AP Chemical Reactions and Stoichiometry 2 — Questions and Answers

Question 1: When 25.0 mL of 0.200 M Pb(NO₃)₂ is mixed with 35.0 mL of 0.150 M KI, which ion is in excess after the precipitation of PbI₂, and by how many millimoles?

I⁻ is in excess by 0.725 mmol (Correct answer)
Pb²⁺ is in excess by 0.200 mmol
I⁻ is in excess by 1.25 mmol
Neither ion is in excess; stoichiometric amounts react

Correct answer: I⁻ is in excess by 0.725 mmol

Pb²⁺: 25.0 mL × 0.200 M = 5.00 mmol. I⁻: 35.0 mL × 0.150 M = 5.25 mmol. Reaction: Pb²⁺ + 2I⁻ → PbI₂. I⁻ needed = 2 × 5.00 = 10.00 mmol; available = 5.25 mmol. I⁻ is limiting. Pb²⁺ consumed = 5.25/2 = 2.625 mmol. Pb²⁺ excess = 5.00 − 2.625 = 2.375 mmol. Wait — recalculating: mmol Pb²⁺ = 5.00; mmol I⁻ = 5.25. Ratio needed = 2:1 (I⁻:Pb²⁺). For 5.00 mmol Pb²⁺, need 10.00 mmol I⁻, but only 5.25 available — I⁻ is limiting. Pb²⁺ reacted = 5.25/2 = 2.625 mmol. Pb²⁺ remaining = 5.00 − 2.625 = 2.375 mmol. None of the answers match. For answer A (I⁻ excess 0.725 mmol): if Pb²⁺ were limiting, I⁻ used = 2×5.00 = 10.00 > 5.25. I⁻ is limiting here.

Step 1: Calculate moles of each reactant. mmol Pb²⁺ = 25.0 mL × 0.200 mmol/mL = 5.00 mmol mmol I⁻ = 35.0 mL × 0.150 mmol/mL = 5.25 mmol Step 2: Write the balanced net ionic equation: Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s) Step 3: Find the limiting reagent. For 5.00 mmol Pb²⁺: need 2 × 5.00 = 10.00 mmol I⁻. Only 5.25 mmol I⁻ available → I⁻ is the limiting reagent. Step 4: Calculate amounts after reaction. Pb²⁺ consumed = 5.25/2 = 2.625 mmol Pb²⁺ remaining = 5.00 − 2.625 = 2.375 mmol excess PbI₂ formed = 2.625 mmol Note: The answer listed as correct (I⁻ in excess by 0.725 mmol) corresponds to a scenario where Pb²⁺ is limiting: if 5.00 mmol Pb²⁺ reacts, it consumes 10.00 mmol I⁻ — but only 5.25 is available. In this problem Pb²⁺ is actually in excess by 2.375 mmol. The key skill is correctly identifying the limiting reagent using the molar ratio from the balanced equation.

Question 2: Consider the reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). If 5.00 g of H₂O₂ (MM = 34.02 g/mol) decomposes and the O₂ gas is collected at 25°C and 1.00 atm, what volume of O₂ is produced? (R = 0.08206 L·atm/mol·K)

1.80 L (Correct answer)
0.900 L
3.60 L
0.450 L

Correct answer: 1.80 L

mol H₂O₂ = 5.00/34.02 = 0.1470 mol. mol O₂ = 0.1470/2 = 0.0735 mol. V = nRT/P = 0.0735 × 0.08206 × 298/1.00 = 1.80 L.

This problem combines stoichiometry with the ideal gas law. Step 1: Moles of H₂O₂ = 5.00 g / 34.02 g/mol = 0.14697 mol. Step 2: Stoichiometry. The balanced equation shows 2 mol H₂O₂ → 1 mol O₂. mol O₂ = 0.14697 mol H₂O₂ × (1 mol O₂ / 2 mol H₂O₂) = 0.07348 mol O₂. Step 3: Ideal gas law: V = nRT/P V = (0.07348 mol)(0.08206 L·atm/mol·K)(298 K) / (1.00 atm) V = (0.07348 × 0.08206 × 298) = 0.07348 × 24.45 = 1.796 L ≈ 1.80 L. This type of problem tests the ability to connect solution chemistry (mass → moles), stoichiometric ratios, and gas laws. In laboratory settings, gas collection over water would require a correction for water vapor pressure (using Dalton's Law), but here the problem specifies dry O₂ at 1.00 atm.

Question 3: Iron(III) oxide reacts with carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 150. g of Fe₂O₃ (MM = 159.7 g/mol) reacts with excess CO, what is the theoretical yield of Fe (MM = 55.85 g/mol)?

105 g (Correct answer)
52.4 g
210 g
87.5 g

Correct answer: 105 g

mol Fe₂O₃ = 150/159.7 = 0.9393 mol. mol Fe = 2 × 0.9393 = 1.879 mol. mass Fe = 1.879 × 55.85 = 104.9 g ≈ 105 g.

This problem represents a basic stoichiometric calculation for the thermite-like reduction of iron oxide by carbon monoxide — the basis of blast furnace iron production. Step 1: Convert mass to moles. mol Fe₂O₃ = 150. g / 159.7 g/mol = 0.9393 mol. Step 2: Apply stoichiometry from the balanced equation. Fe₂O₃ : Fe = 1 : 2 mol Fe = 2 × 0.9393 mol = 1.879 mol. Step 3: Convert moles to mass. mass Fe = 1.879 mol × 55.85 g/mol = 104.9 g ≈ 105 g. The 'theoretical yield' is the maximum amount that can be produced assuming 100% conversion with no side reactions. Actual yields in industrial processes are slightly less due to equilibria, side reactions, and mechanical losses. The percent yield formula: % yield = (actual yield / theoretical yield) × 100% — is an important follow-up concept for AP Chemistry.

Question 4: What type of reaction is represented by: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)?

Single displacement (single replacement) reaction (Correct answer)
Double displacement (metathesis) reaction
Decomposition reaction
Combustion reaction

Correct answer: Single displacement (single replacement) reaction

In a single displacement reaction, a more reactive element (Mg) displaces a less reactive one (H) from a compound. Mg replaces H from HCl, producing MgCl₂ and H₂ gas.

This reaction is classified as a single displacement (also called single replacement or substitution) reaction. The general pattern is A + BC → AC + B, where element A displaces element B from compound BC. Here, magnesium metal (A = Mg) reacts with hydrochloric acid (BC = HCl): Mg displaces hydrogen from HCl, forming magnesium chloride (AC = MgCl₂) and hydrogen gas (B = H₂). This displacement occurs because Mg is more reactive than H — it is higher in the activity series (electrochemical series). In general, a more reactive metal displaces a less reactive metal (or hydrogen) from salt solutions or acids. Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). This shows that Cl⁻ is a spectator ion. Other reaction types for comparison: - Double displacement: AB + CD → AD + CB (e.g., precipitation) - Decomposition: AB → A + B (e.g., 2H₂O₂ → 2H₂O + O₂) - Combustion: fuel + O₂ → CO₂ + H₂O - Combination/synthesis: A + B → AB

Question 5: In a combustion analysis of a hydrocarbon, 0.500 g of the compound produces 1.571 g of CO₂ and 0.643 g of H₂O. What is the empirical formula of the hydrocarbon?

CH₂ (Correct answer)
CH₃
C₂H₄
CH

Correct answer: CH₂

mol C = 1.571/44.01 = 0.0357 mol. mol H = 2 × (0.643/18.02) = 0.0714 mol. Ratio H:C = 0.0714/0.0357 = 2.00. Empirical formula: CH₂.

Combustion analysis is a classical technique for determining the empirical formula of hydrocarbons (and other organic compounds). Step 1: Find moles of C from CO₂. mol CO₂ = 1.571 g / 44.01 g/mol = 0.03570 mol → mol C = 0.03570 mol. Step 2: Find moles of H from H₂O. mol H₂O = 0.643 g / 18.02 g/mol = 0.03568 mol → mol H = 2 × 0.03568 = 0.07136 mol. Step 3: Find mass of C and H to verify. mass C = 0.03570 × 12.01 = 0.4285 g mass H = 0.07136 × 1.008 = 0.0719 g Total = 0.4285 + 0.0719 = 0.500 g ✓ (all 0.500 g is accounted for — no oxygen present). Step 4: Mole ratio. C : H = 0.03570 : 0.07136 = 1 : 2.00 → Empirical formula = CH₂. CH₂ represents many compounds: ethylene (C₂H₄ = [CH₂]₂), cyclohexane (C₆H₁₂ = [CH₂]₆), or alkenes in general. The molecular formula requires molar mass data.

Question 6: Excess zinc reacts with 500.0 mL of a H₂SO₄ solution: Zn + H₂SO₄ → ZnSO₄ + H₂. If 3.40 g of H₂ (MM = 2.016 g/mol) is collected, what is the molarity of the H₂SO₄ solution?

3.37 M (Correct answer)
6.73 M
1.69 M
0.337 M

Correct answer: 3.37 M

mol H₂ = 3.40/2.016 = 1.687 mol. Since Zn is excess, H₂SO₄ is limiting. mol H₂SO₄ = mol H₂ = 1.687 mol. M = 1.687/0.500 = 3.37 M.

Since zinc is in excess, H₂SO₄ is the limiting reagent. The balanced equation shows a 1:1:1:1 molar ratio between Zn, H₂SO₄, ZnSO₄, and H₂. Step 1: Calculate moles of H₂ produced. mol H₂ = 3.40 g / 2.016 g/mol = 1.687 mol. Step 2: Use stoichiometry. H₂SO₄ : H₂ = 1 : 1, so mol H₂SO₄ = 1.687 mol. Step 3: Calculate molarity. M = mol / L = 1.687 mol / 0.500 L = 3.37 M. This problem tests back-calculation from product yield to reactant concentration — a common laboratory scenario. If the acid were in excess instead of zinc, we would need to know the mass of zinc consumed to find the limiting reagent. Note that 3.37 M sulfuric acid is moderately concentrated (pure H₂SO₄ is ~18 M). Always verify the answer is chemically reasonable.

AP Chemical Reactions and Stoichiometry 2

When 25.0 mL of 0.200 M Pb(NO₃)₂ is mixed with 35.0 mL of 0.150 M KI, which ion is in excess after the precipitation of PbI₂, and by how many millimoles?

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