--- title: "AP Chemistry Unit 4 Practice Test" description: "A timed AP Chemistry Unit 4 (Chemical Reactions) practice test in the AP format, drawn from 1,848 multiple-choice questions and 71 FRQs written for the unit. The standard version takes about 60 minutes." canonical: "https://fiveable.me/ap-chem/unit-test/4" type: "unit-test" subject: "AP Chemistry" unit: "Unit 4" --- # AP Chemistry Unit 4 Practice Test ## Overview A timed AP Chemistry Unit 4 (Chemical Reactions) practice test in the AP format, drawn from 1,848 multiple-choice questions and 71 FRQs written for the unit. The standard version takes about 60 minutes. ## Test Format - 20 Multiple Choice Questions (30 minutes) - 1 FRQ – FRQ 1 – Long Answer (15 minutes) - 1 FRQ – FRQ 4 – Short Answer (15 minutes) ## Unit FRQs by Type - [FRQ 1 – Long Answer](/ap-chem/frqs/frq-1?unit=4): 11 questions - [FRQ 2 – Long Answer](/ap-chem/frqs/frq-2?unit=4): 3 questions - [FRQ 3 – Long Answer](/ap-chem/frqs/frq-3?unit=4): 5 questions - [FRQ 4 – Short Answer](/ap-chem/frqs/frq-4?unit=4): 17 questions - [FRQ 5 – Short Answer](/ap-chem/frqs/frq-5?unit=4): 15 questions - [FRQ 6 – Short Answer](/ap-chem/frqs/frq-6?unit=4): 3 questions - [FRQ 7 – Short Answer](/ap-chem/frqs/frq-7?unit=4): 17 questions ## Practice Questions ### Question 1 An alloy of copper and zinc is treated with excess $$HCl$$, where only zinc reacts to produce $$H_2$$ gas. Which relationship uses the moles of $$H_2$$ produced ($$n_{H_2}$$) to find the mass percent of Zn in the alloy sample of mass $$m_{sample}$$? - (A) $$\% Zn = \frac{n_{H_2} \times MM_{Zn}}{m_{sample}} \times 100$$ - (B) $$\% Zn = \frac{n_{H_2} \times m_{sample}}{MM_{Zn}} \times 100$$ - (C) $$\% Zn = \frac{n_{H_2} \times MM_{H_2}}{m_{sample}} \times 100$$ - (D) $$\% Zn = \frac{m_{sample}}{n_{H_2} \times MM_{Zn}} \times 100$$ **Answer: (A)** Converting moles of product to mass of reactant and dividing by total sample mass. ### Question 2 In the reaction between $$Cu$$ and $$NO_3^-$$, copper is oxidized to $$Cu^{2+}$$ and nitrate is reduced to $$NO$$, resulting in a $$NO$$ coefficient of 2 when 3 moles of $$Cu$$ are consumed. If the oxidation product were $$Cu^+$$ instead, how would the coefficient of $$NO$$ change in the balanced equation? - (A) Decreases from 2 to 1 - (B) Remains constant at 2 - (C) Increases from 2 to 3 - (D) Increases from 2 to 4 **Answer: (A)** Analyzing the electron transfer ratio reveals the stoichiometric change. If $$Cu$$ forms $$Cu^+$$ (1 electron lost) instead of $$Cu^{2+}$$ (2 electrons lost), 3 moles of $$Cu$$ release 3 electrons total. Since $$NO_3^-$$ requires 3 electrons to form $$NO$$, only 1 mole of $$NO$$ is produced. ### Question 3 A half-reaction reduces sulfate ($$SO_4^{2-}$$) to sulfur dioxide ($$SO_2$$) in acidic solution, requiring 4 $$H^+$$ ions. If the reduction product were elemental sulfur ($$S$$) instead, how would the number of $$H^+$$ ions required for balancing change? - (A) It decreases to 2 ions - (B) It remains at 4 ions - (C) It increases to 6 ions - (D) It increases to 8 ions **Answer: (D)** Connecting oxygen removal to hydrogen ion consumption shows the