What is a limiting reagent?

The limiting reagent (also called limiting reactant) is the reactant that runs out first and therefore caps how much product a reaction can produce. Every other reactant is 'in excess' — some of it will still be sitting in the flask when the reaction stops. Identifying the limiting reagent is the first step of any theoretical-yield calculation: the balanced equation tells you the mole ratios, but only the limiting reagent tells you how far the reaction actually gets to go.

How do you find the limiting reagent from grams?

Convert each reactant's mass to moles by dividing by its molecular weight. Then divide each mole count by that reactant's coefficient in the balanced equation. The reactant with the smallest quotient is the limiting reagent. This 'moles-over-coefficient' comparison works because it normalizes for the fact that the coefficients set how fast each reactant is consumed — a reactant with a coefficient of 3 is used up three times as fast as one with a coefficient of 1.

Why divide moles by the coefficient instead of comparing moles directly?

Because the balanced equation's coefficients set the consumption ratio. In N₂ + 3 H₂ → 2 NH₃, one mole of H₂ is consumed for every one-third mole of N₂, so 3 mol of H₂ is stoichiometrically equivalent to 1 mol of N₂, not more. Dividing moles by the coefficient converts each reactant to the same currency — 'moles of reaction event' — and then the smallest number wins.

Limiting reagent vs theoretical yield vs percent yield — how do they connect?

They're three stages of the same calculation. The limiting reagent tells you which reactant runs out. The theoretical yield is how much product the limiting reagent could make if the reaction went to 100 % completion (this calculator computes that). The percent yield is what fraction of that theoretical yield you actually isolated in the lab — for that step use the linked percent-yield calculator.

Limiting Reagent Calculator — Theoretical Yield

The limiting reagent (or limiting reactant) is whichever reactant runs out first in a chemical reaction. It caps the amount of product you can produce, no matter how much of the other reactants you add. Every stoichiometry problem that gives you amounts of two or more reactants starts with the same question: which one is limiting? This calculator answers that and then reports the theoretical yield of the target product — the maximum mass and moles you could isolate if the reaction went to completion — plus how much of the excess reagent is left over.

Limiting Reagent Calculator

Enter the balanced coefficients (a A + b B → c C) and the amount of each reactant. Sig figs default to the lowest-precision input.

Formulas are rendered in Show the work above after a successful compute.

The mole-ratio method

For a balanced reaction $a\,A + b\,B \to c\,C + \ldots$ , the recipe is three lines of arithmetic:

Convert the amount of each reactant to moles (divide grams by molecular weight).
Divide each mole count by that reactant’s coefficient: $r_A = n_A / a$ , $r_B = n_B / b$ .
The smaller ratio is the limiting reagent; call that ratio $\xi$ , the extent of reaction. Theoretical product moles are $n_C = c \cdot \xi$ ; theoretical product mass is $m_C = n_C \cdot MW_C$ .

The critical step is step 2. A common mistake is to compare raw mole counts and pick the reactant with fewer moles — that’s only correct when the coefficients are equal. In N₂ + 3 H₂ → 2 NH₃, 5 mol of H₂ is chemically less than 5 mol of N₂ because H₂ is consumed three times as fast. Dividing by the coefficient normalizes for that.

Worked example — Haber process (Zumdahl §3.9)

You mix 25.0 g of N₂ with 5.00 g of H₂ and run them over a catalyst (excess of neither). Which reactant is limiting, and how much NH₃ (theoretical) can you make?

Worked example — 25.0 g N₂ + 5.00 g H₂ → NH₃

Balanced equation: N₂ + 3 H₂ → 2 NH₃ (a=1, b=3, c=2)
m(N₂), MW(N₂): 25.0 g / 28.014 g·mol⁻¹ (PubChem CID 947)
m(H₂), MW(H₂): 5.00 g / 2.016 g·mol⁻¹ (PubChem CID 783)
MW(NH₃): 17.031 g·mol⁻¹ (PubChem CID 222)

Formula
$ξ = min (n_{A} / a, n_{B} / b), m_{C} = c ξ \cdot M W_{C}$
Substitute
$ξ = min (25.0/28.014, (5.00/2.016) /3) = min (0.8924, 0.8267) = 0.8267$
Note
H₂ is limiting even though there is five times more N₂ by mass — because H₂ has a coefficient of 3 in the balanced equation, each mole of it is used up three times as fast as a mole of N₂. After the reaction stops, 0.066 mol of N₂ (about 1.84 g) is left unreacted, and 28.2 g of NH₃ is the maximum possible product mass.
Result
$Limiting: H_{2}; m_{NH_{3}} = 2 \cdot 0.8267 \cdot 17.031 = 28.2 g$

Common pitfalls

Comparing moles without dividing by coefficients. Only valid when the coefficients are equal. Always divide first.
Unbalanced equation. Coefficients must come from a balanced equation. A wrong coefficient invisibly changes which reactant is limiting.
Grams vs moles mix-up. If one amount is a mass and the other is a volume of solution, convert both to moles first (mass ÷ MW, or volume × molarity).
Hydrates. If a reagent is a hydrate like CuSO₄·5H₂O, use its full hydrate MW (249.68 g/mol), not the anhydrous MW — otherwise your mole count is off by the water of crystallization.
Assuming the limiting reagent is the one with less mass. In the Haber example above, H₂ (5 g) is limiting even though there is five times more N₂ (25 g) by mass. Coefficients decide, not the balance reading.

Once you have the theoretical yield from this tool, compare it against the mass you actually isolated using the percent yield calculator to get the full lab-report number.

Practice problems

Attempt each on paper, then expand the worked solution to check your arithmetic and sig-figs.

