Audited ·Last updated 29 Jul 2026·6 citations·Tier 1·0 uses

Limiting Reagent Calculator

Find which of two reactants runs out first. Enter formulas, coefficients and masses to get the limiting reagent, theoretical yield and leftover excess reagent.

Limiting Reagent Calculator

Hill/IUPAC notation, case sensitive: Co is cobalt, CO is carbon monoxide. Parentheses work, e.g. Al2(SO4)3. No spaces, charges, state labels or hydrate dots — write CuSO4·5H2O as CuSO4(H2O)5.
The number in front of reactant A in the BALANCED equation. For 3 Si + 2 N2 -> Si3N4 this is 3. The calculator cannot check that your equation is balanced; if it is not, every result below is wrong.
Grams of reactant A actually put into the flask.
g
The competing reactant. Gases are entered as a mass in grams, so no temperature or pressure basis is needed.
The balanced-equation coefficient on reactant B. Non-integer coefficients are allowed for half-equations such as C2H6 + 3.5 O2.
Grams of reactant B actually put into the flask.
g
The product whose theoretical yield you want. If the reaction makes several products, enter the one you care about and use its own coefficient below.
The balanced-equation coefficient on that product. For 2 H2 + O2 -> 2 H2O with H2O as the product, this is 2.
Limiting reagent
Si is limiting — it runs out first. N2 is in excess.
Which reactant is consumed first, decided by comparing moles divided by coefficient at full precision. An exact stoichiometric mixture is reported as a tie rather than assigned arbitrarily.
Moles of A charged
0.0712 mol
Moles of B charged
0.0535 mol
Mole ratio you charged (A : B)
1.33
Mole ratio the equation needs (A : B)
1.5
Extent of reaction (ξ)
0.0237 mol
Theoretical yield of product
3.33 g
Excess reagent left over
0.0061 mol
Excess reagent left over (mass)
0.17 g
Molar mass of product
140.283 g/mol

Background.

When two reactants are mixed, one of them almost always runs out first. That reactant is the limiting reagent, and it alone sets how much product the reaction can make; everything you charged beyond it is excess and will still be sitting in the flask at the end. This calculator takes two reactants — formula, balanced-equation coefficient, and the mass you actually weighed out — plus the product you care about, and returns which reactant is limiting, how far the reaction can run, the theoretical yield of product, and exactly how much of the excess reagent is left over.

The rule people memorise is 'divide moles by coefficient and take the smaller'. That is correct, and it is also a slightly disguised version of a quantity IUPAC actually defines: the extent of reaction, symbol ξ, measured in moles. For a balanced equation aA + bB → pP, running the reaction as far as the charge allows gives ξ = min(nA/a, nB/b). Whichever reactant achieves that minimum is limiting. Every other amount then follows from a single multiplication: the product formed is ξ × p moles, the amount of A consumed is ξ × a moles, and the leftover of the excess reagent is simply what you charged minus what was consumed. This page reports ξ explicitly because once you have it, the rest of the stoichiometry is arithmetic rather than reasoning.

UNITS AND CONVENTIONS, stated up front rather than hidden in an FAQ. Masses go in as grams, molar masses come out in g/mol, amounts and ξ are in moles. There is no temperature or pressure basis anywhere on this page — a mole balance is valid at any T and p, and gases are entered as masses precisely so that no STP-versus-SATP question arises. Molar masses are computed from the CIAAW Abridged Standard Atomic Weights 2024 table, which is the IUPAC Atomic Weights 2021 recommendation with the 2024 revisions to gadolinium, lutetium and zirconium. Elements with no IUPAC standard atomic weight — technetium, promethium, polonium, astatine, radon, francium, radium, actinium and everything heavier than uranium — are not supported, because a natural-abundance molar mass does not exist for them.

