Stoichiometry calculator

Type the reaction (it is balanced automatically) and the reactant amounts in grams, moles or liters of gas: you get the limiting reactant, the moles and grams of each product, what is left of the excess reactant and, if you give the mass obtained, the percent yield. The limiting reactant is found by the mole ratio and by the product formed, and the products by conversion factors.

What you can do

• Limiting reactant by two methods.
• Conversion factors from grams to moles to grams.
• Gases with the ideal gas law.
• Table of initial, reacting and final amounts.

Sources

Atomic masses, ionization energies, melting points and names from Wikidata, public-domain data (CC0 1.0), checked one by one by Simuveliq. Atomic masses are the IUPAC standard ones.

Guide: stoichiometry and the limiting reactant

Stoichiometry answers questions such as "how much product do I get from these reactants?" or "how much reactant do I need?". It rests on a simple idea: the coefficients of the balanced equation give the ratio of moles in which substances react and form. This guide explains the method and works a complete limiting reactant problem.

The method in three steps

1. Balance the equation. Without correct coefficients everything else is wrong. The calculator balances the equation by itself.
2. Convert the amounts to moles. Mass: n = m/M. Solution: n = c·V. Ideal gas: n = PV/(RT).
3. Use the mole ratio from the coefficients to go from one substance to another and, finally, back to grams, liters or whatever the problem asks for.

Written as a chain of conversion factors, to get the mass of a product P from the mass of a reactant R:

m_P = m_R × (1 mol R / M_R) × (ν_P mol P / ν_R mol R) × (M_P / 1 mol P)

Limiting reactant and excess reactant

When the amounts of two or more reactants are given, they are almost never in the exact ratio of the equation: one runs out first (the limiting reactant) and sets how much product forms; part of the other is left over (the excess reactant). There are two equivalent ways to find it:

• Mole ratio: divide the moles of each reactant by its coefficient; the smallest quotient belongs to the limiting reactant.
• Product formed: calculate how much product each reactant would give if it were used up; the one that gives less is the limiting reactant.

A typical mistake is choosing as limiting the reactant with the fewest grams, or the fewest moles, without accounting for the coefficients.

Worked example: zinc and hydrochloric acid

10.0 g of Zn react with 10.0 g of HCl. Which reactant limits, how much ZnCl₂ forms and how much of the other reactant is left?

1. Balanced equation: Zn + 2 HCl → ZnCl₂ + H₂.
2. Moles: Zn: 10.0/65.38 = 0.1530 mol. HCl: 10.0/36.458 = 0.2743 mol.
3. Limiting reactant: Zn: 0.1530/1 = 0.1530; HCl: 0.2743/2 = 0.1371. HCl's is smaller: HCl is the limiting reactant, even though there are more moles of HCl than of Zn.
4. Product: 0.1371 mol of ZnCl₂ × 136.28 g/mol = 18.69 g form, plus 0.1371 mol of H₂ × 2.016 g/mol = 0.2765 g.
5. Excess: 0.1371 mol of Zn react; 0.1530 − 0.1371 = 0.0158 mol = 1.034 g of Zn is left over.

These are the same values the calculator gives with these data; it also shows both methods for finding the limiting reactant.

Percent yield

The theoretical yield is the amount of product predicted by stoichiometry (18.69 g in the example). In the lab you get less, because of incomplete reactions, side reactions or losses while isolating the product. The percent yield is

percent yield = 100 × actual yield / theoretical yield

If 15.0 g of ZnCl₂ were recovered in the example, the yield would be 100 × 15.0/18.69 ≈ 80.3 %.

How to use the calculator

1. Type the equation without coefficients, for example Zn + HCl -> ZnCl2 + H2; it balances itself.
2. Type the amount of one or more reactants and choose its unit. With a single value the other reactants are assumed in excess; with several, the limiting reactant is found.
3. If you have it, type the actual yield to get the percent yield.
4. Read the amounts table (initial, reacts, final) and the step-by-step solution; the detailed PDF solution has the full derivation.

Gases are treated as ideal (PV = nRT, with R = 0.082057 L·atm/(mol·K)).

Common mistakes

• Using grams in the mole ratio. Coefficients relate moles, not masses.
• Choosing the limiting reactant by the smallest amount without dividing by the coefficient.
• Starting from an unbalanced equation.
• Calculating the product from the excess reactant instead of the limiting one.

Further reading

R. Chang and K. A. Goldsby, Chemistry (McGraw-Hill); T. L. Brown, H. E. LeMay, B. E. Bursten et al., Chemistry: The Central Science (Pearson). Simuveliq is not affiliated with these authors or publishers.

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