reactions-stoichiometry

# 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:**

- Grams to moles: n = mass / molar mass
- Moles to particles: N = n x 6.022 x 10^23
- Moles to volume (gas at STP): V = n x 22.4 L

## Stoichiometric Calculations

Stoichiometry uses balanced equations as conversion factors. The coefficients give mole ratios.

### Worked Example: Mass-to-Mass Calculation

**Problem.** How many grams of CO2 are produced by burning 44.1 g of propane (C3H8)?

**Balanced equation:** C3H8 + 5 O2 -> 3 CO2 + 4 H2O

**Step 1.** Moles of propane: 44.1 g / 44.10 g/mol = 1.000 mol C3H8.

**Step 2.** Mole ratio: 1 mol C3H8 produces 3 mol CO2.
Moles CO2 = 1.000 x 3 = 3.000 mol.

**Step 3.** Mass of CO2: 3.000 mol x 44.01 g/mol = 132.0 g CO2.

### Limiting Reagent and Percent Yield

**Limiting reagent.** The reactant that runs out first, determining the maximum product. The other reactant(s) are in excess.

**Worked example.** *10.0 g of hydrogen reacts with 10.0 g of oxygen to form water. Which is limiting?*

2 H2 + O2 -> 2 H2O

Moles H2: 10.0 / 2.016 = 4.96 mol. Moles O2: 10.0 / 32.00 = 0.3125 mol.

From stoichiometry: 4.96 mol H2 requires 4.96/2 = 2.48 mol O2. We only have 0.3125 mol O2. Oxygen is limiting.

Moles H2O produced: 0.3125 mol O2 x (2 mol H2O / 1 mol O2) = 0.625 mol H2O.

Mass H2O: 0.625 x 18.02 = 11.3 g.

**Percent yield** = (actual yield / theoretical yield) x 100%. If the experiment produced 10.5 g: (10.5 / 11.3) x 100% = 92.9%.

## Reaction Types

### The Five Classical Types

| Type | Pattern | Example |
|---|---|---|
| Synthesis (combination) | A + B -> AB | 2 Na + Cl2 -> 2 NaCl |
| Decomposition | AB -> A + B | 2 HgO -> 2 Hg + O2 |
| Single replacement | A + BC -> AC + B | Zn + CuSO4 -> ZnSO4 + Cu |
| Double replacement (metathesis) | AB + CD -> AD + CB | AgNO3 + NaCl -> AgCl + NaNO3 |
| Combustion | CxHy + O2 -> CO2 + H2O | CH4 + 2 O2 -> CO2 + 2 H2O |

**Activity series for single replacement.** A metal replaces another in solution only if it is more active (higher on the activity series). Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Pt > Au. Zinc replaces copper; copper does not replace zinc.

### Precipitation Reactions

A double replacement reaction where an insoluble product (precipitate) forms. Use solubility rules:

**Soluble:** All Na+, K+, NH4+ salts. All nitrates. Most chlorides (except AgCl, PbCl2).

**Insoluble:** Most carbonates, phosphates, sulfides (except Group 1 and NH4+).

**Worked example.** *Write the net ionic equation for mixing AgNO3(aq) and NaCl(aq).*

Full molecular: AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)

Full ionic: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)

Net ionic (cancel spectators Na+ and NO3-): Ag+(aq) + Cl-(aq) -> AgCl(s)

The net ionic equation captures the chemistry — silver and chloride ions combine to form the insoluble precipitate.

## Acids and Bases

### Three Definitions

| Theory | Acid | Base |
|---|---|---|
| Arrhenius | Produces H+ in water | Produces OH- in water |
| Bronsted-Lowry | Proton (H+) donor | Proton (H+) acceptor |
| Lewis | Electron pair acceptor | Electron pair donor |

Each definition is progressively more general. Bronsted-Lowry is the workhorse for aqueous chemistry. Lewis acid-base theory extends to non-aqueous and coordination chemistry.

### Conjugate Pairs

Every Bronsted-Lowry acid has a conjugate base (what remains after donating H+), and every base has a conjugate acid (what forms after accepting H+).

HCl + H2O -> Cl- + H3O+

Acid: HCl. Conjugate base: Cl-. Base: H2O. Conjugate acid: H3O+.

**Strong acids** (completely dissociate): HCl, HBr, HI, HNO3, H2SO4, HClO4.
**Strong bases** (completely dissociate): Group 1 hydroxides (NaOH, KOH), Ba(OH)2, Ca(OH)2.

### pH Scale

pH =