# Limiting Reagent Calculator - Stoichiometry, Theoretical & Percent Yield

> Balances the equation, finds the limiting reactant from grams, moles, gas volume or molarity, and gives theoretical yield, excess and percent yield with steps.

- URL: https://qilingtools.com/en/math/stoichiometry
- Category: Chemistry Tools (https://qilingtools.com/en/chemistry)
- Tags: [Chemistry](https://qilingtools.com/en/tags/chemistry)
- Last updated: 2026-10-26
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

In a balanced chemical equation the coefficients give the mole ratio of every species. When amounts of two or more reactants are given, the limiting reagent (limiting reactant) is the one that runs out first: divide each reactant's moles by its coefficient, and the smallest result wins. It fixes the theoretical yield of every product, and the other reactants are in excess. Example — 10 g H₂ with 64 g O₂ (2H₂ + O₂ → 2H₂O): n(H₂) = 10 ÷ 2.016 = 4.960 mol, ÷ 2 = 2.480; n(O₂) = 64 ÷ 31.998 = 2.000 mol, ÷ 1 = 2.000, so O₂ is limiting. That gives 4.000 mol or 72.06 g of water, with 0.960 mol (1.94 g) of H₂ left over; if you actually collect 65 g, the percent yield is 65 ÷ 72.06 = 90.2%. This calculator balances the equation for you (unbalanced or wrongly balanced input is fine), lists each species with its coefficient and molar mass, and accepts reactant amounts in g, mg, kg, mol, mmol, gas volume in L or mL (22.414, 22.711 or 24.465 L/mol) or solution volume × molarity, with an optional purity for masses. It reports the limiting reagent, the extent of reaction, the theoretical yield of every product in grams and moles, how much of each excess reactant is consumed and left over, and the percent yield from your actual yield. A reverse mode works out how much of every reactant you need for a target amount of product, optionally allowing for an expected yield. Every mole-ratio step is written out for homework and lab prep.

## How to use

1. Type a chemical equation (balanced or not) or pick an example such as 2H2 + O2 = 2H2O
2. Enter amounts for the reactants and choose units — grams, moles, gas litres or mL × mol/L; leave a reactant blank if it is in excess
3. Read the limiting reagent, theoretical yields and leftovers, add your actual yield for percent yield, or switch modes to find the reactants needed for a target

## Features

- Automatic balancing, with clear messages for impossible or non-unique equations
- Units: g, mg, kg, mol, mmol, gas L / mL and solution volume × molarity
- Choice of gas molar volume (22.414, 22.711 or 24.465 L/mol) and purity for masses
- Identifies the limiting reagent, including exact stoichiometric mixtures
- Theoretical yield of every product plus excess reactant consumed and remaining
- Percent yield from actual yield, with a warning above 100%
- Reverse mode: reactants needed for a target product amount and expected yield
- Full mole-ratio working using IUPAC standard atomic weights

## FAQ

### How do I find the limiting reagent?

Convert each reactant to moles and divide by its coefficient in the balanced equation; the smallest value is the limiting reagent. Comparing grams or raw moles does not work because the coefficients differ. In 2H₂ + O₂ → 2H₂O, 4.96 mol H₂ looks like more than 2.00 mol O₂, but 4.96 ÷ 2 = 2.48 is larger than 2.00 ÷ 1, so O₂ runs out first.

### How are theoretical yield and percent yield calculated?

Theoretical yield = moles of limiting reagent × (product coefficient ÷ limiting reagent coefficient) × molar mass of the product — the most product possible if the limiting reagent reacts completely. Percent yield = actual yield ÷ theoretical yield × 100%. A result over 100% usually means the product is still wet or impure.

### Does the equation need to be balanced first?

No. Any coefficients you type are ignored and the equation is rebalanced exactly from atom conservation, and the balanced form is shown; you are told if your coefficients were in the wrong ratio. If the equation cannot be balanced you are asked to check the formulas, and if several independent balancings exist (more than one reaction) the smallest set is used with a note.

### How are gas volumes and solutions converted to moles?

Gases use the ideal-gas molar volume, n = V ÷ Vm: 22.414 L/mol at 0 °C and 1 atm, 22.711 L/mol at the current IUPAC STP (0 °C, 100 kPa), or 24.465 L/mol at 25 °C and 1 atm — you can switch between them. Solutions use n = c × V with the volume entered in mL. For impure solids, enter a purity and the mass is multiplied by it first.

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