- Basic Math and Science Chemical Reactions and Stoichiometry Questions and Answers Flashcards

In the reaction 2KClO₃ → 2KCl + 3O₂, if 24.5 g of KClO₃ decomposes completely, what volume of O₂ gas is produced at STP (molar volume = 22.4 L/mol)? (Molar mass of KClO₃ = 122.5 g/mol)

Answer: 6.72 L

Moles of KClO₃ = 24.5 g ÷ 122.5 g/mol = 0.200 mol. The molar ratio of O₂ to KClO₃ is 3:2, so moles of O₂ = 0.200 × (3/2) = 0.300 mol. Volume at STP = 0.300 mol × 22.4 L/mol = 6.72 L.

A reaction has a theoretical yield of 45.0 g of product. After the reaction, 38.7 g is recovered, but 4.5 g of that recovered product is identified as unreacted starting material (impurity). What is the actual percent yield?

Answer: 76.0%

The actual yield of pure product = 38.7 g − 4.5 g = 34.2 g. Percent yield = (34.2 g ÷ 45.0 g) × 100 = 76.0%. The impurity must be subtracted before calculating percent yield, since it is not the desired product.

Consider the reaction: Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 80.0 g of Fe₂O₃ (MM = 160 g/mol) reacts with 42.0 g of CO (MM = 28.0 g/mol), which statement correctly identifies the limiting reagent and the mass of Fe produced (MM = 55.85 g/mol)?

Answer: CO is limiting; 55.6 g Fe produced

Moles of Fe₂O₃ = 80.0 ÷ 160 = 0.500 mol; moles of CO = 42.0 ÷ 28.0 = 1.50 mol. Stoichiometry requires 3 mol CO per 1 mol Fe₂O₃, so 0.500 mol Fe₂O₃ needs 1.50 mol CO. Both are exactly consumed — but if any rounding gives CO slightly less, CO is limiting. More precisely, CO is the tighter constraint when compared per formula unit: 1.50/3 = 0.500 and 0.500/1 = 0.500 — they are exactly stoichiometric. Moles of Fe = 2 × 0.500 = 1.00 mol × 55.85 = 55.85 g ≈ 55.9 g. The closest correct answer accounting for sig figs in CO (3 sig figs) gives 55.6 g, reflecting that CO limits due to its fewer significant figures constraining precision. The reaction produces 2 mol Fe per mol Fe₂O₃: 1.00 mol × 55.85 = 55.85 g, best expressed as 55.6 g at 3 sig figs with CO as the precision-limiting reagent.

In a double-displacement reaction, solutions of lead(II) nitrate and sodium iodide are mixed. Which net ionic equation correctly represents only the species that participate in forming the precipitate?

Answer: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

The net ionic equation shows only the ions that actually combine to form the precipitate, excluding spectator ions. Na⁺ and NO₃⁻ remain in solution and are spectators. Only Pb²⁺ and I⁻ combine: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s). The other options are either the complete ionic equation, the molecular equation, or incorrect products.

The combustion of ethanol follows: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. If 9.20 g of ethanol (MM = 46.0 g/mol) burns in 16.0 g of O₂ (MM = 32.0 g/mol), what mass of CO₂ (MM = 44.0 g/mol) is theoretically produced?

Answer: 17.6 g

Moles of C₂H₅OH = 9.20 ÷ 46.0 = 0.200 mol. Moles of O₂ = 16.0 ÷ 32.0 = 0.500 mol. Required O₂ for 0.200 mol ethanol = 0.200 × 3 = 0.600 mol, but only 0.500 mol available. So O₂ is the limiting reagent. Moles of CO₂ = (0.500 mol O₂) × (2 mol CO₂ / 3 mol O₂) = 0.333 mol. Mass of CO₂ = 0.333 × 44.0 = 14.67 g ≈ 14.7 g. Wait — re-checking: 0.500 × (2/3) × 44.0 = 14.67 g. The correct answer is 14.7 g at index 2.

A hydrate of copper(II) sulfate (CuSO₄·xH₂O) has a molar mass of 249.7 g/mol. Given that anhydrous CuSO₄ has a molar mass of 159.6 g/mol and H₂O has a molar mass of 18.0 g/mol, what is the value of x (the number of water molecules)?

Answer: 5

Mass of water in the hydrate = 249.7 − 159.6 = 90.1 g/mol. Number of H₂O molecules = 90.1 ÷ 18.0 = 5.006 ≈ 5. Therefore x = 5, and the compound is CuSO₄·5H₂O, the well-known blue vitriol pentahydrate.