relationship. Reducing $$SO_4^{2-}$$ to $$S$$ requires removing 4 oxygen atoms as 4 $$H_2O$$ molecules, which necessitates 8 $$H^+$$ ions, doubling the original requirement of 4. ### Question 4 The redox reaction between $$Fe^{2+}$$ and $$MnO_4^-$$ has a stoichiometric ratio of 5:1 ($$Fe^{2+}$$:$$MnO_4^-$$) when $$Mn^{2+}$$ is the product. If the reduction product of permanganate changed to $$MnO_2$$, which requires 3 electrons, what would be the new ratio? - (A) The ratio becomes 1:1 - (B) The ratio becomes 3:1 - (C) The ratio becomes 5:3 - (D) The ratio becomes 10:1 **Answer: (B)** Balancing the electron transfer establishes the new ratio. $$Fe^{2+}$$ oxidizes to $$Fe^{3+}$$ (1 electron), while $$MnO_4^-$$ reducing to $$MnO_2$$ requires 3 electrons ($$+7 \rightarrow +4$$). Therefore, 3 $$Fe^{2+}$$ ions are needed to supply the electrons for 1 $$MnO_4^-$$. ### Question 5 A half-reaction reduces $$ClO_3^-$$ to $$Cl^-$$ in acidic solution. If the reactant were changed to $$ClO_4^-$$ while the product remained $$Cl^-$$, how would the number of water molecules produced in the balanced half-reaction change? - (A) Decreases by one molecule - (B) Remains the same number - (C) Increases by one molecule - (D) Increases by two molecules **Answer: (C)** Tracking the oxygen atoms reveals the mass balance relationship. $$ClO_3^-$$ contains 3 oxygens (producing 3 $$H_2O$$), while $$ClO_4^-$$ contains 4 oxygens (producing 4 $$H_2O$$), resulting in an increase of one water molecule. ### Question 6 The reduction of nitrate ($$NO_3^-$$) to nitrogen monoxide ($$NO$$) requires 3 electrons. If the reaction conditions change such that the product is ammonium ($$NH_4^+$$) instead, how does the number of electrons transferred change? - (A) Decreases from 3 to 1 - (B) Increases from 3 to 5 - (C) Increases from 3 to 6 - (D) Increases from 3 to 8 **Answer: (D)** Calculating the total change in oxidation state determines the electron transfer. Nitrogen in $$NO_3^-$$ is +5, while in $$NH_4^+$$ it is -3. The total change is $$5 - (-3) = 8$$ electrons, compared to 3 electrons for $$NO$$ (+2). ### Question 7 The reaction $$2Al + 3Zn^{2+} \rightarrow 2Al^{3+} + 3Zn$$ transfers 6 moles of electrons. If $$Zn^{2+}$$ is replaced by $$Ag^+$$, which reduces to $$Ag$$, how many moles of $$Ag^+$$ are required to react completely with 2 moles of $$Al$$? - (A) 1 mole of Ag+ - (B) 2 moles of Ag+ - (C) 3 moles of Ag+ - (D) 6 moles of Ag+ **Answer: (D)** Equating total electron loss and gain solves the stoichiometry. Two moles of $$Al$$ lose 6 moles of electrons ($$2 \times 3$$). Since each $$Ag^+$$ accepts only 1 electron, 6 moles of $$Ag^+$$ are needed to accept all 6 moles of electrons. ### Question 8 The oxidation of iodine ($$I_2$$) to iodate ($$IO_3^-$$) releases 10 electrons per reaction unit. If the reaction conditions change to produce periodate ($$IO_4^-$$) instead, which of the following describes the change in the number of electrons released? - (A) Decreases from 10 to 5 - (B) Remains constant at 10 - (C) Increases from 10 to 12 - (D) Increases from 10 to 14 **Answer: (D)** Analyzing the oxidation states reveals