WHAT THE TOOL CANNOT CHECK, and this is the important limit. It cannot tell whether the equation you typed is balanced. It takes your coefficients on trust, and if they are wrong every number below is wrong in a way that looks perfectly plausible. Balance the equation on paper first, then read the coefficients off it. The tool also assumes the reaction proceeds to completion through a single channel and that recovery is complete — real reactions reach equilibrium, run side reactions, and lose material in work-up, which is why the theoretical yield is a ceiling rather than a prediction. To convert that ceiling into a real figure, feed it into the percent yield calculator with the mass you actually isolated.

HOW THE DECISION IS MADE, and why that matters at the boundary. The comparison between nA/a and nB/b is performed at arbitrary decimal precision, not on rounded display values. A limiting-reagent verdict that flips because a number was rounded to four decimals before comparison is a real class of bug, and this module avoids it by rounding exactly once, at the end. When the two extents are exactly equal — a genuinely stoichiometric mixture — the result says 'neither', with zero excess remaining, instead of silently assigning the label to whichever reactant floating-point noise happened to favour. That case is tested immediately before, at, and immediately after the stoichiometric point.

READING THE TWO RATIOS. Alongside the verdict the calculator shows the mole ratio you actually charged (nA/nB) next to the ratio the equation requires (a/b). This is the comparison textbooks make and it is worth internalising: if the ratio you charged is SMALLER than the ratio required, you are short of A and A is limiting; if it is LARGER, you have more A than the equation can use and B is limiting. In the worked example below, 2.00 g of silicon and 1.50 g of nitrogen give a charged ratio of 1.330 against a required ratio of 1.500 — short on silicon, so silicon is limiting, and 0.170 g of nitrogen is still there when the silicon is gone.

A practical note on why chemists deliberately use an excess. Charging one reactant in excess is normal and often correct: it drives an equilibrium toward the product, compensates for a reagent that decomposes, or ensures the expensive component is the one fully consumed. The cost is separation — the leftover mass reported here is exactly what your work-up has to remove. Seeing that number in grams before you run the reaction is usually more informative than seeing it as a percentage afterwards.

What is limiting reagent calculator?

The limiting reagent (or limiting reactant) is the reactant that is entirely consumed first and therefore limits the amount of product that can be generated. Every other reactant present is an excess reagent: some of it survives the reaction and has to be separated from the product afterwards. Identifying the limiting reagent is the first step of essentially every quantitative stoichiometry problem, because the theoretical yield is computed from the limiting reagent and from nothing else. The formal statement uses the IUPAC extent of reaction ξ. For a balanced equation aA + bB → pP, the stoichiometric numbers are ν = −a for A, ν = −b for B and ν = +p for P — negative for reactants, positive for products, by IUPAC convention — and any change in amount satisfies Δn(X) = ν(X) × ξ. Because no amount may go negative, the reaction can run only as far as ξmax = min(nA/a, nB/b), and the reactant that sets that minimum is the limiting one. The 'divide by the coefficient' rule taught in first-year chemistry is exactly this, with the units of ξ hidden. Two common misreadings are worth naming. The limiting reagent is NOT simply the reactant present in the smallest mass — silicon at 2.00 g outweighs nitrogen at 1.50 g in the worked example and is still limiting, because the comparison is on moles per coefficient, not grams. And it is NOT necessarily the reactant present in the fewest moles either: coefficients matter, so a reactant present in more moles can still be limiting if the equation demands proportionally more of it.

How to use this calculator.

  1. Balance your chemical equation on paper first. Everything on this page depends on the coefficients being right, and no calculator can verify them for you.
  2. Enter the formula of the first reactant using Hill/IUPAC notation. Capitalisation is meaningful: Co is cobalt, CO is carbon monoxide.
  3. Enter its balanced-equation coefficient and the mass in grams you actually weighed out.
  4. Do the same for the second reactant. Gases go in as grams, so you do not need a temperature or pressure.
  5. Enter the product you want a yield for, with its own balanced-equation coefficient. If several products form, pick the one you are isolating.
  6. Read the verdict at the top. It names the limiting reagent and the excess one in plain language, or reports a tie for an exactly stoichiometric mixture.
  7. Compare the two mole ratios in the breakdown: if the ratio you charged is below the ratio the equation needs, reactant A is limiting; if above, reactant B is.
  8. Take the theoretical yield into the percent yield calculator together with the mass you actually isolated, to get your percent yield.
  9. Check the leftover mass of the excess reagent before you run the reaction — that is the material your work-up will have to remove.

