Limiting reactant calculator

What is a limiting reactant?

The reactant that runs out first. Once it is used up the reaction stops, so it limits how much product can form; the other reactants are left over in excess (OpenStax Chemistry 2e).

How do you find the limiting reactant?

Balance the equation, convert each amount to moles (grams ÷ molar mass), and divide by that reactant’s coefficient. The smallest result is the limiting reactant, and it also tells you how many “batches” of the reaction can run. Each product’s theoretical yield is its coefficient times that number, in moles; multiply by its molar mass for grams.

Example: making silicon nitride

OpenStax: 3Si + 2N₂ → Si₃N₄ with 2.00 g of Si and 1.50 g of N₂. Silicon is 0.0712 mol and would make 0.0237 mol of Si₃N₄; nitrogen is 0.0535 mol and would make 0.0268 mol. Silicon gives less, so it is the limiting reactant — about 3.33 g of Si₃N₄ can form, and about 0.17 g of N₂ is left over.

What if one reactant is “in excess”?

Leave its amount blank. The calculator treats it as unlimited and shows how much of it the reaction uses. OpenStax’s copper example: 1.274 g of CuSO₄ with excess zinc can give at most 0.5072 g of copper.

How do I type the reaction?

Formulas with + between them and = or -> between the sides, like H2 + O2 = H2O. You don’t need coefficients — the reaction is balanced for you. Molar masses use IUPAC standard atomic weights. Once you have an actual yield, the percent yield calculator compares it with the theoretical one.

Is the reactant with the smaller mass the limiting one?

Not necessarily. The limiting reactant depends on moles and on the coefficients, not on grams. In the water example below, 4.00 g of hydrogen is the smaller mass and is limiting, but with 4.00 g of hydrogen and only 16.0 g of oxygen the oxygen runs out first.

What if two reactants run out together?

Then the amounts are in exact stoichiometric ratio and both are limiting, with nothing left over. The calculator flags every reactant whose moles ÷ coefficient equals the smallest value, to within rounding.

Why is the theoretical yield higher than what I got in the lab?

Theoretical yield is the most product the limiting reactant could make if everything reacted as written. Real reactions lose some to side reactions, incomplete reaction and handling, so the actual yield is usually lower. The ratio of the two is the percent yield.

The method in steps

First the equation is balanced, so each substance has a coefficient. Then each amount in grams is divided by its molar mass to get moles. Dividing the moles of each reactant by its coefficient gives a number that says how many times the reaction could run on that reactant alone; the smallest number belongs to the limiting reactant. Every other reactant’s consumed amount is its coefficient times that smallest number, so what is left over is the amount given minus the amount used. The moles of each product equal its coefficient times the same number, and multiplying by the molar mass gives grams.

Worked example: making water

Balanced equation: 2H₂ + O₂ → 2H₂O. You have 4.00 g of H₂ and 32.0 g of O₂. Hydrogen: 1.984 mol, divided by 2 is 0.992. Oxygen: 1.000 mol, divided by 1 is 1.000. Hydrogen gives the smaller number, so it is the limiting reactant. The reaction runs 0.992 times, forming 1.984 mol of water, which is 35.74 g. Oxygen used is 0.992 mol, so 0.008 mol (0.26 g) is left over. With only 16.0 g of O₂ the numbers change: oxygen gives 0.500 and hydrogen 0.992, so oxygen is limiting and 18.02 g of water forms.

Leaving the hydrogen box blank treats it as unlimited: 8.0 g of O₂ then makes at most 9.01 g of water and uses 1.01 g of hydrogen.

Things to watch

• Do not compare grams or even moles directly; divide by the coefficient first. Two moles of H₂ react with only one mole of O₂, so equal moles of the two would leave oxygen in excess.
• Check the formulas for typos and capital letters. CO is carbon monoxide; Co is cobalt. The molar mass comes from the formula as typed.
• The theoretical yield is an upper limit for the product and assumes the reaction goes to completion. It is not a prediction of what the experiment will give.
• Choose grams or moles in the unit selector for each reactant. Entering a mass in the mole field is a common cause of nonsense answers.

Sources: OpenStax Chemistry 2e, section 4.4 (Reaction Yields), with its worked examples; atomic weights from IUPAC CIAAW. Related tools: stoichiometry calculator, percent yield calculator, molar mass calculator, chemical equation balancer.