AP Chemistry

AP Chemistry Unit 4 review: Chemical Reactions

Unit 4 represents reactions with equations, particle diagrams and measured amounts. It is 7–9% of multiple choice. Expect to identify reacting species, balance atoms and charge, and use mole ratios to calculate limiting reactants, yields and titration results.

Unit 4 on the May 2027 exam

AP Chemistry Unit 4 exam facts
2027 exam
Weight7–9% of the multiple-choice section
Exam dateThursday, May 6, 2027, afternoon session
Multiple choice60 questions in 90 minutes, worth 50% of the exam score
Free response7 questions in 105 minutes, worth 50%: three long questions (10 points each) and four short questions (4 points each).
CalculatorBoth sections allow scientific or approved graphing calculators. Four-function calculators are allowed but not recommended. Bluebook includes both Desmos scientific and graphing calculators.
Class timeAbout 14–15 class periods in College Board's pacing guide
2026 results15% of students scored a 5 on the whole exam, and about 76% scored 3 or higher

Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.

What Unit 4 covers

College Board's course description splits the unit into 9 topics. The second column is what you should be able to do for each one.

AP Chemistry Unit 4 topics
TopicWhat you should be able to do
4.1 Introduction for ReactionsUse observations to propose a reaction and identify evidence for chemical change.
4.2 Net Ionic EquationsWrite complete and net ionic equations while retaining weak electrolytes, solids and liquids intact.
4.3 Representations of ReactionsTranslate between balanced equations and particle diagrams with conserved atoms.
4.4 Physical and Chemical ChangesDistinguish changes in particle arrangement from changes in chemical identity.
4.5 StoichiometryCalculate limiting reactants, excess reactants, theoretical yield and percent yield.
4.6 Introduction to TitrationUse a balanced reaction and titrant volume to find the amount of an analyte.
4.7 Types of Chemical ReactionsIdentify precipitation, acid–base and oxidation–reduction reactions from the species involved.
4.8 Introduction to Acid-Base ReactionsIdentify proton donors and acceptors and represent their reaction.
4.9 Oxidation-Reduction (Redox) ReactionsAssign oxidation numbers and balance electron transfer in redox reactions.

A stoichiometry route that works beyond 1:1 reactions

Balance first. Coefficients compare numbers of particles and moles, not grams.

A stoichiometry route that works beyond 1:1 reactions
TaskMethodCheck
Reactant mass → product massMass / molar mass × coefficient ratio × product molar massUnits should end in grams of product.
Limiting reactantCompare available moles / coefficient for each reactantThe smallest value limits reaction extent.
Excess left overInitial moles − moles consumedUse the limiting reactant to calculate consumption.
TitrationM × volume in liters, then apply mole ratioAt equivalence, amounts match coefficients, not necessarily each other.
Percent yieldActual yield / theoretical yield × 100%Use the same units for both masses.
Net ionic equationSplit strong aqueous electrolytes; cancel unchanged ionsAtoms and total charge must balance.
RedoxMatch electrons lost and gainedThe oxidizing agent is reduced.

Common mistakes

Each of these loses points on questions you otherwise understand.

  • Choosing the limiting reactant by smaller mass. Convert to moles and compare with coefficients. The CED identifies mass-to-mass comparisons as an error on printed p. 75.
  • Assuming every reaction is 1:1. A balanced equation supplies the mole ratio for both synthesis and titration. CED p. 75.
  • Splitting a precipitate or weak acid into ions in a net ionic equation. Keep those species intact. CED topic 4.2.
  • Changing subscripts to balance an equation. Change coefficients; changing a subscript changes the substance. CED topics 4.2–4.3.

6 practice questions with answers

Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.

  1. Question 1 · 4.5

    For N₂ + 3H₂ → 2NH₃, a vessel contains 0.40 mol N₂ and 0.90 mol H₂. What is the maximum NH₃ amount?

    • A0.30 mol
    • B0.60 mol
    • C0.80 mol
    • D1.30 mol
    Show the answer

    B. H₂ limits the reaction: 0.90/3 = 0.30 mol reaction extent, compared with 0.40/1 for N₂. NH₃ produced is 2(0.30) = 0.60 mol.

  2. Question 2 · 4.5

    A reaction has a theoretical product yield of 8.40 g and an actual yield of 6.30 g. What is the percent yield?

    • A25.0%
    • B66.7%
    • C133%
    • D75.0%
    Show the answer

    D. Percent yield = 6.30/8.40 × 100 = 75.0%. The theoretical amount is the denominator because it is the predicted maximum for the stated reactants.

  3. Question 3 · 4.6

    A 25.0 mL sample of a diprotic acid H₂A requires 30.0 mL of 0.200 M NaOH for complete neutralization. What is the acid concentration?

    • A0.120 M
    • B0.240 M
    • C0.0600 M
    • D0.480 M
    Show the answer

    A. NaOH amount is 0.0300(0.200) = 0.00600 mol. Two moles of OH⁻ neutralize one mole of H₂A, so acid amount is 0.00300 mol. Divide by 0.0250 L to obtain 0.120 M.

  4. Question 4 · 4.2

    AgNO₃(aq) reacts with NaCl(aq) to form AgCl(s). Which net ionic equation is correct?

    • ANa⁺ + NO₃⁻ → NaNO₃(s)
    • BAgCl(s) → Ag⁺ + Cl⁻
    • CAg⁺ + Cl⁻ → AgCl(s)
    • DAg⁺ + NO₃⁻ → AgNO₃(s)
    Show the answer

    C. Ag⁺ and Cl⁻ form the precipitate. Na⁺ and NO₃⁻ remain dissolved and cancel from the complete ionic equation. Both total charge and atom counts balance.

  5. Question 5 · 4.9

    How many moles of electrons are released when 0.0800 mol Al is oxidized to Al³⁺?

    • A0.0267 mol
    • B0.240 mol
    • C0.0800 mol
    • D0.160 mol
    Show the answer

    B. The oxidation half-reaction is Al → Al³⁺ + 3e⁻. Thus 0.0800 mol Al releases 3(0.0800) = 0.240 mol electrons.

  6. Question 6 · 4.5

    A 0.150 mol sample of CaCO₃ decomposes completely: CaCO₃ → CaO + CO₂. Using CO₂ = 44.0 g mol⁻¹, what mass of CO₂ forms?

    • A2.93 g
    • B3.30 g
    • C11.0 g
    • D6.60 g
    Show the answer

    D. The coefficient ratio is 1:1, so 0.150 mol CO₂ forms. Its mass is 0.150(44.0) = 6.60 g. The calculation uses the product molar mass after the mole conversion.

How to study Unit 4

Solve one limiting-reactant problem by calculating product from each reactant separately. Then solve it using moles divided by coefficients. Compare the results and calculate the excess remaining as a second check.

Translate one precipitation reaction into molecular, complete ionic and net ionic equations. Draw a particle diagram for the final mixture that includes the spectator ions. For titrations, write the balanced neutralization equation before multiplying molarity by volume.

Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.

Questions

Still have a question? support@tryashore.com

It is 7–9% of the multiple-choice section, which is half the exam score. This is not a fixed share of the whole exam: free-response questions can combine units. Suggested pacing is 14–15 class periods.

Learn Chemistry one short lesson at a time

Free to start, no card. A short lesson for every topic in all 9 units.

Starting check

SAT Math · 1 of 5

Calculator allowed

A jacket's price is increased by 20% and then the new price is decreased by 20%. The final price is what percent of the original price?
  • A96%96\%
  • B100%100\%
  • C104%104\%
  • D80%80\%