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:
- Write the unbalanced equation with correct formulas.
- Balance one element at a time, starting with the most complex molecule.
- Balance hydrogen and oxygen last (they appear in many compounds).
- Use the smallest whole-number coefficients.
- 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 = -log[H3O+]. At 25 C: pH 7 is neutral, pH < 7 is acidic, pH > 7 is basic.
pOH = -log[OH-]. pH + pOH = 14.00 at 25 C.
Worked example. Calculate the pH of 0.025 M HCl.
HCl is a strong acid — complete dissociation: [H3O+] = 0.025 M.
pH = -log(0.025) = -log(2.5 x 10^-2) = -(log 2.5 + log 10^-2) = -(0.398 - 2) = 1.60.
Worked Example: Weak Acid Equilibrium
Problem. Calculate the pH of 0.10 M acetic acid (Ka = 1.8 x 10^-5).
CH3COOH <=> CH3COO- + H+
Let x = [H+] at equilibrium. Ka = x^2 / (0.10 - x). Since Ka is small, assume 0.10 - x is approximately 0.10.
x^2 = 1.8 x 10^-5 x 0.10 = 1.8 x 10^-6.
x = 1.34 x 10^-3 M. Check assumption: 1.34 x 10^-3 / 0.10 = 1.3% < 5%. Valid.
pH = -log(1.34 x 10^-3) = 2.87.
Titration and Equivalence Point
At the equivalence point, moles of acid = moles of base. For a strong acid + strong base titration, the equivalence point pH is 7.00. For a weak acid + strong base, the equivalence point pH is above 7 (conjugate base in solution is basic).
Buffer solutions. Mixtures of a weak acid and its conjugate base (or weak base and conjugate acid) resist pH changes. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).
Oxidation-Reduction (Redox)
Oxidation States
Rules for assigning oxidation states (priority order):
- Free elements: 0 (Na, O2, P4 all have oxidation state 0)
- Monatomic ions: equal to charge (Na+ = +1, Cl- = -1)
- Hydrogen: +1 (except in metal hydrides: -1)
- Oxygen: -2 (except in peroxides: -1, and OF2: +2)
- Fluorine: always -1
- Sum of oxidation states = charge of species
Oxidation = increase in oxidation state (loss of electrons). Reduction = decrease in oxidation state (gain of electrons). Mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain).
Half-Reaction Method (Acidic Solution)
Worked example. Balance: MnO4- + Fe^2+ -> Mn^2+ + Fe^3+ in acidic solution.
Step 1. Write half-reactions:
- Oxidation: Fe^2+ -> Fe^3+
- Reduction: MnO4- -> Mn^2+
**St