Stoichiometry Calculator
Balance a reaction, find the limiting reactant, and get the theoretical and percent yield.
It is balanced automatically — you do not need to balance it first.
2H₂ + O₂ → 2H₂O
How much of each reactant?
Leave a reactant blank to treat it as being in excess.
Fill this in to get the percent yield.
Limiting reactant
O2
It runs out first, so it sets how much product can form. Theoretical yield: 36.0323 g of H2O.
Reactants
| Reactant | Mass | Moles | n ÷ coefficient | Left over |
|---|---|---|---|---|
| H2 | 10 g | 4.9603 | 2.4802 | 5.9677 g |
| O2limiting | 32 g | 1.0001 | 1.0001 | 0 g |
Products
- H2O — theoretical yield
- 36.0323 g (2.0001 mol)
Working
- Balanced: 2H₂ + O₂ → 2H₂O
- H2: n = 10 g ÷ 2.016 g/mol = 4.96032 mol; n ÷ coefficient = 4.96032 ÷ 2 = 2.48016
- O2: n = 32 g ÷ 31.998 g/mol = 1.00006 mol; n ÷ coefficient = 1.00006 ÷ 1 = 1.00006
- Smallest ratio is O2, so it is the limiting reactant.
What stoichiometry is
A balanced equation is a recipe given in moles. Stoichiometry is the arithmetic of scaling that recipe to the amounts you actually have — which is why every problem takes the same shape: grams in, moles across the equation, grams out.
Where the marks are lost
Two mistakes account for most wrong answers. The first is comparing moles without dividing by the coefficients. The second is forgetting to balance the equation at all, which makes every ratio wrong from the start. This calculator balances first and shows the division, so both are visible.
Related: molar mass calculator, chemical equation balancer, periodic table.
How it works
A balanced equation is a recipe written in moles. Stoichiometry scales that recipe to the amounts you actually have — which is why every problem takes the same shape: grams in, moles across the equation, grams out.
Limiting reactant and yield
n = m ÷ M reaction extent = n ÷ coefficient ← smallest value is the limiting reactant theoretical yield = extent × coefficient(product) × M(product) percent yield = (actual ÷ theoretical) × 100
- n
- moles of a reactant
- m
- mass you have, in grams
- M
- molar mass, g/mol
- coefficient
- the number in front of that species in the balanced equation
10 g of H₂ with 32 g of O₂
- 2H₂ + O₂ → 2H₂O
- H₂: 10 g (M = 2.016)
- O₂: 32 g (M = 31.998)
- H₂: 10 ÷ 2.016 = 4.96 mol; 4.96 ÷ 2 = 2.48
- O₂: 32 ÷ 31.998 = 1.00 mol; 1.00 ÷ 1 = 1.00
- O₂ has the smaller ratio, so it is limiting
- H₂O = 1.00 × 2 × 18.015
36.03 g of water, with about 6 g of hydrogen left over.
Good to know
- Divide moles by the coefficient before comparing. Comparing raw moles is the classic error — 2 mol of a reactant with coefficient 4 goes half as far as 1 mol of one with coefficient 1.
- Balance the equation first. An unbalanced equation makes every ratio wrong from the start.
- A percent yield above 100% is a measurement problem, not a chemistry one: usually a wet or impure product, or the wrong species assumed to be limiting.
- The excess reactant is not wasted in principle — it is simply left over, and can be recovered.
Related Calculators
Frequently Asked Questions
What is the limiting reactant?
The reactant that runs out first, and so decides how much product can form. Everything else is in excess and is left over when the reaction stops.
How do you find the limiting reactant?
Convert each reactant’s mass to moles, then divide by that species’ coefficient in the balanced equation. The smallest result is the limiting reactant. Comparing raw moles instead is the classic mistake — 2 mol of something with a coefficient of 4 goes half as far as 1 mol of something with a coefficient of 1.
What is theoretical yield?
The mass of product you would get if the reaction went perfectly to completion with nothing lost. It is calculated from the limiting reactant alone.
How do you calculate percent yield?
Actual yield divided by theoretical yield, times 100. Getting 30 g where 36 g was possible is an 83% yield.
Why is my percent yield over 100%?
It cannot really be. In practice it means the product is still wet or contains impurities, the masses were mistyped, or the wrong species was assumed to be limiting. A yield above 100% is a measurement problem, not a chemistry one.