AP Chemistry Chemical Reactions — Worked Answer Explanations

Unit 4 · 12 questions explained

Below is a complete answer key for our AP Chemistry Chemical Reactions practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Chemical Reactions practice test and come back here to review, or head back to the Chemical Reactions unit overview.

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Question 1 · Easy
When a small piece of sodium metal is added to water, bubbles form vigorously and the solution becomes basic. What type of reaction is occurring?
- A
  Double-displacement (metathesis) reaction
  Why not A: A double-displacement involves two ionic compounds exchanging partners; sodium metal is not an ionic compound.
- B
  Combustion reaction
  Why not B: Combustion requires oxygen as a reactant; this reaction involves water.
- C
  Single-displacement (redox) reactionCorrect
- D
  Acid–base neutralization
  Why not D: Neutralization combines an acid and a base; sodium metal is neither.
Explanation
Sodium displaces hydrogen from water: $2 Na (s) + 2 H_{2} O (l) \to 2 NaOH (aq) + H_{2} (g)$ . Na is oxidized (0 → +1); H is reduced (+1 → 0). This is a single-displacement (also called single-replacement) redox reaction.

Key takeaway
Single-displacement reactions involve one element replacing another in a compound, always involving redox.
Question 2 · Easy
How many grams of $H_{2} O$ are produced when $4.00 g$ of $H_{2}$ reacts completely with excess $O_{2}$ ?

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$

(Molar masses: H $_{2}$ = 2.02 g/mol, H $_{2}$ O = 18.02 g/mol)
- A
  $36.0 g$ Correct
- B
  $18.0 g$
  Why not B: Used a 1:1 mole ratio instead of the 2:2 ratio given in the equation.
- C
  $9.00 g$
  Why not C: Off by a factor of 4 — likely only used 1 g of H $_{2}$ .
- D
  $72.1 g$
  Why not D: Doubled the answer incorrectly.
Explanation
Moles H $_{2}$ $= \frac{4.00}{2.02} = 1.98 mol$

Mole ratio H $_{2}$ :H $_{2}$ O = 2:2 = 1:1, so moles H $_{2}$ O = 1.98 mol

Mass H $_{2}$ O $= 1.98 \times 18.02 \approx 36.0 g$

Key takeaway
Stoichiometry: moles of product = moles of reactant × (mole ratio from balanced equation).
Question 3 · Easy
Assign oxidation numbers and identify the element that is oxidized in the following reaction:

$Zn (s) + 2 HCl (aq) \to ZnC l_{2} (aq) + H_{2} (g)$
- A
  H is oxidized (changes from $+ 1$ to $0$ ).
  Why not A: H decreasing from +1 to 0 is reduction, not oxidation.
- B
  Zn is oxidized (changes from $0$ to $+ 2$ ).Correct
- C
  Cl is oxidized (stays at $- 1$ ).
  Why not C: Chlorine's oxidation number does not change in this reaction.
- D
  Zn is reduced (changes from $0$ to $+ 2$ ).
  Why not D: An increase in oxidation number is oxidation, not reduction.
Explanation
In Zn metal, oxidation number = 0. In ZnCl $_{2}$ , Zn is +2. The oxidation number increases → Zn is oxidized (it is the reducing agent). Hydrogen goes from +1 (in HCl) to 0 (in H $_{2}$ ) — that is reduction.

Key takeaway
Oxidation = increase in oxidation number (LEO). Reduction = decrease in oxidation number (GER). OIL RIG.
Question 4 · Easy
The theoretical yield of aspirin in a synthesis reaction is $2.70 g$ , but a student collects only $2.16 g$ . What is the percent yield?
- A
  $80.0%$ Correct
- B
  $125%$
  Why not B: Divided theoretical by actual rather than actual by theoretical.
- C
  $20.0%$
  Why not C: Calculated the percent lost, not the percent yield.
- D
  $54.0%$
  Why not D: Added or multiplied the masses rather than dividing.
Explanation
Percent yield $= \frac{actual yield}{theoretical yield} \times 100 = \frac{2.16}{2.70} \times 100 = 80.0%$

Key takeaway
Percent yield = (actual / theoretical) × 100%. It can never exceed 100% under ideal conditions.
Question 5 · Medium
In the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ , if $28.0 g$ of $N_{2}$ and $9.00 g$ of $H_{2}$ are mixed, which is the limiting reagent? (Molar masses: N $_{2}$ = 28.0 g/mol, H $_{2}$ = 2.02 g/mol)
- A
  $N_{2}$ , because it has a higher molar mass.
  Why not A: Molar mass alone does not determine limiting reagent.
- B
  $H_{2}$ , because more moles of it are required per mole of product.
  Why not B: Need to compare available moles to required moles; H $_{2}$ has 4.46 mol available.
- C
  $N_{2}$ , because comparing moles available to the stoichiometric requirement shows N $_{2}$ runs out first.Correct
- D
  Neither; both are present in stoichiometric amounts.
  Why not D: 1.00 mol N $_{2}$ requires 3.00 mol H $_{2}$ , but only 4.46 mol are available — they are not stoichiometric.
Explanation
Moles N $_{2}$ = 1.00 mol; moles H $_{2}$ = 9.00/2.02 = 4.46 mol.

