Theoretical Yield from 50 g of Limiting Reagent (1:1)
Starting with 50 g of a 100 g/mol limiting reagent in a 1:1 reaction producing a 100 g/mol product gives a theoretical yield of 50 g.
How to use this tool
- Enter mass of limiting reagent, molar mass of limiting reagent, molar mass of product and stoichiometric ratio (mol product / mol reagent) in the fields above.
- Results update instantly as you type — or click Calculate.
- Read your theoretical yield and the full breakdown beneath it.
When reagent and product share the same molar mass in a 1:1 reaction, the theoretical yield equals the starting mass.
Frequently asked questions
- What is a limiting reagent?
- The reactant that is completely consumed first, limiting how much product can form. Any remaining reactant is in excess.
- Where do I find the stoichiometric ratio?
- From the balanced chemical equation. E.g. 2H₂ + O₂ → 2H₂O means 2 mol H₂ produces 2 mol H₂O, so ratio = 1:1. N₂ + 3H₂ → 2NH₃ means 1 mol N₂ → 2 mol NH₃, ratio = 2.