The formula.

ξ = min(nA / a , nB / b) → m(P) = ξ × p × M(P)

Start from the balanced equation aA + bB → pP.

STEP 1 — moles charged. nA = mA / M(A) and nB = mB / M(B), where the molar masses M are summed from the CIAAW Abridged Standard Atomic Weights 2024 table over the parsed formula. Units: grams divided by grams per mole gives moles, so nA and nB are in mol.

STEP 2 — extent of reaction. IUPAC defines the extent of reaction ξ by Δn(X) = ν(X) × ξ, where the stoichiometric number ν is negative for reactants and positive for products. Since no amount can become negative, reactant A alone would permit ξ up to nA/a, and reactant B alone would permit ξ up to nB/b. The reaction therefore runs to ξmax = min(nA/a, nB/b), and the reactant achieving that minimum is the limiting reagent. ξ has units of mol.

STEP 3 — theoretical yield. n(P) = ξ × p, and m(P) = n(P) × M(P) in grams. Note that p, the coefficient on the product, is easy to forget: for 2 H2 + O2 → 2 H2O, one extent of reaction produces TWO moles of water, not one.

STEP 4 — leftover excess. For the non-limiting reagent X with coefficient νX, the amount consumed is ξ × νX and the amount remaining is nX − ξ × νX. This is exactly zero for a stoichiometric mixture and strictly positive otherwise.

ROUNDING STAGE. There is no intermediate rounding anywhere. Molar masses, moles, both candidate extents and the comparison between them are all carried at arbitrary decimal precision; each returned value is rounded exactly once, at the return boundary, to ten decimal places. This is a correctness requirement, not a nicety: the limiting-reagent verdict is a comparison, and comparing pre-rounded numbers can flip the verdict for mixtures close to stoichiometric. The exactly-stoichiometric case is detected as an exact equality and reported as a tie.

SIGNIFICANT FIGURES. Report the theoretical yield to the significant figures of your least precise input, which is normally the balance reading. In the worked example, 2.00 g and 1.50 g are three-figure measurements, so 3.33 g is the honest theoretical yield and the ten decimal places on screen are there to show you where rounding begins, not to be transcribed.

INVALID DOMAIN. Zero and negative masses are refused — a reaction with none of a reactant does not run, and a negative mass is not a physical charge. Zero and negative coefficients are refused, because a balanced equation never has one. Unknown element symbols, unbalanced parentheses and elements with no IUPAC standard atomic weight (Tc, Pm, Po, At, Rn, Fr, Ra, Ac and heavier) raise a field error naming the problem rather than returning a silently wrong molar mass. Non-integer coefficients ARE accepted, because half-equations such as C2H6 + 3.5 O2 → 2 CO2 + 3 H2O are legitimate.

A worked example.

