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Limiting reactant & theoretical yield

Stoichiometry & Limiting Reactant Calculator

Enter the coefficients and available moles of two reactants (aA + bB → cC) to find the limiting reagent, moles of product formed, and theoretical yield mass.

Reaction — aA + bB → cC
mol
mol
g/mol
Limiting Reactant
Reactant B
Theoretical yield: 72.08 g of product C
Product from A
5 mol
if A reacted completely
Product from B
4 mol
if B reacted completely
Excess reactant remaining
1 mol
unreacted A left over

About the Stoichiometry & Limiting Reactant Calculator

Stoichiometry is the arithmetic of balanced chemical equations: once a reaction is balanced, the coefficients act as a mole ratio that connects any reactant to any product. This calculator applies that ratio to two reactants at once to find which one runs out first — the limiting reactant — and how much product that limit actually allows.

The mole ratio from a balanced equation

For a balanced reaction aA + bB → cC, the coefficients a, b, and c give the exact mole ratio the reaction consumes and produces in. If you know how many moles of A you have, the moles of product C it could produce (if A were the only constraint) is (moles A / a) × c — and the same logic applies to B.

Finding the limiting reactant

Compute how much product each reactant could form on its own, assuming it reacts completely. Whichever reactant produces the smaller amount of product is the limiting reactant — it runs out first and stops the reaction, no matter how much of the other reactant remains. The larger of the two is the excess reactant.

Theoretical yield

Once you know the limiting reactant, the theoretical yield in moles is simply whichever "product from X" value is smaller. Converting to mass just requires multiplying by the product's molar mass: theoretical yield (g) = limiting moles of product × molar mass (g/mol). This is the maximum mass of product the reaction can form — real (actual) yield is always equal to or less than this, and percent yield = actual/theoretical × 100%.

Worked example

Consider 2H₂ + O₂ → 2H₂O (a = 2, b = 1, c = 2), with 5 mol H₂ and 2 mol O₂ available, and water's molar mass 18.02 g/mol. Product from H₂: (5/2)×2 = 5 mol. Product from O₂: (2/1)×2 = 4 mol. O₂ gives the smaller value, so O₂ is the limiting reactant, and the theoretical yield is 4 mol × 18.02 g/mol = 72.08 g of water. H₂ is in excess — checking how much H₂ the limiting O₂ actually consumes, (2/1)×2 = 4 mol H₂ used, leaving 1 mol H₂ unreacted.

Frequently asked questions

What if both reactants give the exact same amount of product?

That means the reactants are present in exactly the stoichiometric ratio the equation calls for — neither is in excess, both are fully consumed, and the calculator will show zero excess for both. This is the ideal case for maximizing atom economy and minimizing waste reactant.

Why does the limiting reactant matter in practice?

In industrial and lab settings, reactants often have very different costs, so a process is often deliberately run with one reactant in excess (usually the cheaper one) to ensure the more expensive or reaction-limiting one is fully consumed, maximizing its conversion to product.

How is percent yield different from theoretical yield?

Theoretical yield is the maximum mass of product predicted purely from stoichiometry, assuming the reaction goes to completion with no losses. Actual yield is what you measure in a real experiment or process, which is almost always lower due to side reactions, incomplete conversion, or product lost during recovery. Percent yield = (actual yield / theoretical yield) × 100%.

Can this calculator handle more than two reactants?

This tool checks two reactants directly. For reactions with three or more reactants, apply the same logic pairwise — compute the product each reactant could form alone, then take the overall minimum across all of them; that reactant is the true limiting one.

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