Chemistry
Stoichiometry: limiting reagent and yield
Type the reaction, balanced or not, and the amounts of the reactants: the calculator balances it, finds the limiting reagent, works out how many grams of product form and what is left of the other reactants.
The limiting reagent
The coefficients of a balanced equation are ratios of moles, not of grams: N₂ + 3 H₂ → 2 NH₃ says one mole of nitrogen reacts with three of hydrogen to give two moles of ammonia. Getting from grams to moles means dividing by the molar mass, and that is where most mistakes happen.
The limiting reagent is the one that runs out first. Divide the available moles of each reactant by its coefficient: the smallest quotient wins. With 28 g of N₂ (about 1 mol) and 10 g of H₂ (4.96 mol), the quotients are 1 and 1.65: nitrogen limits, about 34.04 g of ammonia form and nearly 3.96 g of hydrogen is left over.
The theoretical yield is the product you would get if all the limiting reagent reacted. In the lab you get less, from incomplete reactions or losses: if 25 g of NH₃ is collected from that mixture, the percentage yield is 25 / 34.04 ≈ 73.4%.
Common mistakes
- Comparing the grams of the reactants instead of moles divided by coefficients: the reactant present in fewer grams is not necessarily the limiting one.
- Using an unbalanced equation: the mole ratios come from the coefficients, so without balancing the figures are wrong.
- Forgetting that diatomic elements are written H₂, O₂, N₂, Cl₂: the molar mass of O₂ is about 32 g/mol, not 16.
Frequently asked questions
Why divide by the coefficient?
Because the coefficient says how many moles of that reactant each “round” of the reaction needs. The quotient moles/coefficient counts how many rounds each reactant can supply, and the reaction stops at the smallest.
Do I need to balance the equation first?
No: the calculator balances it. Any coefficients you type are ignored and worked out again.
What happens if I leave a reactant empty?
It is treated as in excess, abundant enough never to limit: the usual case for oxygen in a combustion in air.
How this calculation works
Moles n = m / M, with M the molar mass from the formula. For each reactant with a known amount compute nᵢ / νᵢ, with ν the stoichiometric coefficient: the minimum ξ is the extent of reaction and its reactant is the limiting one. Reactant i consumed: νᵢ·ξ moles; left over: nᵢ − νᵢ·ξ. Product j: νⱼ·ξ moles, i.e. νⱼ·ξ·Mⱼ grams of theoretical yield. Percentage yield = actual yield / theoretical yield × 100.
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