Example

Worked example — silicon nitride from the elements: 3 Si(s) + 2 N2(g) → Si3N4(s), starting from 2.00 g of silicon and 1.50 g of nitrogen gas. Molar masses from the CIAAW abridged 2024 table: Si = 28.085 g/mol, N2 = 2 × 14.007 = 28.014 g/mol, Si3N4 = 3 × 28.085 + 4 × 14.007 = 84.255 + 56.028 = 140.283 g/mol. Moles charged: n(Si) = 2.00 / 28.085 = 0.071212 mol, n(N2) = 1.50 / 28.014 = 0.053545 mol. Note that silicon is the heavier charge in grams but that tells you nothing — the comparison is per coefficient. The ratio actually charged is 0.071212 / 0.053545 = 1.330, while the equation requires 3 / 2 = 1.500. The charged ratio is below the required ratio, so silicon is short and SILICON IS LIMITING; nitrogen is in excess. Extent of reaction: ξ = min(0.071212/3, 0.053545/2) = min(0.023737, 0.026772) = 0.023737 mol, set by silicon. Theoretical yield: n(Si3N4) = ξ × 1 = 0.023737 mol, so m = 0.023737 × 140.283 = 3.3300 g, reported as 3.33 g at the three significant figures the balance justifies. Leftover nitrogen: consumed = ξ × 2 = 0.047475 mol, so 0.053545 − 0.047475 = 0.006070 mol remain, which is 0.006070 × 28.014 = 0.1700 g of N2 still in the vessel when the last of the silicon has reacted. That 0.170 g is what a work-up would need to vent or separate. If you then isolate, say, 2.85 g of silicon nitride, feeding 2.85 g and 3.33 g into the percent yield calculator gives 85.6 percent. As a boundary check, charging 0.84255 g of Si with 0.56028 g of N2 — exactly 0.03 mol and 0.02 mol — gives a charged ratio of exactly 1.500, a verdict of 'neither', ξ = 0.0100 mol, a theoretical yield of 1.40283 g, and exactly zero leftover.

mass A2
product FormulaSi3N4
mass B1.5
coefficient B2
coefficient A3
reactant A FormulaSi
reactant B FormulaN2
coefficient Product1

Frequently asked questions.

