reactions-stoichiometry

name	reactions-stoichiometry
description	Balancing chemical equations, reaction classification, stoichiometric calculations, acid-base chemistry, oxidation-reduction (redox) reactions, and thermochemistry. Covers conservation of mass, mole ratios, limiting reagents, percent yield, Bronsted-Lowry and Lewis acid-base theory, pH calculations, oxidation states, half-reaction balancing, enthalpy, Hess's law, and calorimetry. Use when solving quantitative chemistry problems involving reactions, energy changes, or solution chemistry.
type	skill
category	chemistry
status	stable
origin	tibsfox
modified	false
first_seen	"2026-04-12T00:00:00.000Z"
first_path	examples/skills/chemistry/reactions-stoichiometry/SKILL.md
superseded_by	null

Reactions and Stoichiometry

Chemical reactions transform substances by breaking and forming bonds. Stoichiometry is the quantitative bookkeeping — tracking atoms, moles, and energy through these transformations. This skill covers equation balancing, reaction classification, mole-based calculations, acid-base chemistry, redox reactions, and thermochemistry.

Agent affinity: lavoisier (chair, reactions and conservation of mass, primary)

Concept IDs: chem-balancing-equations, chem-reaction-types, chem-acids-bases, chem-oxidation-reduction, chem-thermochemistry

Conservation of Mass and Balancing Equations

Lavoisier's Law. In a chemical reaction, matter is neither created nor destroyed. Every atom present in the reactants must appear in the products.

Balancing procedure:

1. Write the unbalanced equation with correct formulas.
2. Balance one element at a time, starting with the most complex molecule.
3. Balance hydrogen and oxygen last (they appear in many compounds).
4. Use the smallest whole-number coefficients.
5. Verify: count every element on both sides.

Worked Example: Combustion of Propane

Unbalanced: C3H8 + O2 -> CO2 + H2O

Step 1. Balance C: 3 carbons on left, so 3 CO2 on right. C3H8 + O2 -> 3 CO2 + H2O

Step 2. Balance H: 8 hydrogens on left, so 4 H2O on right. C3H8 + O2 -> 3 CO2 + 4 H2O

Step 3. Balance O: Right side has 3(2) + 4(1) = 10 oxygens. Left needs 10/2 = 5 O2. C3H8 + 5 O2 -> 3 CO2 + 4 H2O

Verify: C: 3 = 3. H: 8 = 8. O: 10 = 10. Balanced.

Worked Example: Balancing a More Complex Equation

Unbalanced: Fe2O3 + CO -> Fe + CO2

Step 1. Balance Fe: 2 Fe on left, so 2 Fe on right. Fe2O3 + CO -> 2 Fe + CO2

Step 2. Balance O: Left has 3 (from Fe2O3) + 1 (from CO) = 4 if 1 CO. Right has 2 from CO2. Try: 3 CO on left gives 3 + 3 = 6 oxygens total on left... Systematic approach: Fe2O3 + 3 CO -> 2 Fe + 3 CO2.

Verify: Fe: 2 = 2. O: 3 + 3 = 6, 3(2) = 6. C: 3 = 3. Balanced.

The Mole Concept

Avogadro's number: 6.022 x 10^23 particles per mole. One mole of any element has a mass in grams equal to its atomic mass in amu.

Molar mass. Sum of atomic masses of all atoms in a formula. H2O: 2(1.008) + 16.00 = 18.02 g/mol.

Three conversions every chemist uses: