How to use this calculator
- Enter or choose reactant A name.
- Enter or choose mass of A.
- Enter or choose molar mass of A.
- Enter or choose coefficient of A (a).
- Enter or choose reactant B name.
- The limiting reagent updates as you type. Use Copy result or Share to keep it.
Method
n = m ÷ M; limiting = smaller of n_A ÷ a and n_B ÷ b; product = c × min(n_A/a, n_B/b); yield = product × M_P
- a, b, c
- coefficients in the balanced equation aA + bB → cP
- n_A, n_B
- moles of each reactant
- M_P
- molar mass of the product
Worked example
- n(H₂) = 10 ÷ 2.016 = 4.9603 mol → ÷ 2 = 2.4802; n(O₂) = 64 ÷ 31.998 = 2.0001 mol → ÷ 1 = 2.0001.
- O₂ gives the smaller ratio, so it is limiting: water = 2 × 2.000125 = 4.00025 mol × 18.015 = 72.06 g.
- H₂ used = 4.00025 mol × 2.016 = 8.0645 g, so 10 − 8.0645 = 1.94 g of H₂ is left.
In 2H₂ + O₂ → 2H₂O with 10 g H₂ and 64 g O₂, oxygen is limiting; at most 72.06 g of water forms and 1.94 g of hydrogen is left over.
What the result means
Actual yield is usually lower than theoretical; percentage yield = actual ÷ theoretical × 100.
Assumptions and limits
- The equation is balanced and the reaction goes to completion.
- Reactants are pure.
- Two reactants and one product of interest; for more reactants, compare every n ÷ coefficient ratio the same way.
Questions
Is the reactant with the smaller mass always limiting?
No. Compare moles divided by coefficients, not grams. Here 10 g of hydrogen is in excess over 64 g of oxygen.
Can I use fractional coefficients?
Yes. Only the ratios between coefficients matter.