How do I find the limiting reagent?
Convert each reactant's mass to moles by dividing by its molar mass, then divide each result by that reactant's coefficient in the BALANCED equation. The smallest quotient identifies the limiting reagent, and that quotient is the extent of reaction ξ in moles. In the worked example, 2.00 g of Si is 0.071212 mol, divided by its coefficient 3 gives 0.023737; 1.50 g of N2 is 0.053545 mol, divided by 2 gives 0.026772. Silicon gives the smaller number, so silicon is limiting. An equivalent shortcut is to compare the mole ratio you actually charged (0.071212/0.053545 = 1.330) with the ratio the equation requires (3/2 = 1.500): charged below required means the first reactant is limiting.
Is the limiting reagent just the one with the smallest mass, or the fewest moles?
Neither, and both shortcuts fail regularly. Mass fails because molar masses differ: in the worked example silicon is the larger charge at 2.00 g against 1.50 g of nitrogen, and silicon is still limiting. Moles alone fails because coefficients differ: silicon is present in MORE moles than nitrogen (0.0712 against 0.0535) and is still limiting, because the equation demands three silicons for every two nitrogens. The only reliable comparison is moles divided by the balanced-equation coefficient. That quantity — the extent of reaction each reactant would permit on its own — is what this calculator compares, and it is reported as ξ in the breakdown.
What is the extent of reaction, and why does this page show it?
The extent of reaction ξ is the IUPAC quantity that measures how far a reaction has progressed, in moles. It is defined by Δn(X) = ν(X) × ξ, where the stoichiometric number ν is negative for reactants and positive for products. Its usefulness is that a single number determines every amount in the system: consume ξ × a moles of A, consume ξ × b moles of B, produce ξ × p moles of P. Once you know ξ you never have to reason about ratios again, only multiply. The limiting-reagent problem in this language is simply 'find the largest ξ for which no amount goes negative', which is min(nA/a, nB/b). This page reports ξ so you can compute any other species in the equation yourself with one multiplication.
What happens if both reactants run out at exactly the same time?
That is a stoichiometric mixture, and the calculator reports 'neither' rather than picking one. Both reactants are consumed completely, the leftover is exactly zero, and the theoretical yield is the same whichever reactant you compute it from. Charging 0.84255 g of Si with 0.56028 g of N2 — exactly 0.03 mol and 0.02 mol, a ratio of exactly 3:2 — produces this case, with ξ = 0.0100 mol and a yield of 1.40283 g. The distinction matters more than it seems: the comparison here is made at full arbitrary precision, so a mixture that really is stoichiometric is recognised as such instead of being assigned to whichever reactant a rounding error happened to favour.
Why would anyone deliberately use an excess reactant?
Several good reasons. An excess of one reactant pushes a reversible reaction toward the products by Le Chatelier's principle, raising the yield of the expensive component. It compensates for a reagent that decomposes or evaporates during the run. It ensures the costly or hard-to-source reactant is the one fully consumed rather than left in the waste stream. And in gas-phase or combustion work, excess oxidant guarantees complete combustion. The cost is separation: whatever you charge in excess must come back out during work-up, which is why this page reports the leftover as a mass in grams. Green-chemistry metrics penalise that excess explicitly — atom economy assumes exact stoichiometric quantities, so any excess makes the real material efficiency worse than the atom economy suggests.
Does this calculator check that my equation is balanced?
No, and that is the most important limitation on the page. It takes your coefficients entirely on trust. If you enter 1, 1 and 1 for a reaction that actually needs 3, 2 and 1, the calculator will return a confidently wrong limiting reagent, a wrong theoretical yield and a wrong leftover, with nothing to indicate the problem. Balance the equation on paper first and read the coefficients off it. A quick sanity check you can do yourself: the total mass entering should equal the total mass leaving. Charge 2.0 g of H2 with 32.0 g of O2 for 2 H2 + O2 → 2 H2O and the tool reports H2 limiting, ξ = 0.496032 mol, 17.8720 g of water, and 16.1280 g of leftover oxygen — 2.0 + 32.0 − 16.1280 = 17.8720, so mass closes exactly. (Note it is 17.87 g, not the 18.0 g a mental estimate suggests: 2.0 g of H2 is 0.99206 mol, slightly under a mole, because M(H2) is 2.016 rather than 2.)
Where do the molar masses come from?
From the CIAAW Abridged Standard Atomic Weights 2024 table, which is the IUPAC Commission on Isotopic Abundances and Atomic Weights recommendation published as Atomic Weights 2021 (Pure Appl. Chem. 93(5), 573) with the 2024 revisions to gadolinium, lutetium and zirconium folded in. These are the conventional single values, not the intervals CIAAW publishes for the fourteen elements whose isotopic composition varies measurably by source. Named revision matters here: lithium is 6.94 in the current table, not the 6.941 that pre-2009 textbooks print; ytterbium is 173.05, not 173.04; zirconium was revised to 91.222 in 2024. Elements with no standard atomic weight at all — technetium, promethium, polonium, astatine, radon, francium, radium, actinium and the transuranics — are rejected with an explicit error rather than given a guessed value.
Can I use this for reactions with three or more reactants?
Not directly — this page compares exactly two. For three or more, run the comparison pairwise: find the limiting one of the first two, then compare that winner against the third, and so on. The extent of reaction makes this rigorous, because ξ for the whole system is simply the minimum of nX/νX over ALL reactants, and the minimum of a set can be found by repeated pairwise comparison. In practice, most preparative reactions have one reactant that is obviously in large excess (a solvent-scale reagent, or an oxidant bubbled through) and the real competition is between two, which is the case this page is built for.
The theoretical yield looks too high. What did I get wrong?
Check three things in this order. First, the product coefficient: for 2 H2 + O2 → 2 H2O one extent of reaction makes TWO moles of water, and entering 1 halves the answer. Second, the product formula and its molar mass, which is shown in the breakdown precisely so you can verify it against a periodic table. Third, whether the equation is balanced at all. Beyond those, remember what the number means: the theoretical yield assumes complete conversion by a single reaction channel and complete recovery of the product. Real runs stop at equilibrium, lose material to side reactions, and leave product in the mother liquor and on the filter. That is why the theoretical yield is a ceiling; the percent yield calculator converts it into the figure you actually report.

