Limiting Reagent Calculator — Find Limiting Reactant
Find the limiting reagent, excess reactant, and theoretical yield for any balanced chemical reaction. Enter masses or moles of two reactants, and get complete stoichiometric working with mole ratios and yield calculations.
Quick Presets
Reactant A
Reactant B
Product
What Is the Limiting Reagent Calculator — Find Limiting Reactant?
This calculator identifies the limiting reagent in a chemical reaction with two or three reactants, then calculates the theoretical yield of the product and the amount of each excess reagent left over — with full step-by-step working.
- ›Up to three reactants — add or remove the third reactant with one click to handle three-component reactions.
- ›Grams or moles input — each reactant can be entered in grams or moles independently; the calculator converts internally.
- ›Full mole comparison table — shows grams, moles, coefficient, moles/coefficient, and role (limiting or excess) for every reactant at once.
- ›Excess reagent analysis — reports exactly how many grams and moles of each excess reactant remain after the limiting reagent is consumed.
- ›Always-visible step-by-step working — every conversion, ratio, and conclusion is shown without needing to expand a panel.
- ›Three real-world presets — H₂ + O₂ → H₂O, Na + Cl₂ → NaCl, and Fe + S → FeS load instantly for verification or learning.
Formula
Mole Ratio Comparison Method
For each reactant: mole ratio equivalent = moles available / stoich. coefficient
Limiting reagent = reactant with the SMALLEST mole ratio equivalent
Moles From Grams
n = m / M (moles = mass / molar mass)
Theoretical Yield
n(product) = (n(limiting) / coeff(limiting)) × coeff(product)
m(product) = n(product) × M(product)
Excess Reagent Remaining
n(used) = (n(limiting) / coeff(limiting)) × coeff(excess reagent)
n(excess) = n(available) − n(used)
m(excess) = n(excess) × M(excess reagent)
| Symbol | Name | Description |
|---|---|---|
| n | Moles | Amount of substance; n = m / M |
| m | Mass | Mass of a reactant or product in grams |
| M | Molar mass | Gram-formula mass in g/mol |
| a, b | Stoichiometric coefficients | Integers from the balanced chemical equation |
| LR | Limiting reagent | The reactant that runs out first; determines maximum yield |
| ER | Excess reagent | Any reactant still present after the limiting reagent is consumed |
| n/a | Mole ratio equivalent | n(reactant) / coefficient — the comparison figure for finding the LR |
Reading Coefficients from a Balanced Equation
2 H₂ + O₂ → 2 H₂O
coeff(H₂) = 2 coeff(O₂) = 1 coeff(H₂O) = 2
N₂ + 3 H₂ → 2 NH₃
coeff(N₂) = 1 coeff(H₂) = 3 coeff(NH₃) = 2
2 Na + Cl₂ → 2 NaCl
coeff(Na) = 2 coeff(Cl₂) = 1 coeff(NaCl) = 2
How to Use
- 1Enter Reactant A: Type the name, molar mass (g/mol), amount, and stoichiometric coefficient from the balanced equation. Select grams or moles.
- 2Enter Reactant B: Fill in the same four fields for the second reactant. The unit (grams/moles) can be different from Reactant A.
- 3Add Reactant C (optional): Click "+ Add third reactant" for three-component reactions. Fill in the same fields. Click Remove to go back to two reactants.
- 4Enter Product details: Enter the product name, molar mass (g/mol), and its stoichiometric coefficient from the balanced equation.
- 5Try a preset: Click H₂ + O₂ → H₂O, Na + Cl₂ → NaCl, or Fe + S → FeS to load a complete worked example instantly.
- 6Click Find Limiting Reagent: Results appear: the limiting reagent highlighted, theoretical yield in grams and moles, excess reagent amounts, and a full mole comparison table.
- 7Review step-by-step working: The full four-step derivation is always shown below the results — mole conversion, ratio comparison, identification, and yield calculation.
Example Calculation
H₂ + O₂ → H₂O: 5 g of H₂ and 40 g of O₂ — which runs out first?