Problem 1 Easy For 2 H₂ + O₂ → 2 H₂O, you mix 4.0 mol H₂ with 1.5 mol O₂. What is the limiting reagent and the theoretical moles of water?

Answer: O₂ is limiting; 3.0 mol H₂O

Solve by moles-over-coefficient

a, b (H₂, O₂): 2, 1
c (H₂O): 2
n(H₂): 4.0 mol
n(O₂): 1.5 mol

Formula
$ξ = min (n_{A} / a, n_{B} / b), n_{C} = c ξ$
Substitute
$ξ = min (4.0/2, 1.5/1) = min (2.0, 1.5) = 1.5$
Note
r(H₂) = 4.0/2 = 2.0 and r(O₂) = 1.5/1 = 1.5. The smaller ratio is O₂, so O₂ is limiting and 3.0 mol of H₂O forms. Comparing raw moles (H₂ has more) is not enough — you must divide by the coefficient to normalize for the fact that H₂ is consumed twice as fast as O₂.
Result
$Limiting: O_{2}; n_{H_{2} O} = 2 \cdot 1.5 = 3.0 mol$

Problem 2 Intermediate For N₂ + 3 H₂ → 2 NH₃, you react 25.0 g N₂ (MW 28.014 g/mol) with 5.00 g H₂ (MW 2.016 g/mol). Which is limiting, and how many grams of NH₃ (MW 17.031 g/mol) can you make?

Answer: H₂ is limiting; 28.2 g NH₃ (theoretical)

Solve from grams via moles-over-coefficient

a, b, c: 1, 3, 2
m(N₂), MW(N₂): 25.0 g / 28.014 g·mol⁻¹
m(H₂), MW(H₂): 5.00 g / 2.016 g·mol⁻¹
MW(NH₃): 17.031 g·mol⁻¹

Formula
$ξ = min (n_{A} / a, n_{B} / b), m_{C} = c ξ \cdot M W_{C}$
Substitute
$ξ = min (0.8924, 2.4802/3) = min (0.8924, 0.8267) = 0.8267$
Note
H₂ has the smaller mole-per-coefficient ratio, so H₂ is limiting. N₂ is in excess by about 0.066 mol (1.84 g). This is the canonical Zumdahl §3.9 problem.
Result
$m_{NH_{3}} = 2 \cdot 0.8267 \cdot 17.031 = 28.16 g$

Problem 3 Hard For CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O, you dissolve 10.0 g CaCO₃ (MW 100.09 g/mol) in 20.0 mL of 3.00 M HCl. Which is limiting, and how many grams of CO₂ (MW 44.01 g/mol) form?

Answer: HCl is limiting; 1.32 g CO₂

Solve when one reactant is a solution

a, b, c (CaCO₃, HCl, CO₂): 1, 2, 1
n(CaCO₃): 10.0 / 100.09 = 0.0999 mol
n(HCl): 0.0200 L × 3.00 M = 0.0600 mol
MW(CO₂): 44.01 g·mol⁻¹

Formula
$ξ = min (n_{A} / a, n_{B} / b), m_{C} = c ξ \cdot M W_{C}$
Substitute
$ξ = min (0.0999, 0.0600/2) = min (0.0999, 0.0300) = 0.0300$
Note
HCl is limiting because its coefficient is 2 — you need twice as many moles of HCl as CaCO₃. Excess CaCO₃ left over: 0.0999 − 0.0300 = 0.0699 mol (7.00 g of solid will remain undissolved).
Result
$m_{CO_{2}} = 1 \cdot 0.0300 \cdot 44.01 = 1.32 g$

Frequently asked questions

What is a limiting reagent?: The limiting reagent (also called limiting reactant) is the reactant that runs out first and therefore caps how much product a reaction can produce. Every other reactant is 'in excess' — some of it will still be sitting in the flask when the reaction stops. Identifying the limiting reagent is the first step of any theoretical-yield calculation: the balanced equation tells you the mole ratios, but only the limiting reagent tells you how far the reaction actually gets to go.
How do you find the limiting reagent from grams?: Convert each reactant's mass to moles by dividing by its molecular weight. Then divide each mole count by that reactant's coefficient in the balanced equation. The reactant with the smallest quotient is the limiting reagent. This 'moles-over-coefficient' comparison works because it normalizes for the fact that the coefficients set how fast each reactant is consumed — a reactant with a coefficient of 3 is used up three times as fast as one with a coefficient of 1.
Why divide moles by the coefficient instead of comparing moles directly?: Because the balanced equation's coefficients set the consumption ratio. In N₂ + 3 H₂ → 2 NH₃, one mole of H₂ is consumed for every one-third mole of N₂, so 3 mol of H₂ is stoichiometrically equivalent to 1 mol of N₂, not more. Dividing moles by the coefficient converts each reactant to the same currency — 'moles of reaction event' — and then the smallest number wins.
Limiting reagent vs theoretical yield vs percent yield — how do they connect?: They're three stages of the same calculation. The limiting reagent tells you which reactant runs out. The theoretical yield is how much product the limiting reagent could make if the reaction went to 100 % completion (this calculator computes that). The percent yield is what fraction of that theoretical yield you actually isolated in the lab — for that step use the linked percent-yield calculator.

Sources

Zumdahl & Zumdahl, Chemistry, 10th ed. (2018), §3.9 — Calculations Involving a Limiting Reactant
Silberberg & Amateis, Chemistry: The Molecular Nature of Matter and Change, 8th ed. (2018), §3.4 — Reaction Yields
PubChem CID 947 — Nitrogen (N₂, MW 28.014 g/mol)
PubChem CID 783 — Hydrogen (H₂, MW 2.016 g/mol)
PubChem CID 222 — Ammonia (NH₃, MW 17.031 g/mol)

Last updated: 2026-09-18.