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AP Chemistry / Topic

Chemical Reactions

Unit 4 covers writing and balancing equations, net ionic equations, stoichiometry with limiting reactants, titrations, and classifying reactions as precipitation, acid-base, or redox. It supplies the quantitative core of the free-response section.

31 practice questions1 FRQs29 skill areas
Practice Chemical ReactionsQuick 5 warm-upQuestion library

What the exam asks

  • Write balanced molecular and net ionic equations using solubility rules
  • Run mole-to-mole, mass-to-mass, and solution stoichiometry, including limiting reactant and percent yield
  • Interpret titration calculations and titration curves at the equivalence point
  • Assign oxidation numbers and identify what is oxidized and reduced
  • Classify reactions and predict products for synthesis, decomposition, and single/double replacement

Key formulas and rules

Stoichiometry path

grams → moles → mole ratio → moles → grams

Titration at equivalence

moles acid × (ratio) = moles base; MaVa = MbVb for 1:1

Percent yield

actual / theoretical × 100%

Always-soluble ions

Group 1 cations, NH4+, NO3−, C2H3O2− salts dissolve

Oxidation shorthand

OIL RIG — oxidation is loss, reduction is gain (of electrons)

Question bank breakdown

Multiple choice

31

Free response

1

Difficulty mix

10 easy · 14 medium · 7 hard

Skills covered

StoichiometryNet ionic equationsTypes of chemical reactionsNet ionic equations and spectator ionsStoichiometry and limiting reagentsPercent yieldTitration calculationsNet Ionic EquationsNet Ionic Equations with Weak AcidsPrecipitation Reactions and Solubility RulesPrecipitation and Spectator IonsBrønsted-Lowry Acids and BasesConjugate Acid-Base PairsOxidation NumbersOxidizing and Reducing AgentsIdentifying Oxidation and ReductionClassifying Redox vs. Non-Redox ReactionsGas StoichiometryLimiting Reactant StoichiometryLimiting Reactant and Excess RemainingPercent YieldTitration CalculationsTitration ConceptsTitration Stoichiometry with Polyprotic Ratios

Every question in the bank comes with a per-choice explanation, so you learn why each wrong answer is wrong — not just the key.

Sample question

From the free tier of the ScoreMint AP Chemistry bank — try it, then check the answer.

Consider the balanced equation: 2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s) If 5.4 g of Al (molar mass 27.0 g/mol) reacts with excess Cl₂, what mass of AlCl₃ (molar mass 133.5 g/mol) is produced?

  • A. 13.35 g
  • B. 26.7 g
  • C. 40.05 g
  • D. 53.4 g
Show answer and explanation

Answer: B. Correct. Step 1: Moles of Al = 5.4 g / 27.0 g/mol = 0.200 mol Al. Step 2: From the balanced equation, 2 mol Al produces 2 mol AlCl₃ (1:1 ratio). So 0.200 mol Al produces 0.200 mol AlCl₃. Step 3: Mass of AlCl₃ = 0.200 mol × 133.5 g/mol = 26.7 g. ✓

Try more AP Chemistry questions

Study tip

For net ionic equations: balance the molecular equation, split only strong electrolytes that are aqueous, then cancel spectators. Solids, liquids, gases, and weak acids stay intact — that is where most points are lost.

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Study guides

Guides for this subject

Free strategy and pacing guides that pair with this topic's practice.

AP Chemistry Exam Tactics: How to Earn the Points You Know

AP Chemistry exam strategy: reasoning from particle models, earning free-response partial credit, using the equations sheet, and pacing.

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How to Write AP Free-Response Answers That Score

How to write AP free-response answers that score: command verbs, showing work for partial credit, and self-scoring against the rubric.

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