Stoichiometry & Limiting Reagent Calculator
Find the limiting reagent, the theoretical yield of every product, the excess left over and your percent yield from any chemical equation. It balances the equation for you and shows each mole ratio.
How to Use
- Type the reaction, for example
N2 + H2 = NH3. It is balanced for you; coefficients you type are kept if they already balance. - Keep Limiting reagent selected and enter the amount of each reactant in grams or moles. Leave a reactant blank if it is in excess, like oxygen from the air.
- Choose the product you are interested in. Add the actual yield you got if you want the percent yield.
- Or choose Target and enter how much product you want, to see how much of each reactant it takes.
- The bars show what is used, left over and formed; Show Work has every mole ratio and the full before-and-after table.
Worked Example
Ammonia from 28 g of N2 and 5 g of H2. The balanced equation is N2 + 3H2 → 2NH3. Moles: 28 ÷ 28.014 = 0.9995 mol N2 and 5 ÷ 2.016 = 2.4802 mol H2. Divide by the coefficients: 0.9995 ÷ 1 = 0.9995 and 2.4802 ÷ 3 = 0.8267. Hydrogen gives the smaller number, so it is limiting. Yield: 2.4802 × 2/3 = 1.6534 mol NH3 × 17.031 g/mol = 28.16 g. Nitrogen used: 0.8267 mol, so 0.1728 mol (4.84 g) is left.
Percent yield for aspirin. C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2, all 1 : 1. 2.00 g of salicylic acid is 0.014480 mol and 5.00 g of acetic anhydride is 0.048977 mol, so salicylic acid is limiting and the theoretical yield is 0.014480 × 180.159 = 2.6087 g. Collecting 2.10 g is 2.10 ÷ 2.6087 = 80.5%.
The common mistake: comparing moles without the coefficients. In the ammonia example N2 has fewer moles (0.9995 against 2.4802), so it looks limiting, and that gives 2 × 0.9995 × 17.031 = 34.04 g of NH3. But each N2 needs three H2, and 0.9995 mol of N2 would need 2.9985 mol of H2, more than there is. Divided by the coefficients, hydrogen is limiting and the right answer is 28.16 g.
Show Work
Formulas
Where Stoichiometry Comes From
The word was coined by the German chemist Jeremias Benjamin Richter in 1792, from the Greek stoicheion (element) and metron (measure), in a book on the proportions in which acids and bases neutralise each other. Joseph Proust’s law of definite proportions in the 1790s and John Dalton’s atomic theory of 1808 explained why those proportions are fixed whole-number ratios.
Amedeo Avogadro proposed in 1811 that equal volumes of gases at the same temperature and pressure hold equal numbers of molecules. The mole became an SI base unit in 1971, and since 2019 it is defined as exactly 6.02214076 × 1023 particles, which is why the mole ratios in a balanced equation are simply ratios of particle counts.
About This Calculator
This calculator balances the equation exactly, works out every molar mass from IUPAC standard atomic weights, and converts your amounts to moles. It then finds the limiting reagent by moles divided by coefficient, and gives the yield of every product, what is left of each excess reactant, and your percent yield. Target mode runs the same ratios backwards to give the reactants a wanted amount of product takes.
Amounts can be in grams, milligrams, kilograms, pounds, moles or millimoles, mixed freely. Everything runs in your browser; nothing is sent anywhere.
Related tools: Chemical Equation Balancer, Molar Mass Calculator, and Dilution Calculator.
Frequently Asked Questions
How do I find the limiting reagent?
Convert each reactant to moles, divide by its coefficient in the balanced equation, and the smallest result is the limiting reagent. For N2 + 3H2 → 2NH3 with 28 g of N2 (0.9995 mol) and 5 g of H2 (2.4802 mol): 0.9995 ÷ 1 = 0.9995 and 2.4802 ÷ 3 = 0.8267, so hydrogen runs out first even though there are more moles of it.
How do I calculate the theoretical yield?
Multiply the moles of the limiting reagent by the mole ratio from the equation, then by the product’s molar mass. In the example, 2.4802 mol H2 × (2 mol NH3 ÷ 3 mol H2) = 1.6534 mol NH3, and 1.6534 mol × 17.031 g/mol = 28.16 g.
How do I calculate percent yield?
Divide what you actually collected by the theoretical yield and multiply by 100. Making aspirin from 2.00 g of salicylic acid has a theoretical yield of 2.6087 g; collecting 2.10 g is 2.10 ÷ 2.6087 × 100 = 80.5%. A result above 100% usually means the product is still wet or impure.
How much of the excess reactant is left over?
Work out how much the limiting reagent uses up and subtract it. In thermite, 2Al + Fe2O3 → Al2O3 + 2Fe, 100 g of Fe2O3 (0.62623 mol) uses 1.2525 mol of aluminium; from 50 g (1.8531 mol) that leaves 0.60064 mol, or 16.21 g of aluminium.
What if I don’t know how much of one reactant there is?
Leave it blank and it is treated as being in excess. Burning 44 g of propane (0.9978 mol) in plenty of air gives 2.9934 mol, or 131.74 g, of CO2, and the working shows it uses 159.64 g of oxygen.
How do I use the Stoichiometry & Limiting Reagent Calculator?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
Ammonia synthesis
To make 100 g of NH3 by N2 + 3H2 → 2NH3 you need at least 82.24 g of nitrogen and 17.76 g of hydrogen.
Aspirin lab report
2.00 g of salicylic acid with 5.00 g of acetic anhydride gives a theoretical 2.6087 g of aspirin; 2.10 g collected is an 80.5% yield.
Thermite
50 g of aluminium with 100 g of iron(III) oxide: the oxide is limiting, giving 69.94 g of iron and leaving 16.21 g of aluminium.
Combustion
44 g of propane burns to 131.74 g of carbon dioxide and 71.90 g of water, using 159.64 g of oxygen.
Fuel cells
10 g of hydrogen with 64 g of oxygen: oxygen runs out first, giving 72.07 g of water with 1.94 g of hydrogen to spare.
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