Introduction to Quantitative Chemistry — complete assessment covering all three inquiry questions across L01–L20. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
How many molecules are in 0.500 mol of CO₂? (Nₐ = 6.022×10²³)
What mass of H₂O contains 2.00 mol? (H = 1.008, O = 15.999)
A compound is 40.0% C, 6.72% H, and 53.3% O by mass. Its empirical formula is:
What volume does 3.00 mol of N₂ occupy at RTP?
0.250 mol of NaOH is dissolved to make 500 mL of solution. What is the concentration?
50.0 mL of 2.00 mol/L HCl is diluted to 400 mL. What is the new concentration?
AgNO₃ + NaCl → AgCl↓ + NaNO₃. If 25.0 mL of AgNO₃ produces 0.717 g of AgCl, what is [AgNO₃]? (MM AgCl = 143.32)
25.0 mL NaOH is titrated with 0.100 mol/L HCl. Titres: Rough = 24.5, T1 = 22.4, T2 = 22.3, T3 = 22.5 mL. What is [NaOH]?
In 2Al + 3Cl₂ → 2AlCl₃, if 6.00 mol of Al reacts completely, how many moles of Cl₂ are consumed?
In 2H₂ + O₂ → 2H₂O, what mass of water forms from 4.00 g of H₂? (H=1.008, O=15.999)
In Fe + S → FeS, 11.2 g of Fe and 8.00 g of S are mixed. Which is the limiting reagent? (Fe=55.845, S=32.06)
Theoretical yield = 25.0 g, actual yield = 19.0 g. What is the percentage yield?
CaCO₃ → CaO + CO₂. What volume of CO₂ at STP forms from 50.05 g of CaCO₃? (Ca=40.078, C=12.011, O=15.999)
In AgNO₃ + NaCl → AgCl + NaNO₃, 40.0 mL of 0.100 mol/L AgNO₃ is used. What mass of AgCl forms? (MM AgCl = 143.32)
25.0 mL H₂SO₄ is titrated with 0.200 mol/L NaOH. Titre = 20.0 mL. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is [H₂SO₄]?
A compound has the following percentage composition: 27.3% C, 72.7% O. (a) Determine the empirical formula. (b) The molar mass of the compound is 88.0 g/mol. Determine the molecular formula. (c) Calculate the percentage by mass of carbon in the compound. (C = 12.011, O = 15.999)
A student has 100.0 mL of concentrated HCl at 12.0 mol/L. (a) She dilutes 20.0 mL of this stock solution to 500.0 mL. Calculate the concentration of the diluted solution. (b) The concentrated HCl is 98.0% pure by mass (density = 1.18 g/mL). What mass of pure HCl is in 20.0 mL of the concentrated solution? (MM HCl = 36.46)
In 2Al + 3Cl₂ → 2AlCl₃, 8.10 g of Al reacts with 14.2 g of Cl₂. (a) Identify the limiting reagent, showing the full comparison. (b) Calculate the theoretical yield of AlCl₃. (c) If 21.3 g of AlCl₃ is collected, calculate the percentage yield. (Al = 26.982, Cl = 35.453)
Barium chloride reacts with sodium sulfate in solution: BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl. 40.0 mL of 0.250 mol/L BaCl₂ is mixed with 60.0 mL of 0.150 mol/L Na₂SO₄. (a) Identify the limiting reagent, showing your full comparison. (b) Calculate the mass of BaSO₄ precipitate. (c) Calculate the concentration of the excess reagent remaining in the 100.0 mL solution. (Ba = 137.33, S = 32.06, O = 15.999)
A student prepares a standard solution by dissolving 0.795 g of anhydrous Na₂CO₃ (MM = 105.99) in water and making up to 150.0 mL. She titrates 25.0 mL aliquots against HCl solution. Titres: Rough = 18.9 mL, T1 = 20.8 mL, T2 = 20.7 mL, T3 = 20.9 mL. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Calculate the average concordant titre. (b) Calculate [HCl].
(a) Discard rough (18.9). T1=20.8, T2=20.7, T3=20.9 — range = 0.2 mL; all concordant. Average = (20.8+20.7+20.9)÷3 = 20.8 mL
(b) n(Na₂CO₃) = 0.795÷105.99 = 7.500×10⁻³ mol c(Na₂CO₃) = 7.500×10⁻³÷0.150 = 0.05000 mol/L n(aliquot) = 0.05000 × 0.0250 = 1.250×10⁻³ mol Ratio 1:2; n(HCl) = 1.250×10⁻³ × 2 = 2.500×10⁻³ mol c(HCl) = 2.500×10⁻³ ÷ 0.02080 = 0.120 mol/L Marks: 1 — correct average titre | 1 — n(Na₂CO₃) and c | 1 — n(aliquot) | 1 — ratio applied (×2) | 1 — correct [HCl]