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HKDSE Chemistry Practice Paper: Stoichiometry, Acids and Electrochemistry (12 Questions, 100 Marks, 75 Minutes)
HKDSE Chemistry: Stoichiometry and Acids
HKDSE Chemistry practice paper: 12 questions, 100 marks, 75 minutes on moles, titration, pH, redox, electrolysis, enthalpy, equilibrium and ionic compounds.
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Paper details Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12
Section A: Stoichiometry and Molar Volume of Gases (Questions 1-4, 32 marks)
Each question carries 8 marks; write the equation or mole relationship, show the working with units and give final answers to three significant figures.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A 25.0 g sample of pure calcium carbonate (CaCO3, Mr = 100.0) is strongly heated in air until it decomposes completely: CaCO3(s) -> CaO(s) + CO2(g). (a) Calculate the amount of CaCO3. (b) Calculate the volume of CO2 produced at room temperature and pressure (molar volume 24.0 dm3 mol-1). (c) Calculate the mass of the solid residue and use conservation of atoms to justify your answer. | (a) n(CaCO3) = 25.0 / 100.0 = 0.250 mol. (b) From the equation n(CO2) = n(CaCO3) = 0.250 mol, so V = 0.250 x 24.0 = 6.00 dm3. (c) n(CaO) = 0.250 mol, mass = 0.250 x 56.0 = 14.0 g; check: 14.0 g of CaO contains 0.250 mol of calcium, the same as in the reactant, so atoms are conserved. [Marks: amount 3; gas volume 3 with 24.0 dm3 mol-1 shown; residue mass and check 2.] |
| 2 | 25.0 cm3 of 0.100 M sodium hydroxide solution is neutralised by 0.0500 M sulphuric acid: 2NaOH + H2SO4 -> Na2SO4 + 2H2O. (a) Calculate the amount of NaOH. (b) Calculate the volume of sulphuric acid needed (cm3). (c) Calculate the mass of sulphuric acid used (Mr of H2SO4 = 98.0). | (a) n(NaOH) = 0.100 x 0.0250 = 2.50×10^-3 mol. (b) From the equation n(H2SO4) = n(NaOH) / 2 = 1.25×10^-3 mol, so V = 1.25×10^-3 / 0.0500 = 0.0250 dm3 = 25.0 cm3. (c) Mass = 1.25×10^-3 x 98.0 = 0.123 g. [Marks: amount of NaOH 2; amount and volume of acid 3; mass 3.] |
| 3 | 4.60 g of a compound containing only carbon, hydrogen and oxygen is burnt completely, giving 8.80 g of carbon dioxide and 5.40 g of water. (a) Calculate the masses of carbon and hydrogen in the compound. (b) Calculate the mass of oxygen and the mole ratio of carbon to hydrogen to oxygen. (c) Write the empirical formula; if the molar mass of the compound is 46.0 g mol-1, write the molecular formula and name the compound. | (a) n(CO2) = 8.80 / 44.0 = 0.200 mol, so mass of carbon = 0.200 x 12.0 = 2.40 g; n(H2O) = 5.40 / 18.0 = 0.300 mol, so mass of hydrogen = 0.600 x 1.0 = 0.600 g. (b) Mass of oxygen = 4.60 – 2.40 – 0.600 = 1.60 g, n(O) = 0.100 mol, so n(C):n(H):n(O) = 0.200:0.600:0.100 = 2:6:1. (c) The empirical formula is C2H6O with formula mass 46.0, equal to the given molar mass, so the molecular formula is C2H6O and the compound is ethanol. [Marks: masses of C and H 3; mass of O and ratio 3; empirical formula, molecular formula and name 2.] |
| 4 | A gas X has a mass of 0.320 g and occupies 240 cm3 at room temperature and pressure (molar volume 24.0 dm3 mol-1; Avogadro constant 6.02×10^23 mol-1). (a) Calculate the amount of X. (b) Calculate the molar mass of X and deduce which elemental gas it is. (c) Calculate the density of X under these conditions (g dm-3) and the number of molecules in 240 cm3. | (a) n(X) = 0.240 / 24.0 = 0.0100 mol. (b) Mr = 0.320 / 0.0100 = 32.0; the elemental gas with molar mass 32.0 g mol-1 is oxygen, O2. (c) Density = 0.320 / 0.240 = 1.33 g dm-3; number of molecules = 0.0100 x 6.02×10^23 = 6.02×10^21. [Marks: amount 2; molar mass and deduction 3; density and molecule count 3.] |
