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.
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
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:
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.
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.
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:
Stoichiometry uses balanced equations as conversion factors. The coefficients give mole ratios.
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. 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%.
| 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.
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.
| 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.
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 = -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.
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.
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]).
Rules for assigning oxidation states (priority order):
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).
Worked example. Balance: MnO4- + Fe^2+ -> Mn^2+ + Fe^3+ in acidic solution.
Step 1. Write half-reactions:
Step 2. Balance atoms other than O and H:
Step 3. Balance O with H2O:
Step 4. Balance H with H+:
Step 5. Balance charge with electrons:
Step 6. Equalize electrons: multiply Fe half-reaction by 5.
Step 7. Add half-reactions (electrons cancel): MnO4- + 5 Fe^2+ + 8 H+ -> Mn^2+ + 5 Fe^3+ + 4 H2O
Verify: Mn: 1 = 1. Fe: 5 = 5. O: 4 = 4. H: 8 = 8. Charge: (-1) + 5(+2) + 8(+1) = +17 on left; (+2) + 5(+3) + 0 = +17 on right.
Standard reduction potentials (E-zero) predict spontaneous redox reactions. A positive cell potential (E-zero-cell = E-zero-cathode - E-zero-anode) means the reaction is spontaneous. This connects stoichiometry to electrical energy — the basis of batteries and electrolysis.
Exothermic: delta-H < 0 (releases heat). Combustion reactions, most neutralizations. Endothermic: delta-H > 0 (absorbs heat). Photosynthesis, dissolving NH4NO3 in water.
If a reaction can be expressed as the sum of two or more steps, the overall delta-H is the sum of the delta-H values of the steps. Enthalpy is a state function — only initial and final states matter, not the path.
Worked example. Calculate delta-H for: C(s) + 1/2 O2(g) -> CO(g), given:
(1) C(s) + O2(g) -> CO2(g), delta-H1 = -393.5 kJ (2) CO(g) + 1/2 O2(g) -> CO2(g), delta-H2 = -283.0 kJ
Solution. We need C + 1/2 O2 -> CO. Take reaction (1) as-is and reverse reaction (2):
(1) C + O2 -> CO2, delta-H = -393.5 kJ (2 reversed) CO2 -> CO + 1/2 O2, delta-H = +283.0 kJ
Sum: C + O2 + CO2 -> CO2 + CO + 1/2 O2. Cancel CO2 and simplify O2: C + 1/2 O2 -> CO, delta-H = -393.5 + 283.0 = -110.5 kJ.
q = m x c x delta-T, where q is heat (J), m is mass (g), c is specific heat capacity (J/g-C), and delta-T is temperature change.
Worked example. 50.0 g of water at 25.0 C absorbs 2,092 J. What is the final temperature?
c(water) = 4.184 J/g-C.
delta-T = q / (m x c) = 2092 / (50.0 x 4.184) = 10.0 C.
T_final = 25.0 + 10.0 = 35.0 C.
delta-H-rxn = sum(delta-Hf-zero products) - sum(delta-Hf-zero reactants).
Elements in their standard states have delta-Hf-zero = 0 by definition.
| Mistake | Why it fails | Fix |
|---|---|---|
| Unbalanced equations in stoichiometry | Wrong mole ratios give wrong answers | Always verify atom and charge balance before calculating |
| Ignoring limiting reagent | Assuming all reactants are fully consumed | Calculate moles of product from each reactant; the smaller result identifies the limiter |
| Forgetting to convert grams to moles | Stoichiometry works in moles, not grams | Always convert mass to moles before using coefficients |
| pH from weak acid = -log(initial concentration) | Weak acids do not fully dissociate | Use Ka equilibrium expression to find [H+] |
| Wrong sign on reversed Hess's law step | Reversing a reaction flips the sign of delta-H | Keep careful track of sign changes |
| Confusing oxidation and reduction | Leads to incorrect electron transfer | OIL RIG: Oxidation Is Loss, Reduction Is Gain |