The formula
- Test for the limiting reactant
compare n(X) / coefficient(X) across reactants; the smallest value is limiting- Moles of a reactant from its mass
n(X) = mass(X) / M(X)- Product from the limiting reactant
n(product) = n(limiting) × (coefficient of product / coefficient of limiting)- Excess reactant left over
n(left) = n(X) - n(consumed)
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
n(X) | Moles of reactant X supplied | mol |
mass(X) | Mass of reactant X weighed out | g |
coefficient(X) | Stoichiometric coefficient of X in the balanced equation | dimensionless |
M(X) | Molar mass of X | g/mol |
n(product) | Moles of product the limiting reactant can make | mol |
n(consumed) | Moles of the excess reactant actually used, set by the limiting reactant | mol |
n(left) | Moles of the excess reactant remaining when the reaction stops | mol |
When it applies
- Amounts of two or more reactants are given. When only one amount appears, that reactant is limiting by default and no comparison is needed.
- The equation is balanced first. The coefficients are the whole point of the test, and an unbalanced equation gives a confidently wrong answer.
- Compare moles divided by coefficients, never raw masses or raw moles. The reactant present in the smallest mass is often not the limiting one.
- Once identified, every later quantity, theoretical yield included, is computed from the limiting reactant alone.
Worked example
Problem. Hydrogen and oxygen react as 2H₂(g) + O₂(g) → 2H₂O(l). Starting from 10.0 g of H₂ and 64.0 g of O₂, which reactant is limiting, how much water forms, and how much of the excess reactant is left?
- Convert both to moles: n(H₂) = 10.0 / 2.016 = 4.9603 mol and n(O₂) = 64.0 / 31.998 = 2.0001 mol.
- Divide each by its coefficient: hydrogen gives 4.9603 / 2 = 2.4802, oxygen gives 2.0001 / 1 = 2.0001. Oxygen is smaller, so oxygen is limiting.
- Use the mole ratio 2 mol H₂O per 1 mol O₂: n(H₂O) = 2 × 2.0001 = 4.0003 mol.
- Convert to mass with M(H₂O) = 18.015 g/mol: 4.0003 × 18.015 = 72.06 g, which is 72.1 g of water.
- The oxygen consumed 2 × 2.0001 = 4.0003 mol of hydrogen, leaving 4.9603 - 4.0003 = 0.9600 mol, which is 0.9600 × 2.016 = 1.94 g of H₂ unreacted.
Answer. Oxygen is limiting, 72.1 g of water forms, and 1.94 g of hydrogen is left over.
Common mistakes
- Calling the reactant with the smaller mass limiting. Here hydrogen weighs far less than oxygen and is still the one in excess.
- Comparing moles without dividing by the coefficients, which ignores the fact that this reaction needs two hydrogens for every oxygen.
- Computing the product from the excess reactant, which overstates the yield.
- Forgetting that the leftover excess reactant is still there at the end, which matters whenever a problem asks what remains in the flask.
Related formulas
- Empirical formula:
%X = (mass X / mass compound) × 100% - Percent yield formula:
percent yield = (actual yield / theoretical yield) × 100% - Molar mass:
M = Σ (atoms of each element × atomic mass)