To react with 1.00 mol N $_{2}$ we need $3 \times 1.00 = 3.00 mol$ H $_{2}$ . We have 4.46 mol H $_{2}$ , so H $_{2}$ is in excess. N $_{2}$ (only 1.00 mol available for 1.00 mol needed) is the limiting reagent.

Key takeaway
Limiting reagent: divide available moles by stoichiometric coefficient and pick the smallest ratio.
Question 6 · Medium
Which of the following is the correct net ionic equation for the reaction between aqueous $Pb (N O_{3})_{2}$ and aqueous $KI$ , which forms a yellow precipitate of $Pb I_{2}$ ?
- A
  $Pb (N O_{3})_{2} (aq) + 2 KI (aq) \to Pb I_{2} (s) + 2 KN O_{3} (aq)$
  Why not A: This is the overall (molecular) equation, not the net ionic equation.
- B
  $P b^{2 +} (aq) + 2 I^{-} (aq) \to Pb I_{2} (s)$ Correct
- C
  $K^{+} (aq) + N O_{3}^{-} (aq) \to KN O_{3} (aq)$
  Why not C: KNO $_{3}$ is soluble and remains as spectator ions; this does not describe the reaction that occurs.
- D
  $P b^{2 +} (aq) + I^{-} (aq) \to Pb I_{2} (s)$
  Why not D: The equation is not balanced; two iodide ions are needed per lead ion.
Explanation
Pb(NO $_{3}$ ) $_{2}$ , KI, and KNO $_{3}$ are all soluble → split into ions. K $^{+}$ and NO $_{3}^{-}$ are spectator ions. Only Pb $^{2 +}$ and I $^{-}$ participate in forming the insoluble precipitate PbI $_{2}$ . Net ionic: $P b^{2 +} (aq) + 2 I^{-} (aq) \to Pb I_{2} (s)$ .

Key takeaway
Net ionic equation: write only the ions that change form; exclude spectator ions.
Question 7 · Medium
When excess $NaOH (aq)$ is added to a solution of $FeC l_{3} (aq)$ , a reddish-brown precipitate forms. The full ionic equation includes $F e^{3 +} (aq) + 3 O H^{-} (aq) \to Fe (OH)_{3} (s)$ . What are the spectator ions?
- A
  $N a^{+}$ and $C l^{-}$ Correct
- B
  $F e^{3 +}$ and $O H^{-}$
  Why not B: These ions participate in forming the precipitate — they are not spectators.
- C
  $N a^{+}$ and $F e^{3 +}$
  Why not C: Fe $^{3 +}$ participates in the net ionic equation and is not a spectator.
- D
  $C l^{-}$ and $O H^{-}$
  Why not D: OH $^{-}$ participates in forming Fe(OH) $_{3}$ — it is not a spectator.
Explanation
The full ionic equation: $F e^{3 +} + 3 C l^{-} + 3 N a^{+} + 3 O H^{-} \to Fe (OH)_{3} (s) + 3 N a^{+} + 3 C l^{-}$ . Na $^{+}$ and Cl $^{-}$ appear unchanged on both sides — they are the spectator ions. Cancel them to get the net ionic equation.

Key takeaway
Spectator ions appear identically on both sides of the complete ionic equation and do not participate in the actual chemical change.
Question 8 · Medium
A student reacts $10.0 g$ of $CaC O_{3}$ with excess HCl and collects $C O_{2}$ gas. The molar mass of $CaC O_{3}$ is $100.1 g/mol$ . How many moles of $C O_{2}$ are produced?

$CaC O_{3} (s) + 2 HCl (aq) \to CaC l_{2} (aq) + H_{2} O (l) + C O_{2} (g)$
- A
  $0.100 mol$ Correct
- B
  $0.200 mol$
  Why not B: Used a 2:1 ratio from HCl to CO $_{2}$ rather than the 1:1 ratio from CaCO $_{3}$ to CO $_{2}$ .
- C
  $1.00 mol$
  Why not C: Did not divide by molar mass.
- D
  $0.0500 mol$
  Why not D: Used molar mass of 200 instead of 100.1 g/mol.
Explanation
Moles CaCO $_{3}$ = $\frac{10.0}{100.1} \approx 0.0999 \approx 0.100 mol$ . The 1:1 mole ratio (CaCO $_{3}$ :CO $_{2}$ ) means 0.100 mol CO $_{2}$ is produced.

Key takeaway
Mole-to-mole ratio from the balanced equation directly gives moles of product from moles of limiting reactant.
Question 9 · Medium
When $Cu (s)$ is placed in $AgN O_{3} (aq)$ , which observation is consistent with this single-displacement reaction, and what does the activity series predict?
- A
  No reaction occurs because Cu is below Ag in the activity series.
  Why not A: Cu is above Ag in the activity series and CAN displace Ag $^{+}$ .
- B
  Silver metal deposits on the copper and the solution turns blue as Cu $^{2 +}$ forms.Correct
- C
  Copper dissolves and forms $C u_{2} O$ , a red solid.
  Why not C: Cu $_{2}$ O forms in different oxidation conditions; in AgNO $_{3}$ solution, Cu $^{2 +}$ forms, not Cu $_{2}$ O.
- D
  Silver nitrate decomposes to release $N O_{2}$ gas and $O_{2}$ .
  Why not D: Thermal decomposition of AgNO $_{3}$ requires high temperatures, not copper metal.
Explanation
Cu is above Ag in the activity series, so Cu displaces Ag $^{+}$ : $Cu (s) + 2 A g^{+} (aq) \to C u^{2 +} (aq) + 2 Ag (s)$ . Solid silver coats the copper while $C u^{2 +} (aq)$ turns the solution pale blue.

Key takeaway
Activity series: a metal higher on the series displaces ions of metals lower on the series from solution.
Question 10 · Hard
In a balanced redox equation in acidic solution, $Mn O_{4}^{-}$ is reduced to $M n^{2 +}$ while $F e^{2 +}$ is oxidized to $F e^{3 +}$ . How many $F e^{2 +}$ ions are required to reduce one $Mn O_{4}^{-}$ ?
- A
  $2$
  Why not A: Mn gains only 5 electrons; each Fe loses 1 electron, so 5 Fe $^{2 +}$ are needed.
- B
  $3$
  Why not B: Mn goes from +7 to +2 — a gain of 5 electrons, not 3.
- C
  $5$ Correct
- D
  $7$
  Why not D: 7 is the oxidation state of Mn in MnO $_{4}^{-}$ , not the number of electrons transferred.
Explanation
Mn in MnO $_{4}^{-}$ is +7; in Mn $^{2 +}$ it is +2. Change = gain of 5 electrons (reduction). Fe $^{2 +}$ → Fe $^{3 +}$ is a loss of 1 electron (oxidation). To balance electrons transferred: 5 Fe $^{2 +}$ each donate 1 $e^{-}$ , providing the 5 $e^{-}$ needed to reduce one MnO $_{4}^{-}$ .

Key takeaway
In redox half-reactions, electrons gained in reduction must equal electrons lost in oxidation.
Question 11 · Hard
Which of the following describes a precipitation reaction?
- A
  $HCl (aq) + NaOH (aq) \to NaCl (aq) + H_{2} O (l)$
  Why not A: This is an acid–base neutralization; all products remain soluble.
- B
  $CaC l_{2} (aq) + N a_{2} S O_{4} (aq) \to CaS O_{4} (s) + 2 NaCl (aq)$ Correct
- C
  $Zn (s) + CuS O_{4} (aq) \to ZnS O_{4} (aq) + Cu (s)$
  Why not C: This is a single-displacement redox reaction.
- D
  $2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$
  Why not D: This is a decomposition reaction, not precipitation.
Explanation
$CaS O_{4}$ is insoluble in water (sparingly soluble), forming a solid precipitate when $C a^{2 +}$ and $S O_{4}^{2 -}$ combine. The formation of an insoluble solid from mixing two aqueous solutions is the hallmark of a precipitation reaction.

Key takeaway
Precipitation reaction: two soluble ionic compounds combine to form an insoluble product (solid precipitate).
Question 12 · Hard
Consider a combustion reaction: $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$ . If $44.0 g$ of propane ( $M = 44.1 g/mol$ ) reacts with $128 g$ of $O_{2}$ ( $M = 32.0 g/mol$ ), which is the limiting reagent?
- A
  $C_{3} H_{8}$ , because its mass is smaller.
  Why not A: Limiting reagent is not determined by mass alone.
- B
  $O_{2}$ , because 5 moles are needed per mole of propane but only 4 moles of O $_{2}$ are available.Correct
- C
  $C_{3} H_{8}$ , because the mole ratio requires more propane than available.
  Why not C: Nearly 1 mol C $_{3}$ H $_{8}$ is available; O $_{2}$ at 4 mol requires 0.8 mol C $_{3}$ H $_{8}$ — propane is in excess.
- D
  Neither; both are present in stoichiometric amounts.
  Why not D: 1:5 ratio requires 5 mol O $_{2}$ per mol propane; 4 mol O $_{2}$ : ~1 mol propane is not stoichiometric.
Explanation
Moles C $_{3}$ H $_{8}$ = 44.0/44.1 ≈ 0.998 mol; moles O $_{2}$ = 128/32.0 = 4.00 mol.

$\frac{0.998}{1} = 0.998$ for propane; $\frac{4.00}{5} = 0.800$ for O $_{2}$ .

Smaller ratio → O $_{2}$ is limiting.

Key takeaway
To find limiting reagent: divide available moles by stoichiometric coefficient. The smallest quotient identifies the limiting reagent.