Calculate reactant and product masses from balanced equations. Identify limiting reagents and theoretical yields with visual ratios.
Enter a balanced equation and the masses of reactants you actually have, and this tool runs the standard stoichiometry pipeline: convert each mass to moles using molar masses, apply the equation's coefficients as a mole ratio, and convert the result back to grams for every product. It also identifies the limiting reagent and the theoretical yield, and shows the ratios visually so you can see how the coefficients distribute. The usual surprise: coefficients compare moles, not grams - "2 H2 + O2" does not mean 2 g of hydrogen reacts with 1 g of oxygen.
A concrete input and expected output from the current implementation.
Input
Equation: 2 H2 + O2 -> 2 H2O. Amounts: 6.0 g H2 and 32.0 g O2.
Expected output
Limiting reagent: O2. Theoretical yield: 36.0 g H2O (2.00 mol).
32.0 g of O2 is 1.00 mol, and 6.0 g of H2 is 2.98 mol. The equation needs 2 mol of H2 per mole of O2, so the 2.00 mol required is less than the 2.98 mol available: oxygen is consumed first and caps the reaction. The water follows the limiting reagent: 1.00 mol of O2 makes 2.00 mol of H2O, which weighs 36.0 g.
Each entered mass is divided by the substance's molar mass to get moles, then multiplied by the coefficient ratio from the equation - product coefficient over reactant coefficient, or reactant over reactant - to reach the target substance's moles, then multiplied by its molar mass to give grams. Only the coefficient ratio carries across the arrow; masses never convert directly to other masses.
For each reactant the tool works out how much product that reactant alone could form if all of it reacted. The reactant that can produce the least product is the limiting reagent, and that smallest amount is the theoretical yield. Any other reactant is in excess and cannot add to the product; that comparison is what the visual ratios make legible.
A balanced equation states the ratio in which substances react, not how much you must supply. If you start with more of one reactant than the ratio demands, the surplus simply cannot react once the other side is used up. That is the definition of excess, not an error. To consume it, add more of the limiting reagent and recalculate; the roles can flip.
Yes, total mass is conserved, so the sum of all product masses equals the sum of all reactant masses. But substance by substance, grams do not follow the coefficients - only moles do. Four grams of H2 with excess O2 gives about 36 g of H2O, more than the 4 g, because oxygen joins in; that is why the tool converts through moles in the middle.
The example and behavioral notes were checked against the browser implementation. Standards and primary references below define the relevant format, formula, or platform behavior.
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