Limiting Reactant Calculator — Theoretical Yield & Excess

Quick Answer

To find the limiting reactant, convert each reactant to moles (grams ÷ molar mass), divide each by its coefficient in the balanced equation, and the smallest result is the limiting reactant — it runs out first and sets the theoretical yield. For 3 g H₂ and 8 g O₂ in 2H₂ + O₂ → 2H₂O, oxygen is limiting even though there are more grams of it, giving a theoretical yield of about 9.01 g of water.

Limiting reactant method: convert each reactant from grams to moles, divide by its balanced coefficient, and the smallest moles-over-coefficient ratio is the limiting reactant that sets the theoretical yield — The whole method in one picture: convert each reactant to **moles**, divide by its **coefficient**, and the **smallest ratio** is the limiting reactant — it runs out first and caps the theoretical yield. This **limiting reactant calculator** does every step and shows the work.

The Limiting Reactant Formula (Moles ÷ Coefficient)

The limiting reactant is the one that runs out first, so it decides how much product you can make. The reliable way to find it — the method this limiting reactant calculator uses — is the moles ÷ coefficient test on a balanced equation:

ratio = moles ÷ coefficient → smallest ratio = limiting reactant
where moles = grams ÷ molar mass. The limiting reactant’s ratio (its reaction extent) then sets the theoretical yield: product moles = ratio × product coefficient.

Do not just compare grams — or even raw moles. A reactant can be present in the most grams and still run out first, because the coefficients set how fast each is consumed. Dividing moles by the coefficient is what makes the comparison fair.

How to Find the Limiting Reactant (Step by Step)

Whether you have two reactants or five, the same four steps work every time:

Balance the equation
Get whole-number coefficients so atoms are conserved on both sides — e.g. 2 H₂ + O₂ → 2 H₂O. The coefficients are the numbers you divide by, so an unbalanced equation gives the wrong answer.
Convert every reactant to moles
If you are given grams, divide by the molar mass: moles = grams ÷ molar mass. If you are already given moles, skip this step. (For solutions, moles = molarity × litres.)
Divide each by its coefficient
This “normalises” the reactants so you compare like with like. The result is the reaction extent each reactant could support on its own.
Pick the smallest — that is the limiting reactant
The reactant with the smallest moles ÷ coefficient ratio is used up first and limits the product. Everything else is in excess.

Reactant	Given	Moles	÷ coefficient	Result
H₂ (coef 2)	3 g	3 ÷ 2.016 = 1.488	1.488 ÷ 2 = 0.744	excess
O₂ (coef 1)	8 g	8 ÷ 31.998 = 0.250	0.250 ÷ 1 = 0.250	limiting

O₂ wins the “smallest ratio” test even though there are more grams of it — this is exactly the trap the calculator saves you from.

How to Use the Limiting Reactant Calculator

Enter the coefficients and formulas
From your balanced equation, type each reactant’s coefficient in the Coef. box and its formula (like H2, O2, Fe2O3) in the formula box. The calculator shows the molar mass it read (e.g. M = 18.015 g/mol) so you can check it. Capital letters matter: CO is carbon monoxide, Co is cobalt. For the particle counts behind a formula, the atom calculator maps atomic number, mass number and charge to protons, neutrons and electrons.
Type each amount and choose g or mol
Enter how much of each reactant you have and pick grams or moles from the little unit picker. You can mix — grams for one, moles for the other.
Add the product to get theoretical yield
Type the product’s coefficient and formula (like 2 and H2O). The result panel names the limiting reactant, the theoretical yield in grams and moles, and how much excess reactant is left over.
Optional: enter your actual yield for percent yield
If you ran the reaction and weighed the product, type that mass in the Actual yield box to get percent yield = actual ÷ theoretical × 100.

Worked Examples

Each row is a balanced reaction you can reproduce in the calculator above — the limiting reactant, theoretical yield and leftover excess all come straight from the moles ÷ coefficient method:

Reaction	You have	Limiting	Theoretical yield	Excess left
2 H₂ + O₂ → 2 H₂O	3 g H₂ + 8 g O₂	O₂	9.01 g H₂O	1.99 g H₂
N₂ + 3 H₂ → 2 NH₃	14 g N₂ + 5 g H₂	N₂	17.0 g NH₃	1.98 g H₂
2 Na + Cl₂ → 2 NaCl	10 g Na + 20 g Cl₂	Na	25.4 g NaCl	4.58 g Cl₂
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe	5 mol Al + 2 mol Fe₂O₃	Fe₂O₃	204 g Al₂O₃	27.0 g Al

Notice the pattern in the first row: hydrogen is the smaller mass (3 g vs 8 g) but oxygen is limiting. Grams alone never tell you the answer — only moles ÷ coefficient does.

Excess Reactant, Theoretical Yield & Percent Yield

Once you know the limiting reactant, three more numbers fall out of the same calculation:

Quantity	How it’s found
Theoretical yield	limiting-reactant extent × product coefficient × product molar mass — the most product the reaction can make
Excess reactant left	starting moles of the excess reactant − (extent × its coefficient), converted back to grams
Percent yield	actual yield ÷ theoretical yield × 100 — how efficient the real reaction was

A percent yield above 100% is a red flag, not a bonus — it usually means the product is still wet or impure. Dry it fully and re-weigh before trusting the number.

Common Mistakes (and the Grams Trap)

The biggest error is comparing grams instead of moles ÷ coefficient. A heavy reactant can still be the one that runs out. Two more to avoid: forgetting to balance the equation first (the coefficients are half the calculation), and dividing moles by the wrong coefficient or skipping the division entirely.

Frequently Asked Questions

Balance the equation, convert each reactant to moles (grams ÷ molar mass), then divide each mole value by that reactant’s coefficient. The smallest result is the limiting reactant — it runs out first and sets the maximum product. This calculator does all three steps and shows the division for each reactant.

Convert the grams to moles first: moles = grams ÷ molar mass. Then divide by the coefficient and compare. You cannot compare grams directly — for 3 g H₂ and 8 g O₂ in 2H₂ + O₂ → 2H₂O, oxygen is limiting even though it is the heavier amount, because 0.250 ÷ 1 is smaller than 1.488 ÷ 2.

The limiting reactant is completely used up and stops the reaction, capping how much product forms. The excess reactant is whatever is left over after the limiting one runs out. The calculator reports both — it names the limiting reactant and tells you how many grams of the excess reactant remain.

Take the limiting reactant’s moles ÷ coefficient value (its reaction extent), multiply by the product’s coefficient to get product moles, then multiply by the product’s molar mass for grams. Theoretical yield is the maximum a perfect reaction could produce; the actual yield in the lab is almost always lower.

Yes — that is the whole point of using moles ÷ coefficient. A reactant with a high molar mass has fewer moles per gram, and a large coefficient means it is consumed faster. Either can make a heavier reactant run out first, which is why grams alone never decide the limiting reactant.

The same moles ÷ coefficient rule extends to any number of reactants — the smallest ratio is always limiting. This calculator is built for the two-reactant case that covers most homework and lab problems; for three or more, compute each ratio and pick the smallest.