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Limiting Reactant Calculator:
Theoretical Yield & Excess.
Enter a chemical equation and the amount of each reactant — the calculator balances it, finds the limiting reactant, computes the theoretical yield of any product, and shows the step-by-step working you can copy.
Balanced: CH₄ + ₂O₂ → CO₂ + ₂H₂O
Quick examples
Reactant amounts
Limiting reactant
O₂
ξ = 0.75 · runs out first
Theoretical yield of CO₂
33.01 g
0.75 mol · M = 44.01 g/mol
Reactant breakdown
| Reactant | ν | M (g/mol) | n₀ (mol) | ξ | Left over |
|---|---|---|---|---|---|
| CH₄ | 1 | 16.04 | 0.9973 | 0.9973 | 0.2473 mol (3.967 g) |
| O₂limiting | 2 | 32 | 1.5 | 0.75 | — |
Step-by-step working
Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
Step 1 — convert each reactant to moles (n = mass ÷ molar mass).
• CH4: 16 g = 16 g ÷ 16.04 g/mol = 0.9973 mol
• O2: 48 g = 48 g ÷ 32 g/mol = 1.5 mol
Step 2 — compute the reaction extent ξᵢ = nᵢ ÷ coefficientᵢ for each reactant.
• CH4: ξ = 0.9973 ÷ 1 = 0.9973
• O2: ξ = 1.5 ÷ 2 = 0.75
Step 3 — the smallest ξ wins. Limiting reactant: O2 (ξ = 0.75).
Step 4 — theoretical yield of CO2: n = ξ_min × ν = 0.75 × 1 = 0.75 mol
mass = 0.75 mol × 44.01 g/mol = 33.01 g
Step 5 — excess reactants left over (nᵢ_rem = nᵢ − ξ_min × νᵢ).
• CH4: 0.9973 − 0.75 = 0.2473 mol left (3.967 g)
• O2: limiting — fully consumed.Field guide
Limiting reactants, in plain terms.
In any real reaction you almost never combine reactants in perfect stoichiometric ratios. One of them runs out before the others — and once it's gone, the reaction stops. That reactant is the limiting reactant. The others are excess reactants, and a chunk of them will be left over when the reaction is done.
The reaction extent ξ
The cleanest way to find the limiting reactant is to compute a reaction extent for each one. For a reactant i with initial moles nᵢ and stoichiometric coefficient νᵢ:
ξᵢ is how far each reactant alone could drive the reaction. The actual extent the reaction can reach is the smallest of those values — that's the limit, set by whichever reactant runs out first.
Theoretical yield of a product
Once ξ_min is known, every product's theoretical yield falls out of the same number. For a product P with coefficient ν_P and molar mass M_P:
This is the maximum amount of P you could get in an ideal world with no losses, side reactions, or impurities. Real actual yield is usually lower; the ratio (actual ÷ theoretical) × 100% is the percent yield.
Excess reactants left over
For each excess reactant, multiply ξ_min by its coefficient to find how many moles got consumed, then subtract:
Multiply n_remaining by the reactant's molar mass for the leftover mass in grams. The reactant breakdown table shows all of this for every reactant at once.
Worked example
Burn 16 g of methane in 48 g of oxygen:
- n(CH₄) = 16 ÷ 16.04 ≈ 0.998 mol · ξ = 0.998 ÷ 1 = 0.998
- n(O₂) = 48 ÷ 32.00 = 1.500 mol · ξ = 1.500 ÷ 2 = 0.750
O₂ has the smaller extent, so O₂ is limiting. The theoretical yield of CO₂ is 0.750 × 1 = 0.750 mol = 33.0 g, and the leftover CH₄ is (0.998 − 0.750) × 16.04 ≈ 3.97 g. The calculator above does these steps for you and shows the working.
Why mass alone misleads you
It's tempting to assume the reactant with the smallest mass is limiting — but molar masses and coefficients differ, so masses on their own can't be compared. The fair comparison is moles divided by coefficient, which puts every reactant on the same scale of “how much reaction can I support?”.
About the data: Atomic weights used by this tool follow the IUPAC 2021 standard atomic weights table, cross-checked against the NIST reference data for individual elements. Results are intended as a study aid for stoichiometry homework and lab planning — not a substitute for measured experimental values or professional lab-grade analysis.