AP Chemical Reactions and Stoichiometry 2

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?

Accepted Answer

I⁻ is in excess by 0.725 mmol

Answer

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.

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)

Accepted Answer

1.80 L

Answer

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.

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)?

Accepted Answer

105 g

Answer

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.

Question 4

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

Accepted Answer

Single displacement (single replacement) reaction

Answer

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.

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?

Accepted Answer

CH₂

Answer

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.

AP Practice Test

AP Practice Test

AP Chemical Reactions and Stoichiometry 2