Stoichiometry Calculator (Limiting Reagent and Theoretical Yield)

Mass-to-mass stoichiometry from a balanced equation, with the limiting reagent, excess left over and theoretical yield.

Convert grams to moles (÷ molar mass), divide by the coefficient to find the limiting reagent, then multiply by the product's coefficient and molar mass. For 2 H₂ + O₂ → 2 H₂O with 10 g H₂ and 64 g O₂: H₂ gives 4.96 mol ÷ 2 = 2.48, O₂ gives 2.00 mol ÷ 1 = 2.00, so O₂ is limiting and 4.00 mol (72.1 g) of water forms.

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How to use it

  1. Enter each reactant's formula, its coefficient in the balanced equation and its mass in grams.
  2. Enter the product formula and its coefficient.
  3. Read the theoretical yield, the limiting reagent and the excess left over.

What you enter

What you get

Formula

n = m / M | extent = n / coefficient | n(product) = extent(limiting) × coefficient(product)

Worked example: making water

Input: 2 H₂ + O₂ → 2 H₂O, 10 g H₂, 64 g O₂

Result: 72.1 g H₂O; O₂ limiting; 1.94 g H₂ left over

Even though there is less hydrogen by mass, oxygen runs out first.

Questions

How do you find the limiting reagent?

Convert each reactant to moles and divide by its coefficient. The smallest result is the limiting reagent; it runs out first and sets how much product forms.

What are the steps of a stoichiometry problem?

Grams of A → moles of A (÷ molar mass) → moles of B (× coefficient ratio) → grams of B (× molar mass).

What is theoretical yield?

The most product the limiting reagent can make. Percent yield is actual yield ÷ theoretical yield × 100.

Does the equation have to be balanced?

Yes. The coefficients are the mole ratios, so an unbalanced equation gives wrong answers. Use the chemical equation balancer first.

Limitations

Related reading

More Chemistry

All free tools · By Dr. Zubair Khalid, DVM, MS, PhD.