Chemical Reactions — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Reaction classification, net ionic equations, stoichiometric calculations including limiting reagent and percent yield, oxidation-reduction identification, and reactions in solution — AP Chemistry Unit 4.

You should already know: Atomic structure (Unit 1), molecular structure (Unit 2), intermolecular forces (Unit 3).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. Why Chemical Reactions Matter

Unit 4 is the operational heart of AP Chemistry — every later unit (Kinetics, Thermodynamics, Equilibrium, Acids & Bases, Electrochemistry) starts from a balanced reaction. About 7–9% of the AP score comes directly from Unit 4 questions, but every multi-unit FRQ silently assumes you can write and balance equations correctly.

The four core skills you must be able to do under timed exam conditions:

1. Classify a reaction (synthesis, decomposition, single replacement, double replacement, combustion, redox, acid-base).
2. Write a balanced molecular equation.
3. Convert it to net ionic equation by removing spectator ions.
4. Compute stoichiometry — moles, mass, gas volume, solution concentration — including limiting reagent and percent yield.

2. Reaction Classification

Synthesis (combination): A + B → AB. Example: $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$.

Decomposition: AB → A + B. Example: $2{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}+{\text{O}}_2$.

Single replacement: A + BC → AC + B. The replaced metal/halogen must be more reactive (use the activity series).

Double replacement: AB + CD → AD + CB. Driven by formation of (a) a precipitate, (b) a gas, or (c) water (acid-base neutralisation). If none of these forms, no reaction occurs.

Combustion: hydrocarbon + O₂ → CO₂ + H₂O. Memorise: complete combustion gives CO₂ and water; incomplete combustion yields CO and/or soot.

Redox: oxidation states change. Test with an oxidation-state assignment.

Acid-base (Brønsted-Lowry): proton transfer. Strong acid + strong base → salt + water.

3. Writing Net Ionic Equations

Three steps:

Step 1. Write the balanced molecular equation with phases (s, l, g, aq).

Step 2. Expand all aqueous strong electrolytes into their ions. Don't dissociate solids, liquids, gases, or weak electrolytes.

Step 3. Remove spectator ions (those appearing identically on both sides).

Example: ${\text{AgNO}}_3(aq)+\text{NaCl}(aq)\to \text{AgCl}(s)+{\text{NaNO}}_3(aq)$ Total ionic: ${\text{Ag}}^{+}+{\text{NO}}_3^{-}+{\text{Na}}^{+}+{\text{Cl}}^{-}\to \text{AgCl}(s)+{\text{Na}}^{+}+{\text{NO}}_3^{-}$ Net ionic: ${\text{Ag}}^{+}(aq)+{\text{Cl}}^{-}(aq)\to \text{AgCl}(s)$

Memorise the solubility rules (chlorides soluble except Ag, Pb, Hg; sulfates soluble except Ba, Pb, Sr, Ca; carbonates and phosphates mostly insoluble) — they tell you which products precipitate.

4. Stoichiometry

The mole-ratio backbone: $$n_{\text{product}}=n_{\text{reactant}}\times \frac{\text{coefficient of product}}{\text{coefficient of reactant}}$$

For mass: $m=n\cdot M$. For gas at STP: $V=n\cdot 22.4\text{ L}$. For solution: $n=M\cdot V$.

Limiting reagent: the reactant that runs out first. Compute moles of product each reactant would produce — the smaller answer is the actual yield, the corresponding reactant is limiting.

Percent yield: $$\%\text{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\%$$

5. Redox Identification

Oxidation = loss of electrons (oxidation state increases). Reduction = gain of electrons (oxidation state decreases). Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Common rules for assigning oxidation states:

• Free element: 0 (e.g. Na, O₂).
• Simple ion: charge of ion (e.g. Na⁺ is +1, Cl⁻ is −1).
• Oxygen: usually −2; in peroxides (H₂O₂) it is −1; with F it is +2.
• Hydrogen: usually +1; with metals (NaH) it is −1.
• Sum of oxidation states: equal to the species' charge.

A redox reaction has at least one element whose oxidation state changes. Identify the oxidising agent (gets reduced — accepts electrons) and the reducing agent (gets oxidised — donates electrons).

