This limiting reactant calculator settles the question every reaction with two measured reactants asks: which one runs out first? It compares each reactant’s moles-per-coefficient ratio, names the limiting reactant, computes the product formed (moles and grams), and reports how much of the excess reactant survives.

Reactant A

Reactant B

Product P

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Compare Ratios, Not Grams

The beginner's instinct — “whichever reactant has fewer grams is limiting” — fails the moment molar masses and coefficients differ, which is always. The correct comparison is the mole ratio test: divide each reactant's moles by its stoichiometric coefficient; the smaller quotient is the extent of reaction ξ, and its owner is limiting. The defaults run the canonical 2H₂ + O₂ → 2H₂O with 10 g H₂ (4.960 mol, ratio 2.480) against 80 g O₂ (2.500 mol, ratio 2.500): despite oxygen outweighing hydrogen eight to one on the scale, hydrogen is limiting by a whisker — producing 4.960 mol = 89.36 g of water and leaving 0.64 g of O₂ unburned. Grams mislead; ratios decide. That near-tie also teaches the practical convention: industry runs one reactant (the cheap, safe, or easily-separated one) in deliberate excess to drive the limiting (expensive, precious) reactant to complete consumption — excess is a strategy, chosen with exactly this arithmetic.

The calculator's audit line deserves its role: consumed reactant mass must equal total product mass (conservation), so for multi-product reactions you run each product through its own coefficient from the same ξ — ammonia synthesis leftovers, combustion with CO/CO₂ splits, and precipitation yields all reduce to ξ times coefficients, plus bookkeeping. And the theoretical yield it reports is the ceiling the Percent Yield page then grades reality against (actual ÷ theoretical × 100) — the two pages are one workflow.

Worked Examples (Verified)

  • Defaults (2H₂ + O₂ → 2H₂O): ratios 2.480 vs 2.500 → H₂ limiting; water 89.36 g; O₂ left 0.64 g; H₂ left 0.
  • Precipitation classic: AgNO₃ (M 169.87, coef 1) 17.0 g = 0.100 mol vs NaCl (58.44, coef 1) 5.0 g = 0.0856 mol → NaCl limits; AgCl (143.32) formed = 0.0856 mol = 12.26 g, with 2.47 g AgNO₃ left in solution.
  • The deliberate excess: esterification runs the cheap alcohol at 3–5× the acid's ratio to push the valuable acid to completion — read backwards from this calculator's leftover line: the excess you PLAN is the leftover you get.

Limiting Reactant Discipline

  • Balance the equation FIRST: coefficients are inputs here, and unbalanced coefficients manufacture wrong limiting reactants with total confidence.
  • Purity matters: 10 g of 90%-pure reactant is 9 g of reactant — multiply masses by purity before the mole conversion in real lab work.
  • Solutions enter by moles (M × V), gases by PV/RT or molar volume — convert everything to moles before ratios; the mass fields here assume you've done that step for non-weighed reactants.
  • Hydrated salts carry their waters in the molar mass (CuSO₄·5H₂O = 249.68, not 159.61) — the classic silent error in precipitation problems.
  • “Excess” is about the OTHER reactant: the limiting one is (ideally) consumed entirely; leftovers reported here assume complete reaction of the limiting reactant — kinetics and equilibrium may still refuse in practice.

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Frequently Asked Questions

How do you find the limiting reactant?

Convert each reactant to moles, divide by its balanced-equation coefficient, and compare: the smaller ratio limits. Defaults: H₂ 4.960/2 = 2.480 vs O₂ 2.500/1 = 2.500 → hydrogen limits, setting the extent ξ = 2.480. Every product and leftover follows from ξ × coefficients. This calculator runs the test and the full bookkeeping.

Why isn't the reactant with the smallest mass limiting?

Because reactions consume MOLECULES in coefficient ratios, not grams. 10 g of H₂ is 4.96 mol of tiny molecules; 80 g of O₂ is only 2.50 mol of heavy ones — and each O₂ needs two H₂ partners. The defaults were chosen to make the point brutally: the reactant with 8× LESS mass is the limiting one. Mass comparisons work only when molar masses and coefficients happen to match, which is almost never.

How do you calculate the amount of excess reactant left?

Excess consumed = ξ × (its coefficient) moles; leftover moles = initial − consumed; back-convert with its molar mass. Here O₂ consumed = 2.480 × 1 = 2.480 mol (79.36 g), leaving 0.020 mol = 0.64 g. The limiting reactant's leftover is zero by definition (complete-reaction assumption).

What is the extent of reaction (ξ)?

The number of “reaction units” that occur — the moles of the reaction-as-written that proceed. ξ = min(moles/coefficient) across reactants, and every species' change is ξ × its coefficient (products gain, reactants lose). It's the single variable the whole stoichiometry of the mixture hangs on, and the quantity this calculator's ratio test actually computes.

How does limiting reactant relate to theoretical yield?

The limiting reactant DETERMINES the theoretical yield: product moles = ξ × product coefficient. Theoretical yield calculators that take one reactant as given are implicitly assuming it's limiting (or in excess-adjusted purity). Percent yield then compares the actual isolated product against this ceiling — a yield over 100% signals wet product, impurities or a limiting-reactant misidentification upstream.

Can a reaction have two limiting reactants?

Only in the exact-tie sense: if two ratios are precisely equal, both exhaust simultaneously (a perfectly stoichiometric mixture, like our defaults' near-miss made exact). Practically, ties are measure-zero events; labs deliberately break them with a planned excess of one reactant. If your calculation shows a tie, suspect rounded molar masses before celebrating the precision.

What if a reactant is a solution or a gas?

Convert to moles first: solutions via molarity × volume (0.100 M × 0.250 L = 0.0250 mol), gases via PV/RT (or 22.4 L/mol at STP as an approximation) — then enter the equivalent “mass” bookkeeping through moles: this page's fields take masses and molar masses, so multiply your moles by the molar mass to get the grams field, or equivalently enter mass = moles × M. The ratio test itself is unit-agnostic once everything is moles.

Is this limiting reactant calculator free?

Yes — free, instant, no sign-up. Enter both reactants (mass, molar mass, coefficient) and the product line for the limiting reactant, theoretical yield and leftovers.