Balanced equation: 2 H₂ + O₂ → 2 H₂O
Given: 5 g H₂ (M = 2.016), 40 g O₂ (M = 32.00)
Step 1 — Convert to moles
n(H₂) = 5.000 g ÷ 2.016 g/mol = 2.4802 mol
n(O₂) = 40.00 g ÷ 32.00 g/mol = 1.2500 mol
Step 2 — Divide by stoichiometric coefficient
H₂: 2.4802 / 2 = 1.2401
O₂: 1.2500 / 1 = 1.2500
Step 3 — Smallest ratio = limiting reagent
H₂: 1.2401 ← LIMITING (smaller)
O₂: 1.2500 (excess)
Step 4 — Theoretical yield of H₂O
n(H₂O) = 1.2401 × 2 = 2.4802 mol
m(H₂O) = 2.4802 mol × 18.015 g/mol = 44.69 g
| Reactant | Grams | Moles | Coeff | Moles/Coeff | Role |
|---|---|---|---|---|---|
| H₂ | 5.0000 | 2.480159 | 2 | 1.240079 | Limiting |
| O₂ | 40.000 | 1.250000 | 1 | 1.250000 | Excess |
Excess O₂ remaining
O₂ consumed = 1.2401 mol × 1 = 1.2401 mol (= 39.68 g). O₂ remaining = 1.2500 − 1.2401 = 0.0099 mol (= 0.317 g). Most of the O₂ is used, but H₂ runs out fractionally earlier — confirming H₂ is the limiting reagent.
Understanding Limiting Reagent — Find Limiting Reactant
What Is a Limiting Reagent?
In any chemical reaction, the limiting reagent (also called the limiting reactant) is the substance that is completely consumed first, stopping the reaction from continuing. Once the limiting reagent is gone, no more product can form — regardless of how much of the other reactants remain.
The concept is easiest to visualise with a non-chemistry analogy: if you are making sandwiches and you have 10 slices of bread but only 3 slices of cheese, you can make at most 3 sandwiches (assuming 1 slice of cheese each). The cheese is the "limiting reagent"; the bread is in excess. You will have 4 slices of bread left over.
How to Identify the Limiting Reagent
The formal method uses stoichiometric mole ratios and works for any number of reactants:
- ›Write the balanced equation. Read the stoichiometric coefficients for every reactant and the product. These must come from a correctly balanced equation — they are not the masses you used.
- ›Convert all amounts to moles. If you know grams, divide by molar mass: n = m / M. If you are given moles directly, use them as-is.
- ›Divide each reactant's moles by its coefficient. This normalises each reactant to a common "per formula unit" basis.
- ›The smallest ratio is the limiting reagent. That reactant would be consumed first if the reaction ran to completion with equal proportions.
Why divide by the coefficient?
The balanced equation 2 H₂ + O₂ → 2 H₂O tells you that H₂ and O₂ react in a 2:1 mole ratio. If you have 4 mol H₂ and 3 mol O₂, it is not enough to compare raw moles (4 vs 3). You need 2 mol H₂ for every 1 mol O₂. Dividing: H₂ gives 4/2 = 2; O₂ gives 3/1 = 3. H₂ is limiting (smaller ratio). This is the mole ratio equivalent method.
The Mole Ratio Method
The mole ratio method is the most systematic approach and works for any balanced equation. The "mole ratio equivalent" for each reactant is:
mole ratio equivalent = n(reactant) / stoichiometric coefficient
Example: 2 Na + Cl₂ → 2 NaCl
10 g Na ÷ 22.99 g/mol = 0.4350 mol Na, 0.4350 / 2 = 0.2175
20 g Cl₂ ÷ 70.90 g/mol = 0.2821 mol Cl₂, 0.2821 / 1 = 0.2821
Na: 0.2175 ← LIMITING (smallest)
Cl₂: 0.2821 (excess)
Once the limiting reagent is identified, theoretical yield flows directly from it: multiply the mole ratio equivalent of the limiting reagent by the product's coefficient to get moles of product, then by product molar mass to get grams.
Excess Reagent and How Much Is Left
The excess reagent is any reactant that is not completely consumed. After the reaction stops (because the limiting reagent ran out), some of the excess reagent remains in the reaction mixture. To find how much:
- ›Moles consumed = (n(limiting) / coeff(limiting)) × coeff(excess reagent)
- ›Moles remaining = n(available) − n(consumed)
- ›Grams remaining = moles remaining × M(excess reagent)
In industrial settings, chemists deliberately use one reagent in slight excess (5–20%) to ensure the more expensive or harder-to-obtain reactant is fully consumed. This excess must then be separated from the product and either recycled or disposed of — a cost that must be weighed against the benefit of complete conversion.