6. Worked Example

Aluminium metal reacts with copper(II) sulfate solution: $2\text{Al}(s)+3{\text{CuSO}}_4(aq)\to {\text{Al}}_2({\text{SO}}_4{)}_3(aq)+3\text{Cu}(s)$. Given 5.4 g Al and 50.0 mL of 1.50 M CuSO₄ solution: (a) Identify the limiting reagent. (b) Calculate the theoretical mass of Cu produced. (c) Identify oxidation/reduction for Al and Cu. (d) Write the net ionic equation.

Solution.

(a) Moles Al $=5.4/27=0.20$ mol. Moles CuSO₄ $=0.050\cdot 1.50=0.075$ mol.

From stoichiometry, 2 mol Al reacts with 3 mol CuSO₄. Mole ratio of Al:CuSO₄ available = $0.20:0.075=2.67$, but the reaction needs $2:3=0.67$. So Al is in excess; CuSO₄ is the limiting reagent.

(b) Moles Cu = (3/3) × 0.075 = 0.075 mol. Mass Cu = 0.075 × 63.55 = 4.77 g.

(c) Al goes from 0 (free element) to +3 (in Al₂(SO₄)₃) — Al is oxidised (loses electrons), Al is the reducing agent. Cu goes from +2 to 0 — Cu²⁺ is reduced, Cu²⁺ is the oxidising agent.

(d) Net ionic: $2\text{Al}(s)+3{\text{Cu}}^{2+}(aq)\to 2{\text{Al}}^{3+}(aq)+3\text{Cu}(s)$. (Sulfate is a spectator ion.)

7. Common Pitfalls

• Forgetting phase labels: AP graders deduct points for missing (s), (l), (g), (aq). Always add them.
• Dissociating weak electrolytes: weak acids (CH₃COOH), weak bases (NH₃), and water do not dissociate. Leave them as molecular formulas in the ionic equation.
• Wrong limiting-reagent logic: don't compare moles directly — compare moles weighted by stoichiometric coefficients. The reagent that produces less product is limiting.
• Misassigning oxidation states: in polyatomic ions like SO₄²⁻ and ClO₃⁻, work out S or Cl from the constraint that the ion's charges sum correctly.

8. Practice Questions (CED Style)

1. Balance and write the net ionic equation for the reaction of Pb(NO₃)₂(aq) with KI(aq), and identify the precipitate.
2. 2.00 g of N₂(g) reacts with 1.00 g of H₂(g) to form NH₃(g). Calculate the limiting reagent, theoretical mass of NH₃, and the mass of excess reactant remaining.
3. Identify the oxidising and reducing agents in: ${\text{MnO}}_4^{-}+5{\text{Fe}}^{2+}+8{\text{H}}^{+}\to {\text{Mn}}^{2+}+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$.

9. Quick Reference Cheatsheet

• Reaction types: synthesis, decomposition, single replacement, double replacement, combustion, redox, acid-base.
• Net ionic: write balanced molecular → expand strong electrolytes → cancel spectators.
• Stoichiometry chain: mass → mol → mol → mass (use coefficients in the middle).
• Limiting reagent: reactant producing less product.
• % yield = actual / theoretical × 100.
• OIL RIG: Oxidation Is Loss; Reduction Is Gain (of electrons).
• Solubility quick rules: NO₃⁻, group 1, NH₄⁺ all soluble; halides soluble except Ag/Pb/Hg; sulfates soluble except Ba/Sr/Pb/Ca.

10. What's Next

Chemical Reactions feeds directly into Unit 5 (Kinetics) — once you can balance a reaction, you can ask how fast it goes. Unit 6 (Thermodynamics) asks how much energy it releases. Unit 7 (Equilibrium) asks where it stops. Unit 8 (Acids & Bases) is a special case of equilibrium for proton-transfer reactions. Unit 9 (Apps of Thermodynamics) combines redox with energy → electrochemistry. Use Ollie to walk through any single FRQ: "Help me identify the limiting reagent in this multi-step synthesis" or "Why isn't this an oxidation-reduction reaction even though atoms move around?".