Section B: Acids and Bases, Redox, Electrolysis and Enthalpy Change (Questions 5-8, 36 marks)
Each question carries 9 marks; use the correct electron transfer numbers and the Faraday constant, and keep signs and units consistent with what the question asks.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 25.0 cm3 of 0.100 M hydrochloric acid is mixed with 15.0 cm3 of 0.100 M sodium hydroxide solution; assume the volumes are additive. (a) Calculate the amounts of H+ and OH- before mixing. (b) State whether the mixture is acidic, alkaline or neutral and calculate the concentration of the excess ion. (c) Calculate the pH of the mixture (log 2.5 = 0.40). | (a) n(H+) = 0.100 x 0.0250 = 2.50×10^-3 mol; n(OH-) = 0.100 x 0.0150 = 1.50×10^-3 mol. (b) H+ is in excess by 1.00×10^-3 mol, so the mixture is acidic; total volume 40.0 cm3, so [H+] = 1.00×10^-3 / 0.0400 = 0.0250 M. (c) pH = -log(0.0250) = 2 – 0.40 = 1.60. [Marks: the two amounts 3; acidic with excess concentration 3; pH 3.] |
| 6 | In acidic conditions, 25.0 cm3 of iron(II) sulphate solution is titrated with 0.0200 M potassium permanganate solution and 20.0 cm3 is required. MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O. (a) Calculate the amount of MnO4- used. (b) Calculate the concentration of Fe2+ in the solution. (c) Calculate the mass of iron present in 250 cm3 of the same solution (Fe = 55.8). | (a) n(MnO4-) = 0.0200 x 0.0200 = 4.00×10^-4 mol. (b) From the equation n(Fe2+) = 5 x 4.00×10^-4 = 2.00×10^-3 mol, so [Fe2+] = 2.00×10^-3 / 0.0250 = 0.0800 M. (c) In 250 cm3, n(Fe2+) = 0.0800 x 0.250 = 0.0200 mol, so the mass of iron is 0.0200 x 55.8 = 1.12 g. [Marks: amount of MnO4- 3; concentration of Fe2+ 3 using the 1:5 ratio; mass of iron 3.] |
| 7 | A current of 2.00 A is passed for 30.0 minutes through copper(II) sulphate solution (Faraday constant 96500 C mol-1, Cu = 63.5). (a) Calculate the quantity of charge passed. (b) Calculate the mass of copper deposited at the cathode. (c) Calculate the amount of oxygen liberated at the anode and its volume at room temperature and pressure (24.0 dm3 mol-1). | (a) Q = I x t = 2.00 x 1800 = 3600 C. (b) n(e-) = 3600 / 96500 = 0.0373 mol, so n(Cu) = 0.0373 / 2 = 0.0187 mol and mass = 0.0187 x 63.5 = 1.18 g. (c) At the anode 2H2O -> O2 + 4H+ + 4e-, so n(O2) = 0.0373 / 4 = 9.33×10^-3 mol and V = 9.33×10^-3 x 24.0 = 0.224 dm3, that is 224 cm3. [Marks: charge 2; mass of copper 4; amount and volume of oxygen 3.] |
| 8 | Burning 0.500 g of methanol (CH3OH, Mr = 32.0) raises the temperature of 200 g of water from 20.0 C to 47.0 C (specific heat capacity of water 4.2 J g-1 K-1). (a) Calculate the heat absorbed by the water in kJ. (b) Calculate the amount of methanol burnt. (c) Calculate the enthalpy change of combustion per mole of methanol in kJ mol-1 with its sign, and give one reason why the experimental value differs from the literature value. | (a) Heat = 200 x 4.2 x 27.0 = 22680 J = 22.68 kJ. (b) n(CH3OH) = 0.500 / 32.0 = 0.0156 mol. (c) dH = -22.68 / 0.015625 = -1452 kJ mol-1, so about 1452 kJ is released per mole of methanol burnt. A reason for the difference: some heat is lost to the surroundings and the container, or the methanol is not completely burnt and carbon monoxide forms. [Marks: heat 3; amount 3; enthalpy change with sign and reason 3.] |
