AP - Advanced Placement Chemical Reactions and Stoichiometry 1 — Questions and Answers
Question 1: In the reaction 2Al(s) + 3Cl₂(g) → 2AlCl₃(s), 5.40 g of Al (MM = 26.98 g/mol) reacts with excess Cl₂, yielding 21.3 g of AlCl₃ (MM = 133.33 g/mol). What is the percent yield of this reaction?
- 62.3%
- 79.8% (Correct answer)
- 83.4%
- 95.2%
Correct answer: 79.8%
Moles of Al = 5.40/26.98 = 0.2002 mol. The 1:1 molar ratio gives 0.2002 mol AlCl₃ theoretical, or 0.2002 × 133.33 = 26.7 g. Percent yield = (21.3/26.7) × 100 = 79.8%.
Question 2: Which net ionic equation correctly represents the reaction between Na₂SO₄(aq) and BaCl₂(aq)?
- 2Na⁺(aq) + SO₄²⁻(aq) → Na₂SO₄(s)
- Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) (Correct answer)
- 2Cl⁻(aq) + 2Na⁺(aq) → 2NaCl(s)
- Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2NaCl(s)
Correct answer: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Na⁺ and Cl⁻ are spectator ions and cancel out. Only the insoluble precipitate BaSO₄ forms, so the net ionic equation involves only Ba²⁺ and SO₄²⁻. The last option is the complete ionic equation, not the net ionic equation.
Question 3: In a titration, 24.5 mL of 0.150 M NaOH is required to neutralize 15.0 mL of H₂SO₄ solution. Using the reaction 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, what is the molarity of the H₂SO₄?
- 0.0613 M
- 0.123 M (Correct answer)
- 0.245 M
- 0.300 M
Correct answer: 0.123 M
Moles NaOH = 0.0245 L × 0.150 mol/L = 3.675 × 10⁻³ mol. From the 2:1 mole ratio, moles H₂SO₄ = 3.675 × 10⁻³ / 2 = 1.8375 × 10⁻³ mol. Molarity = 1.8375 × 10⁻³ mol / 0.0150 L = 0.123 M.
Question 4: Propane burns completely according to: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). If 11.0 g of C₃H₈ (MM = 44.11 g/mol) is combusted with excess O₂, what mass of CO₂ (MM = 44.01 g/mol) is produced?
- 11.0 g
- 22.0 g
- 32.9 g (Correct answer)
- 44.0 g
Correct answer: 32.9 g
Moles C₃H₈ = 11.0/44.11 = 0.2493 mol. Using the 1:3 mole ratio, moles CO₂ = 0.2493 × 3 = 0.7479 mol. Mass CO₂ = 0.7479 × 44.01 = 32.9 g.
Question 5: Which of the following represents a single-displacement (single-replacement) reaction?
- CaCO₃(s) → CaO(s) + CO₂(g)
- 2Na(s) + Cl₂(g) → 2NaCl(s)
- Zn(s) + 2AgNO₃(aq) → Zn(NO₃)₂(aq) + 2Ag(s) (Correct answer)
- HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Correct answer: Zn(s) + 2AgNO₃(aq) → Zn(NO₃)₂(aq) + 2Ag(s)
In a single-displacement reaction, one element replaces another in a compound. Here, Zn replaces Ag⁺ ions. Option A is decomposition, option B is synthesis (combination), and option D is double displacement (neutralization).
Question 6: In the reaction 3CuCl₂ + 2Al → 3Cu + 2AlCl₃, 25.4 g of CuCl₂ (MM = 134.45 g/mol) is combined with 5.40 g of Al (MM = 26.98 g/mol). Which reagent is limiting, and what is the theoretical yield of Cu (MM = 63.55 g/mol)?
- CuCl₂ is limiting; 12.0 g Cu (Correct answer)
- Al is limiting; 19.1 g Cu
- CuCl₂ is limiting; 8.01 g Cu
- Al is limiting; 12.0 g Cu
Correct answer: CuCl₂ is limiting; 12.0 g Cu
Moles CuCl₂ = 25.4/134.45 = 0.1889 mol; moles Al = 5.40/26.98 = 0.2002 mol. CuCl₂ requires Al in a 3:2 ratio, so 0.1889 mol CuCl₂ needs only 0.1259 mol Al. Since 0.2002 mol Al is available, CuCl₂ is limiting. Moles Cu = 0.1889 mol (1:1 ratio with CuCl₂); mass Cu = 0.1889 × 63.55 = 12.0 g.
AP - Advanced Placement Chemical Reactions and Stoichiometry 1
In the reaction 2Al(s) + 3Cl₂(g) → 2AlCl₃(s), 5.40 g of Al (MM = 26.98 g/mol) reacts with excess Cl₂, yielding 21.3 g of AlCl₃ (MM = 133.33 g/mol).
What is the percent yield of this reaction?