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
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.
- 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.
- Enter the formula of the first reactant using Hill/IUPAC notation. Capitalisation is meaningful: Co is cobalt, CO is carbon monoxide.
- Enter its balanced-equation coefficient and the mass in grams you actually weighed out.
- Do the same for the second reactant. Gases go in as grams, so you do not need a temperature or pressure.
- Enter the product you want a yield for, with its own balanced-equation coefficient. If several products form, pick the one you are isolating.
- 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.
- 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.
- Take the theoretical yield into the percent yield calculator together with the mass you actually isolated, to get your percent yield.
- 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.
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.
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.
Frequently asked questions.
How do I find the limiting reagent?
Is the limiting reagent just the one with the smallest mass, or the fewest moles?
What is the extent of reaction, and why does this page show it?
What happens if both reactants run out at exactly the same time?
Why would anyone deliberately use an excess reactant?
Does this calculator check that my equation is balanced?
Where do the molar masses come from?
Can I use this for reactions with three or more reactants?
The theoretical yield looks too high. What did I get wrong?
References& sources.
- [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]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]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]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]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]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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