Limiting Reactant & Theoretical Yield Calculator

Understanding the limiting reactant calculator — concept and principle

In a reaction, the reactants are rarely present in exactly the ratio the equation requires. The one that runs out first is the limiting reactant; it decides how much product can form. This calculator identifies the limiting reactant from the masses and coefficients of two reactants, computes the theoretical yield of the product and, if you enter the mass you actually obtained, the percent yield.

Count how many recipe sets each reactant can make; O₂ allows only 2.00, so it sets the 72.06 g yield.

How it is calculated

Moles per coefficient: n_A = m_A ÷ M_A ÷ a and n_B = m_B ÷ M_B ÷ b, with masses in g, molar masses in g/mol and a, b the coefficients in the balanced equation.
The smaller of n_A and n_B marks the limiting reactant.
Theoretical yield = min(n_A, n_B) × p × M_p, where p is the product coefficient and M_p its molar mass.
Percent yield = actual yield ÷ theoretical yield × 100 (shown only when actual yield > 0).

The principle

Dividing moles by the coefficient tells you how many "reaction units" each reactant could supply. In 2H₂ + O₂ → 2H₂O, one reaction unit uses 2 mol H₂ and 1 mol O₂ and makes 2 mol H₂O. Whichever reactant can supply fewer units stops the reaction, and each unit makes p moles of product. This is the law of conservation of mass expressed in moles.

Worked example

Defaults: reaction 2H₂ + O₂ → 2H₂O; 10 g H₂ (2.016 g/mol), 64 g O₂ (31.998 g/mol), product 18.015 g/mol.

H₂: 10 ÷ 2.016 = 4.96032 mol; ÷ 2 = 2.48016 units.
O₂: 64 ÷ 31.998 = 2.000125 mol; ÷ 1 = 2.000125 units.
O₂ gives fewer units, so the limiting reactant is B (O₂).
Theoretical yield: 2.000125 × 2 × 18.015 = 72.06450403 g of water.
Actual yield is 0, so percent yield shows —.

How to read the result

All 64 g of O₂ is consumed, along with 4.00025 mol (8.0645 g) of H₂, leaving about 1.94 g of H₂ in excess. The mass balance checks out: 64 + 8.0645 = 72.0645 g. Real reactions usually give a percent yield below 100% because of incomplete reaction, side products and losses during isolation; a yield above 100% normally means the product is wet or impure.

Common mistakes

Comparing masses or raw moles instead of moles divided by coefficients.
Using an unbalanced equation.
Entering the atomic mass of H (1.008) instead of the molecular H₂ (2.016).
Calculating yield from the excess reactant.

Keep in mind: The theoretical yield assumes the reaction goes to completion with no side reactions. Reactions that reach equilibrium, or reagents that are impure, will give less. Never mix reactive gases like hydrogen and oxygen outside a controlled setting.

Frequently asked questions

How do I find the excess reactant left over?

Multiply the limiting units by the excess reactant's coefficient and molar mass, then subtract from the starting mass.

Can this handle three reactants?

Not directly; compute moles ÷ coefficient for each and pick the smallest by hand.

What is a good percent yield?

It depends on the reaction; many teaching-lab syntheses land between roughly 50% and 90%.

Sources: Atkins, P., Jones, L., Chemical Principles, W. H. Freeman · Zumdahl, S. S., Zumdahl, S. A., Chemistry, Cengage

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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