Section C: Equilibrium, Reaction Rate, an Alcohol and an Ionic Compound (Questions 9-12, 32 marks)
Each question carries 8 marks; show the working for equilibrium and rate calculations, and write structural formulae and ion symbols clearly.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | In a 2.00 dm3 vessel, 1.00 mol of hydrogen and 1.00 mol of iodine vapour react at constant temperature: H2(g) + I2(g) ⇌ 2HI(g). At equilibrium the vessel contains 1.60 mol of HI. (a) Calculate the amount of each substance at equilibrium. (b) Calculate the equilibrium concentration of each substance. (c) Calculate the equilibrium constant Kc at this temperature and write the expression for Kc. | (a) 1.60 / 2 = 0.800 mol of iodine reacts, so at equilibrium n(H2) = 0.200 mol, n(I2) = 0.200 mol and n(HI) = 1.60 mol. (b) With a volume of 2.00 dm3, [H2] = [I2] = 0.200 / 2.00 = 0.100 M and [HI] = 1.60 / 2.00 = 0.800 M. (c) Kc = [HI]^2 / ([H2][I2]) = 0.800^2 / (0.100 x 0.100) = 0.640 / 0.0100 = 64.0. [Marks: equilibrium amounts 3; concentrations 2; Kc expression and value 3.] |
| 10 | 0.500 g of zinc granules is added to excess 1.00 M hydrochloric acid: Zn + 2HCl -> ZnCl2 + H2 (Zn = 65.4, molar volume 24.0 dm3 mol-1). (a) Calculate the amount of zinc. (b) Calculate the volume of hydrogen released on complete reaction (cm3). (c) If 50.0 cm3 of hydrogen is collected in the first 25.0 s, calculate the average rate (cm3 s-1) and explain why the rate falls with time. | (a) n(Zn) = 0.500 / 65.4 = 7.65×10^-3 mol. (b) From the equation n(H2) = n(Zn), so V = 7.65×10^-3 x 24.0 = 0.1835 dm3, that is 183.5 cm3. (c) Average rate = 50.0 / 25.0 = 2.00 cm3 s-1; the rate falls because the acid becomes less concentrated and the surface area of the zinc decreases, so collisions are less frequent. [Marks: amount of zinc 2; volume of hydrogen 3; rate and explanation 3.] |
| 11 | 0.740 g of a saturated monohydric alcohol reacts with excess sodium to give 120 cm3 of hydrogen at room temperature and pressure (molar volume 24.0 dm3 mol-1): 2ROH + 2Na -> 2RONa + H2. (a) Calculate the amount of hydrogen. (b) Calculate the amount and the molar mass of the alcohol. (c) Deduce the molecular formula and the structural formula of the alcohol. | (a) n(H2) = 0.120 / 24.0 = 5.00×10^-3 mol. (b) From the equation n(ROH) = 2 x 5.00×10^-3 = 0.0100 mol, so Mr = 0.740 / 0.0100 = 74.0. (c) A saturated monohydric alcohol has the general formula CnH2n+2O, so 14n + 18 = 74.0 and n = 4; the molecular formula is C4H10O and the structural formula is CH3CH2CH2CH2OH (butan-1-ol). [Marks: amount of hydrogen 2; amount and molar mass of the alcohol 4; molecular and structural formula 2.] |
| 12 | Element X has atomic number 12 and forms an ionic compound with oxygen. (a) Write the electron arrangement of X and the symbol of the ion it forms. (b) 0.243 g of X reacts completely with oxygen to give 0.403 g of an oxide; deduce the empirical formula of the oxide (X = 24.3, O = 16.0). (c) Calculate the molar mass of the oxide and write the equation for its reaction with water. | (a) The electron arrangement of X is 2, 8, 2; it loses the two outermost electrons to form an X2+ ion. (b) n(X) = 0.243 / 24.3 = 0.0100 mol; mass of oxygen = 0.403 – 0.243 = 0.160 g, so n(O) = 0.0100 mol and n(X):n(O) = 1:1, giving the empirical formula XO. (c) Molar mass = 24.3 + 16.0 = 40.3 g mol-1; the oxide is magnesium oxide, which reacts with water as MgO + H2O -> Mg(OH)2. [Marks: electron arrangement and ion 3; empirical formula 3; molar mass and equation 2.] |
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