Limiting Reagent Calculator
Find limiting reagent, excess reagent and theoretical yield from a balanced equation.
How to enter the problem
Enter the reaction, list each reactant amount, and choose the product whose theoretical yield you want. The calculator balances the equation first, converts grams to moles when needed, then compares reactants using the balanced coefficients.
| Input | Example | Meaning |
|---|---|---|
| Equation | H2 + O2 -> H2O | Unbalanced is accepted; it becomes 2H2 + O2 -> 2H2O |
| Amounts | H2:2:mol,O2:16:g | formula:value:type, separated by commas |
| Amount type | mol or g | Grams are converted using molar mass |
| Product | H2O | The product used for theoretical yield |
Formula used
Worked examples
Water formation: 2H2 + O2 -> 2H2O. If H2 = 2.00 mol and O2 = 16.00 g, O2 is 0.500 mol. Compare H2: 2.00/2 = 1.00 and O2: 0.500/1 = 0.500. Oxygen limits the reaction and can form 1.00 mol H2O.
Methane combustion: CH4 + 2O2 -> CO2 + 2H2O. If CH4 = 16 g and O2 = 64 g, both are near their exact stoichiometric ratio, so neither has meaningful excess after rounding.
How the bars help
The visual bars compare each reactant's mole-per-coefficient value. The shortest bar is the bottleneck for the reaction. Leftover bars and the table show how much excess reactant remains after the limiting reagent has been fully consumed.
Where this calculator is useful
- Finding the limiting reagent before theoretical yield calculations.
- Converting reactant masses into moles for stoichiometry.
- Checking balanced equation coefficients in multi-reactant problems.
- Estimating leftover excess reagent after a reaction.
- Preparing lab reports where percent yield is calculated from theoretical yield.
Common mistakes
- Comparing grams directly instead of converting to moles first.
- Ignoring coefficients from the balanced equation.
- Choosing a product formula that is not on the product side.
- Leaving one reactant out of the amount list.
- Using the limiting reagent mass as theoretical product mass without stoichiometry.
Rounding and result checking
Keep extra digits while converting grams to moles and comparing ratios. The limiting reagent is the one with the smallest available ratio, not always the smallest mass. The theoretical yield should be reported for a specific product, and percent yield should use the same product unit as the experimental yield.
Related Chemistry Tools
How To Find Limiting Reactant
To answer how to find limiting reactant, balance the equation, convert each reactant to moles, divide each mole amount by its coefficient, and choose the smallest result. That reactant runs out first and controls the theoretical yield.
FAQs
How to express limiting reactant in chemical formula?
The limiting reactant is the species that runs out first; the formula does not change but it is normally identified next to the equation when working a problem. Identification procedure: balance the equation, then for each reactant divide moles available by its stoichiometric coefficient. The reactant with the smallest ratio is the limiting reactant. Worked example: 2 H2 + O2 → 2 H2O with 4.0 mol H2 and 1.0 mol O2. H2: 4.0/2 = 2.0; O2: 1.0/1 = 1.0. O2 is limiting. Notational conventions vary by textbook — common forms are underlining the limiting reactant in the equation, marking it 'LR' or 'L.R.' in the work, or noting it in the answer line. The convention matters only for clarity; the chemical formula itself is unchanged.
What is the limiting reactant?
The limiting reactant (also called limiting reagent) is the reactant that is consumed completely first; once it is gone, the reaction stops, even if other reactants are still available. The amount of product formed is determined by the moles of the limiting reactant and the stoichiometric ratios in the balanced equation. The other reactant(s) remaining at the end are the excess reactants. Example: N2 + 3 H2 → 2 NH3. Starting with 2.0 mol N2 and 3.0 mol H2: H2 needs 6.0 mol to consume all the N2, so H2 runs out first — H2 is limiting. NH3 produced = (3.0/3) × 2 = 2.0 mol. Leftover N2 = 2.0 - 1.0 = 1.0 mol (excess). Bench analogy: if a recipe calls for 2 cups of flour and 1 egg per batch, ten cups of flour and two eggs only let you make 2 batches — the eggs are the limiting reagent.
How to find excess reactant?
Identify the limiting reactant first, then compute how much of the other reactant the limiting reactant actually consumes; whatever is left over is the excess. Procedure: (1) Balance the equation. (2) Convert each reactant to moles. (3) Divide moles by stoichiometric coefficient — the smaller ratio is the limiting reactant. (4) From the limiting reactant's moles, use the equation's ratio to calculate how many moles of the other reactant were consumed. (5) Subtract from the initial amount to get the excess. Example: 2 H2 + O2 → 2 H2O. Start with 5.0 mol H2 and 4.0 mol O2. H2: 5.0/2 = 2.5; O2: 4.0/1 = 4.0. H2 is limiting. O2 consumed = (5.0/2) × 1 = 2.5 mol. Excess O2 = 4.0 - 2.5 = 1.5 mol.
What statements are always true about limiting reactants?
(1) The limiting reactant is consumed completely — no moles remain at the end. (2) It alone determines the theoretical yield; the moles of product follow directly from the moles of limiting reactant and the equation's stoichiometric ratio. (3) The limiting reactant is not always the one with the smallest mass or even the smallest moles — it depends on the stoichiometric coefficients. (4) When the limiting reactant is exhausted, the reaction stops, even if other reactants remain. (5) Other reactants are excess; some quantity of each remains after the reaction. Worked example illustrating point (3): 2 Na + Cl2 → 2 NaCl. With 100 g Na (≈ 4.35 mol) and 1.0 g Cl2 (≈ 0.014 mol), Cl2 is the limiting reactant despite having the smaller mass — because the stoichiometry needs only 0.028 mol of Na to react with that Cl2.
How to find limiting reactant from balanced equation?
Use the four-step ratio method. (1) Balance the equation. (2) Convert each reactant's mass to moles using moles = mass ÷ molar mass. (3) Divide each reactant's moles by its stoichiometric coefficient in the balanced equation. (4) The reactant with the smallest ratio is the limiting reactant; use its moles to compute the yield. Worked example: N2 + 3 H2 → 2 NH3. Given 28 g N2 (= 1.00 mol) and 9.0 g H2 (= 4.46 mol): N2 ratio = 1.00 / 1 = 1.00; H2 ratio = 4.46 / 3 = 1.49. The smaller ratio is N2, so N2 is limiting. NH3 produced = 2 × 1.00 = 2.00 mol = 34.0 g. This method works for any number of reactants — just compute the ratio for each.