Excess Reagent
Calculate Excess Reagent instantly with the exact formula and a worked example.
Excess Reagent
Working out which reactant is in excess is one of the most common stoichiometry problems in chemistry class. This calculator solves it from masses and the balanced equation and tells you how much of the excess reactant remains.
How the calculation works
For reactants A and B you enter the mass in grams, the molar mass in g/mol (the sum of atomic masses from the periodic table) and the stoichiometric coefficient from the balanced equation. Each mass is converted to moles with n = m / M.
Next, each amount is divided by its coefficient: n_A / a and n_B / b. The reactant with the smaller ratio is used up completely — that is the limiting reagent — and the other is in excess. Excess = n(excess) − n(limiting) × coefficient(excess) / coefficient(limiting). Multiplying by the molar mass of the excess reagent gives the leftover mass in grams.
The results show the excess in moles (the headline value), which reactant is limiting, which is in excess, the excess mass and the moles of A and B. If the two ratios are exactly equal, the mixture is stoichiometric and the excess is zero.
Worked example
Zn + 2HCl → ZnCl₂ + H₂ (the default values). Zinc: 6.54 g ÷ 65.38 g/mol = 0.1 mol, coefficient 1. Hydrogen chloride: 10 g ÷ 36.46 g/mol = 0.2743 mol, coefficient 2. Ratios: 0.1 / 1 = 0.1 and 0.2743 / 2 = 0.137, so zinc is limiting. 0.1 mol Zn consumes 0.2 mol HCl, leaving 0.2743 − 0.2 = 0.0742 mol HCl, about 0.0742 × 36.46 ≈ 2.71 g. Product amounts follow the zinc: 0.1 mol of H₂ is released.
Things to keep in mind
- Balance the equation first; wrong coefficients give a wrong answer even with perfect masses.
- Never compare grams or raw moles directly — only the moles-to-coefficient ratio decides which reactant is limiting.
- For solutions, enter the mass of the pure solute: 100 g of 10% hydrochloric acid contains 10 g of HCl.
- Theoretical yield and percent yield are always calculated from the limiting reagent.
- Round molar masses the way your course does; 36.5 vs 36.46 g/mol changes the answer only in the third significant figure.
More about: Excess Reagent
What it calculates
The “Excess Reagent” calculator computes Excess in mol from 6 parameters: mass of reactant a (g), molar mass of a (g/mol), coefficient of a in the equation, mass of reactant b (g), molar mass of b (g/mol), coefficient of b in the equation.
A standard physics formula used in educational and engineering tasks.
Example calculation
With parameters Mass of reactant A = 6.54 g, Molar mass of A = 65.38 g/mol, Coefficient of A in the equation = 1, Mass of reactant B = 10 g, Molar mass of B = 36.46 g/mol, Coefficient of B in the equation = 2 the result is 0.0742 mol.
How to use
- Enter mass of reactant a, molar mass of a, coefficient of a in the equation, mass of reactant b, molar mass of b and coefficient of b in the equation — each field above is adjustable with a slider.
- Excess (mol) is calculated automatically as you type.
- Check the worked example below to see the formula applied to real numbers.
- Copy the result or bookmark this calculator.
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FAQ
What is a limiting reagent?
Can I use it for three reactants?
Why can’t I just compare masses?
Does it work for gases?
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