References& sources.

  1. [1]Flowers, P., Theopold, K., Langley, R. & Robinson, W. R. (2019). Chemistry 2e, Section 4.4 'Reaction Yields', Example 4.12. OpenStax, Rice University. Peer-reviewed, CC BY 4.0. PRIMARY SOURCE for this page: defines the limiting reactant as the substance 'entirely consumed' and thus 'limiting the amount of product that may be generated', and works the exact problem reproduced here — 3 Si(s) + 2 N2(g) -> Si3N4(s) from 2.00 g Si and 1.50 g N2, with a provided Si:N2 mole ratio of 1.33 against a required 1.5, giving Si as limiting. Retrieved 2026-07-29. Open access.
  2. [2]IUPAC. Compendium of Chemical Terminology (the Gold Book), entries 'stoichiometric number, ν' (S06025) and 'extent of reaction, ξ' (E02283). SECOND, INDEPENDENT AUTHORITY consulted for this page. S06025 is sourced to Pure Appl. Chem. 1996, 68, 149 ('A glossary of terms used in chemical kinetics, including reaction dynamics', IUPAC Recommendations 1996) at p. 187, and to the Green Book 2nd ed. p. 42: stoichiometric numbers are positive for products and negative for reactants, and dξ = dn_B / ν_B. This agrees with the textbook 'divide moles by coefficient' rule and supplies the sign convention and the ξ = min(nA/a, nB/b) formulation used by this calculator. Retrieved 2026-07-29. Open access.
  3. [3]IUPAC (2007). Quantities, Units and Symbols in Physical Chemistry ('the Green Book'), 3rd edition, section 2.12 'Chemical reactions' — the formal definitions of stoichiometric number, extent of reaction and amount of substance on which this page's ξ formulation rests. International Union of Pure and Applied Chemistry / RSC Publishing, 2007 (2nd printing 2008). Retrieved 2026-07-29. Bibliographic reference: a print and PDF edition rather than a live web page.
  4. [4]Meija, J. et al. (2021). 'Atomic weights of the elements 2021 (IUPAC Technical Report)'. Pure and Applied Chemistry 93(5), 573–600, doi:10.1515/pac-2019-0603, as maintained in the CIAAW table 'Abridged Standard Atomic Weights 2024' (revision: 2024, incorporating the Gd, Lu and Zr revisions on the Atomic Weights 2021 base). This is the named, versioned source of every molar mass this calculator computes: Si 28.085, N 14.007, Cu 63.546, S 32.06, O 15.999, Zn 65.38. Retrieved 2026-07-29. Open access.
  5. [5]Constable, D. J. C., Curzons, A. D. & Cunningham, V. L. (2002). 'Metrics to "green" chemistry — which are the best?' Green Chemistry 4(6), 521–527, doi:10.1039/B206169B. Royal Society of Chemistry, peer-reviewed. Table 1 tabulates the molar excess of the second reactant actually used across twenty-eight industrial reaction classes (from 120% for an N-alkylation to 2650% for an N-dealkylation), which is the evidence behind this page's statement that deliberate excess is normal industrial practice and that the leftover mass is a real separation cost. Retrieved 2026-07-29. Publisher page paywalled; an open copy was consulted.
  6. [6]Sheldon, R. A. (2018). 'Metrics of Green Chemistry and Sustainability: Past, Present, and Future'. ACS Sustainable Chemistry & Engineering 6(1), 32–48, doi:10.1021/acssuschemeng.7b03505. Peer-reviewed. Source for this page's remark that atom economy 'assumes the use of exact stoichiometric quantities of starting materials', which is why an excess reagent degrades real material efficiency below the atom-economy figure. Retrieved 2026-07-29. Publisher version paywalled; the author's TU Delft repository copy is open.

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