Stoichiometry & Limiting Reactant Calculator
Limiting reactant, excess, theoretical and percent yield — step by step.
Reactants
Products (theoretical yield)
How it was worked out
About the Stoichiometry & Limiting Reactant Calculator
Type a reaction — it is balanced for you — then the amount of each reactant you have, in grams, moles, litres of gas or millilitres of a solution of known molarity. The calculator converts everything to moles, finds the limiting reactant, how much of each excess reactant is left over, and the theoretical yield of every product in moles, grams and, for gases, litres. Add the yield you actually obtained to get the percent yield, and see the atom economy of the product you want.
You can also work backwards: choose A product amount to find how much of each reactant is needed to make, say, 16 g of a product. Every step is shown, from the mole conversions to the extent of reaction ξ that ties all the amounts together, using IUPAC (CIAAW 2024) atomic weights.
How to use it
- Type the equation, for example
Si + N2 = Si3N4; the balanced form (3Si + 2N₂ → Si₃N₄) appears under it. - For each reactant, type the amount you have and choose its unit: g, mg, kg, mol, mmol, L or mL of gas, or mL or L of a solution (then type its molarity). Leave a reactant empty if it is in excess.
- Read the limiting reactant, the theoretical yields and what is left over. To start from a product instead, choose A product amount.
- Optionally open Gas conditions, actual yield and atom economy to set the temperature and pressure for gas volumes, enter your actual yield and pick the desired product.
Examples
3Si + 2N₂ → Si₃N₄ with 2.00 g Si and 1.50 g N₂
Si is limiting; 0.02374 mol (3.330 g) Si₃N₄; 0.00607 mol (0.170 g) N₂ left
OpenStax Chemistry 2e Example 4.12.
CuSO₄ + Zn → Cu + ZnSO₄ with 1.274 g CuSO₄, excess Zn, 0.392 g Cu obtained
Theoretical 0.5072 g Cu; percent yield 77.3 %
OpenStax Example 4.13.
MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl, 16 g Mg(OH)₂ wanted
21.95 g NaOH (and 26.12 g MgCl₂)
OpenStax Example 4.10: 22 g NaOH.
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O with 702 g octane
2,458 g O₂ reacts
OpenStax Example 4.11: 2.46 × 10³ g.
HCl + NaOH → NaCl + H₂O with 25.0 mL of 0.100 M HCl and 20.0 mL of 0.100 M NaOH
NaOH is limiting; 0.002 mol (0.1169 g) NaCl; 0.0005 mol HCl left
Common uses
- Homework and exams: limiting-reagent, theoretical-yield and percent-yield problems with the working shown.
- Lab preparation: how much of each reactant to weigh for a target amount of product.
- Lab reports: percent yield from the mass you actually recovered.
- Green chemistry: comparing the atom economy of different routes to the same product.
How the limiting reactant is found
Every amount is first converted to moles: mass ÷ molar mass, gas volume ÷ molar volume (R·T ÷ p for an ideal gas), or solution volume × molarity. Dividing each reactant’s moles by its coefficient gives how many times the reaction as written could run on that reactant alone. The smallest of these is the extent of reaction ξ (in moles of reaction), and the reactant that gives it is the limiting reactant. Then every product forms coefficient × ξ mol, every reactant is used at coefficient × ξ mol, and the excess reactants keep the rest.
This is the same as the textbook method of comparing mole ratios, or of working out which reactant gives the least product (OpenStax Chemistry 2e §4.4) — just written once for any number of reactants.
Percent yield and atom economy
- Theoretical yield: the amount of product the limiting reactant can form, from the balanced equation.
- Percent yield = actual yield ÷ theoretical yield × 100 %. Above 100 % means the product is not pure or dry, or a measurement is wrong.
- Atom economy = mass of the desired product ÷ mass of all the reactants × 100 %, both from the balanced equation (coefficient × molar mass). It measures how much of the reactants’ mass ends up in the product, however good the yield; it is one of the principles of green chemistry (OpenStax §4.4). A reaction with a single product, such as 2H₂ + O₂ → 2H₂O, has 100 %.
Units you can use
- Mass: g, mg, kg.
- Amount: mol, mmol.
- Gas: L or mL of gas at the chosen conditions — IUPAC STP (0 °C, 100 kPa, 22.711 L/mol) by default, or 0 °C and 1 atm, 25 °C, or any temperature and pressure.
- Solution: mL or L of a solution of the molarity you type (mol/L).
Products written with (g) also get their volume as an ideal gas. Electrons in a half-reaction can be given in mol or mmol.
Sources
- OpenStax, Chemistry 2e (CC BY 4.0): §4.3 Reaction Stoichiometry and §4.4 Reaction Yields (method, worked examples, percent yield and atom economy).
- IUPAC, Quantities, Units and Symbols in Physical Chemistry (Green Book, 3rd ed., 2007), §2.10: extent of reaction ξ and amount concentration.
- IUPAC CIAAW, Standard Atomic Weights 2024.
- NIST CODATA 2022: molar volume of an ideal gas and molar gas constant; IUPAC Gold Book, standard conditions for gases.
Limitations
- The theoretical yield assumes the reaction goes to completion as written, with no side reactions. Equilibria and real-world losses give less.
- Gas volumes use the ideal-gas law; real gases deviate a little, more at high pressure and low temperature.
- Solution volumes need the molarity of the solution itself; for a mass concentration, convert to mol/L first (the molarity calculator does this).
- When an equation can be balanced in several ways, the amounts depend on the coefficients used: type your own in the equation to choose them.
Privacy
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Frequently asked questions
How do I find the limiting reactant?
Convert each reactant to moles and divide by its coefficient in the balanced equation; the smallest result is the limiting reactant. For 3Si + 2N₂ → Si₃N₄ with 2.00 g Si (0.07121 mol ÷ 3 = 0.02374) and 1.50 g N₂ (0.05354 mol ÷ 2 = 0.02677), silicon is limiting.
What is the theoretical yield?
The most product the limiting reactant can make, from the mole ratio in the balanced equation. It is shown for every product in moles, grams and — for gases — litres.
How is percent yield calculated?
Actual yield ÷ theoretical yield × 100 %. Recovering 0.392 g of copper when 0.5072 g was possible is a 77.3 % yield (OpenStax Example 4.13).
Can I use volumes of gases or solutions?
Yes. Choose “L of gas” or “mL of gas” and set the conditions (STP by default), or “mL of solution” and type its molarity: 25.0 mL of 0.100 M HCl is 0.00250 mol.
What if I only know one reactant?
Type just that one and leave the others empty: they are treated as being in excess, and the page shows how much of each would react.
What is atom economy?
The mass of the desired product as a percentage of the mass of all the reactants in the balanced equation. Making CaO from CaCO₃ (CaCO₃ → CaO + CO₂) has an atom economy of 56.0 %, because the CO₂ is not wanted.