questions
How do you find the limiting reactant?
Convert each reactant to moles, then divide by its coefficient in the balanced equation. The reactant with the smallest result runs out first: it is the limiting reactant. This calculator shows that “moles ÷ coefficient” value for each reactant.
Is the limiting reactant the one with the smallest mass?
Not necessarily. In N₂ + 3H₂ → 2NH₃, 28 g of N₂ (1.0 mol) needs 6.0 g of H₂ (3.0 mol). With 8.0 g of H₂ there is more than enough, so N₂ is limiting even though its mass is larger.
What is theoretical yield?
The maximum mass of product that can form when the limiting reactant is completely used up, calculated from the mole ratio. Real experiments give less; the percent yield compares the two.
How do I calculate how much excess reactant is left?
Multiply the moles of limiting reactant by the ratio (coefficient of excess ÷ coefficient of limiting) to get moles of excess used, subtract from moles available, and convert back to grams.
Can there be more than two reactants?
Yes. Enter an amount for each reactant in the equation; the one with the smallest moles-per-coefficient is limiting. Leave a box empty if that reactant is in large excess (like O₂ from air).