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 = -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):

1. Free elements: 0 (Na, O2, P4 all have oxidation state 0)
2. Monatomic ions: equal to charge (Na+ = +1, Cl- = -1)
3. Hydrogen: +1 (except in metal hydrides: -1)
4. Oxygen: -2 (except in peroxides: -1, and OF2: +2)
5. Fluorine: always -1
6. 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+

Step 2. Balance atoms other than O and H:

• Fe^2+ -> Fe^3+ (Fe balanced)
• MnO4- -> Mn^2+ (Mn balanced)

Step 3. Balance O with H2O:

• MnO4- -> Mn^2+ + 4 H2O

Step 4. Balance H with H+:

• 8 H+ + MnO4- -> Mn^2+ + 4 H2O

Step 5. Balance charge with electrons:

• Fe^2+ -> Fe^3+ + e- (charge: +2 -> +3, add 1 e- to right)
• 5 e- + 8 H+ + MnO4- -> Mn^2+ + 4 H2O (charge: 5(-1) + 8(+1) + (-1) = +2 on left; +2 on right)

Step 6. Equalize electrons: multiply Fe half-reaction by 5.

• 5 Fe^2+ -> 5 Fe^3+ + 5 e-
• 5 e- + 8 H+ + MnO4- -> Mn^2+ + 4 H2O

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.

Electrochemistry Connection

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.

Thermochemistry

Enthalpy (delta-H)

Exothermic: delta-H < 0 (releases heat). Combustion reactions, most neutralizations. Endothermic: delta-H > 0 (absorbs heat). Photosynthesis, dissolving NH4NO3 in water.

Hess's Law

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.

Calorimetry

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.

Standard Enthalpy of Formation (delta-Hf-zero)

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.

Common Mistakes

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

Cross-References

• lavoisier agent: Conservation of mass, reaction classification, stoichiometric calculations. Primary agent for this skill.
• atomic-structure skill: Atomic mass and mole concept derive from isotope-weighted averages.
• chemical-bonding skill: Bond energies provide an alternative route to estimating enthalpy changes.
• analytical-methods skill: Titration is a quantitative analytical technique rooted in acid-base and redox stoichiometry.
• organic-chemistry skill: Organic reactions (combustion, substitution, addition) follow the same stoichiometric principles.

References

• Zumdahl, S. S. & Zumdahl, S. A. (2017). Chemistry. 10th edition. Cengage Learning.
• Silberberg, M. S. (2018). Chemistry: The Molecular Nature of Matter and Change. 8th edition. McGraw-Hill.
• Atkins, P. & de Paula, J. (2014). Atkins' Physical Chemistry. 10th edition. Oxford University Press.
• Lavoisier, A. L. (1789). Traite Elementaire de Chimie. (Translated: Elements of Chemistry.)
• Bronsted, J. N. (1923). "Some Remarks on the Concept of Acids and Bases." Recueil des Travaux Chimiques, 42, 718-728.
• Hess, G. H. (1840). "Recherches Thermochimiques." Bulletin de l'Academie des Sciences de Saint-Petersbourg, 8, 257-272.