Why Limiting Reagents Matter in Industry
The limiting reagent concept is central to the economics of chemical manufacturing:
- ›Cost control. The limiting reagent is often the most expensive starting material. Knowing which reagent limits production lets engineers buy exactly the right amount and avoid waste.
- ›Scale-up calculations. To produce a target mass of product, engineers calculate how much limiting reagent is needed (reverse stoichiometry) and size purchasing, reactors, and waste systems accordingly.
- ›Green chemistry. Atom economy — the fraction of reactant mass incorporated into the product — is maximised when stoichiometric amounts are used and excess is minimised.
- ›Reaction optimisation. Knowing the excess reagent and its amount helps design purification steps (washes, extractions) to remove it from the product without loss.
- ›Safety. In energetic or exothermic reactions, excess of one reagent can cause runaway heating. Engineers deliberately make one reagent limiting to control the reaction rate and heat output.
Frequently Asked Questions
What is the limiting reagent in chemistry?
The limiting reagent (or limiting reactant) is the substance that runs out first. Once it is gone, the reaction stops regardless of how much of the other reactants remain.
- ›It determines the theoretical yield — the maximum possible product
- ›Other reactants present in excess are called excess reagents
- ›In a balanced equation aA + bB → cC, the limiting reagent is whichever gives the smaller n/a or n/b ratio
- ›Identifying it correctly is essential before calculating any yield
How do I find the limiting reagent step by step?
The mole ratio method — reliable for any number of reactants:
For aA + bB → cC:
n(A) = m(A) / M(A) n(B) = m(B) / M(B)
ratio_A = n(A) / a ratio_B = n(B) / b
Limiting = reactant with smallest ratio
n(C) = min(ratio_A, ratio_B) × c
This calculator performs all four steps automatically and shows every intermediate value.
What is the difference between limiting reagent and excess reagent?
- ›Limiting reagent — 0 mol remaining after reaction; all consumed; determines theoretical yield
- ›Excess reagent — some mol remaining; amount = n(available) − n(consumed by limiting reagent)
- ›In 2 H₂ + O₂ → 2 H₂O with 4 mol H₂ and 3 mol O₂: H₂ is limiting (ratio 4/2=2 vs 3/1=3); O₂ excess = 3 − 2 = 1 mol left
How do I calculate theoretical yield from the limiting reagent?
For aA + bB → cC, where A is limiting:
n(C) = n(A) / a × c
m(C) = n(C) × M(C)
Example: 2 Al + 3 Cl₂ → 2 AlCl₃ — 5 g Al (M=26.98) is limiting:
- ›n(Al) = 5.00 / 26.98 = 0.1853 mol
- ›n(AlCl₃) = (0.1853 / 2) × 2 = 0.1853 mol
- ›m(AlCl₃) = 0.1853 × 133.34 = 24.71 g (theoretical yield)
Can there be more than one limiting reagent?
When two reactants have identical moles/coefficient ratios, neither is technically "more limiting" — they are consumed simultaneously. This is called a stoichiometric mixture.
- ›Example: exactly 2 mol H₂ and 1 mol O₂ for 2 H₂ + O₂ → 2 H₂O
- ›H₂: 2/2 = 1.0 | O₂: 1/1 = 1.0 — identical ratios, both fully consumed
- ›This is the most efficient use of reactants — no excess
- ›The calculator uses the first-found minimum; both will show Limiting in a tie
What happens to the excess reagent after the reaction?
The excess reagent ends up mixed with the product and must be separated:
- ›Insoluble excess solid + soluble product: filter the solid, collect the filtrate
- ›Aqueous excess reagent: wash the organic product with water
- ›Volatile excess reagent: evaporate or distil it away
- ›Industrial recycling: recover and reuse excess reagent to reduce cost
The amount of excess can be calculated: moles remaining = moles available − moles consumed. This calculator reports excess mass and moles for each non-limiting reactant.
Does this calculator save my inputs?
Yes — inputs are automatically persisted to your browser's localStorage:
- ›All reactant fields (name, molar mass, amount, unit, coefficient) are saved
- ›Product fields are saved
- ›Whether the third reactant panel is open or closed is saved
- ›All data stays in your browser — nothing is sent to any server
- ›Inputs are restored the next time you open the page
Click Reset All to clear both the form and the saved localStorage data.