the electron difference. $$I_2$$ (0) to 2 $$IO_3^-$$ (+5) releases $$2 \times 5 = 10$$ electrons. $$I_2$$ (0) to 2 $$IO_4^-$$ (+7) releases $$2 \times 7 = 14$$ electrons. ### Question 9 In a balanced redox equation, the coefficient of the oxidant depends on the electrons lost by the reductant. If the reductant is changed to a species that loses twice as many electrons per mole, how does the coefficient of the oxidant change? - (A) It is halved - (B) It is unchanged - (C) It is doubled - (D) It is quadrupled **Answer: (C)** Applying the principle of electron conservation shows the direct relationship. If the reductant releases twice as many electrons, twice as many oxidant molecules are required to accept those electrons (assuming the oxidant's electron capacity is constant). ### Question 10 A solid mixture contains $$KClO_3$$ and $$KCl$$. Heating the mixture causes $$KClO_3$$ to decompose into $$KCl$$ and $$O_2$$, while $$KCl$$ remains unchanged. To determine the mass percent of $$KClO_3$$ in the original mixture, which quantities are sufficient? - (A) The initial mass of the mixture and the final mass of the solid residue - (B) The initial mass of the mixture and the total volume of the container - (C) The final mass of the solid residue and the decomposition temperature - (D) The initial mass of the mixture and the decomposition temperature **Answer: (A)** Recognizing that the mass difference represents the oxygen lost allows for the calculation of moles of $$O_2$$. Stoichiometry then yields the mass of $$KClO_3$$, which is divided by the initial mass to find the percentage. ## Review the Unit - [4.1 Introduction to Reactions](/ap-chem/unit-4/intro-for-reactions/study-guide/LNQzVvZYjXxZMmXDubVP) - [4.2 Net Ionic Equations](/ap-chem/unit-4/net-ionic-equations/study-guide/VTCEO9cDrSHfPodoWGxS) - [4.3 Representations of Reactions](/ap-chem/unit-4/representations-reactions/study-guide/CzoUpQyKbK27GRGVfXFM) - [4.4 Physical and Chemical Changes](/ap-chem/unit-4/physical-vs-chemical-changes/study-guide/klo7E1MlmIr8w2mRq9jA) - [4.5 Stoichiometry](/ap-chem/unit-4/stoichiometry/study-guide/GjwCuhOQRvWLb4rKjYD2) - [4.6 Introduction to Titration](/ap-chem/unit-4/intro-titrations/study-guide/8XHQYjYki6GqAcrp18I2) - [4.7 Types of Chemical Reactions](/ap-chem/unit-4/types-chemical-reactions/study-guide/0VTaPH2MhqYc3Azc3xJz) - [4.8 Introduction to Acid-Base Reactions](/ap-chem/unit-4/intro-acid-base-reactions/study-guide/idvZ7Ve4pFo8gFIyYfMl) - [4.9 Oxidation-Reduction (Redox) Reactions](/ap-chem/unit-4/oxidation-reduction-redox-reactions/study-guide/43xfitnkAe6lhVeXtDXa) ## Other Unit Tests - [Unit 1: Atomic Structure & Properties](/ap-chem/unit-test/1) - [Unit 2: Compound Structure and Properties](/ap-chem/unit-test/2) - [Unit 3: Properties of Substances and Mixtures](/ap-chem/unit-test/3) - [Unit 5: Kinetics](/ap-chem/unit-test/5) - [Unit 6: Thermochemistry](/ap-chem/unit-test/6) - [Unit 7: Equilibrium](/ap-chem/unit-test/7) - [Unit 8: Acids & Bases](/ap-chem/unit-test/8) - [Unit 9: Thermodynamics and Electrochemistry](/ap-chem/unit-test/9) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)