S5 Chemistry Practice Paper: Stoichiometry, Titration, Enthalpy And Organic Chemistry (100 Marks)

S5 Chemistry Practice Paper: Stoichiometry, Titration, Enthalpy and Organic Chemistry (100 marks)

2026-11-12

S5 Chemistry – Calculations and Practical Work

S5 chemistry practice paper: 6 questions, 24 parts, 100 marks, 60 minutes. Covers moles, titration and pH, enthalpy, rates and equilibrium, redox and organic chemistry.

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Paper details　S5 chemistry practice paper on moles, titration, enthalpy, equilibrium and redox, with full answers.

Part 1: Stoichiometry (18 marks)

Four questions. Show all working and units.

#	Question	Answer and marking notes
1	(a) 4.8 g of magnesium (relative atomic mass 24.0) burns completely in oxygen to form magnesium oxide. Find the mass of magnesium oxide formed and the increase in mass of the solid.	Answer: moles of Mg = 4.8 / 24.0 = 0.200 mol. From 2Mg + O2 = 2MgO the moles of MgO equal 0.200, so mass = 0.200 x 40.0 = 8.00 g. The solid gains 3.20 g, the mass of oxygen combined.
2	(b) A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Find its empirical formula.	Answer: in a 100 g sample, C = 85.7 / 12 = 7.14 mol and H = 14.3 / 1.0 = 14.30 mol. The mole ratio is 1 : 2.00, so the empirical formula is CH2 with formula mass 14.
3	(c) 25.0 cm3 of 0.100 M sodium carbonate solution reacts completely with excess dilute hydrochloric acid. Find the moles of carbon dioxide, its volume at room temperature (molar volume 24.0 dm3/mol) and its mass.	Answer: moles of Na2CO3 = 25.0 / 1000 x 0.100 = 0.003 mol. From Na2CO3 + 2HCl = 2NaCl + H2O + CO2 the moles of CO2 are 0.003; volume = 0.003 x 24000 = 60.00 cm3 and mass = 0.003 x 44.0 = 0.11 g.
4	(d) A compound has molar mass 180 g/mol and contains 40.0% carbon and 6.67% hydrogen by mass, the rest being oxygen. Find its molecular formula.	Answer: each mole contains 6 mol C (180 x 0.400 / 12), 12 mol H and 6 mol O from the remaining 96 g. The molecular formula is C6H12O6, glucose.

Part 2: Acid-base titration and pH (18 marks)

Four questions. Give pH values to two decimal places.

#	Question	Answer and marking notes
5	(a) 25.0 cm3 of sodium hydroxide solution is titrated with 0.100 M hydrochloric acid; the mean titre is 22.50 cm3. Find the concentration of the alkali.	Answer: moles of HCl = 22.50 / 1000 x 0.100 = 0.002 mol. NaOH reacts 1:1, so concentration = 0.090 M.
6	(b) The alkali in (a) is made up to 250 cm3. Find the mass of sodium hydroxide present (relative formula mass 40.0).	Answer: moles = 0.090 x 250 / 1000 = 0.022 mol; mass = 0.022 x 40.0 = 0.90 g.
7	(c) At 25 °C (Kw = 1.0 x 10^-14) find the pH of 0.010 M hydrochloric acid and of 0.010 M sodium hydroxide.	Answer: the acid is strong, so [H+] = 0.010 M and pH = 2.00. For the alkali [OH-] = 0.010 M, pOH = 2.00, pH = 14.00 – 2.00 = 12.00.
8	(d) Ammonia has Kb = 1.8 x 10^-5. Find the pH of 0.100 M aqueous ammonia and the volume of 0.100 M hydrochloric acid needed to reach the equivalence point with 25.0 cm3 of it.	Answer: [OH-] = sqrt(Kb x c) = 0.001 M, pOH = 2.87, pH = 11.13. At the equivalence point moles are equal, so acid volume = 0.100 x 25.0 / 0.100 = 25.00 cm3.

Part 3: Enthalpy change and calorimetry (16 marks)

Take the specific heat capacity of water as 4.18 J g^-1 K^-1 and assume no heat loss, with solution density 1.00 g/cm3.

#	Question	Answer and marking notes
9	(a) 50 g of water is heated from 20.0 °C to 32.0 °C. Find the heat absorbed in J and in kJ.	Answer: q = mcDT = 50 x 4.18 x 12.00 = 2,508 J = 2.508 kJ.
10	(b) If that heat came from burning 0.02 mol of a fuel completely, find its enthalpy of combustion in kJ/mol, including sign.	Answer: dH = -q / n = -2.508 / 0.02 = -125.40 kJ/mol, that is about 125 kJ released per mole.
11	(c) Given bond enthalpies H-H 436, O=O 496 and O-H 463 kJ/mol, use them to find the enthalpy change of 2H2 + O2 = 2H2O and the value per mole of water.	Answer: bonds broken absorb 2 x 436 + 496 = 1,368 kJ; bonds formed release 4 x 463 = 1,852 kJ. dH = -484 kJ per equation, that is -242 kJ per mole of water.
12	(d) 50.0 cm3 of 1.00 M hydrochloric acid is mixed with 50.0 cm3 of 1.00 M sodium hydroxide and the temperature rises by 6.70 °C. Find the enthalpy of neutralisation in kJ/mol of water.	Answer: the mixture mass is 100 g, so q = 100 x 4.18 x 6.70 = 2,801 J. Moles of water formed = 0.05, giving dH = -2.801 / 0.05 = -56.01 kJ/mol.

Part 4: Reaction rate and equilibrium (17 marks)

Four questions. State the formula used in each calculation.

#	Question	Answer and marking notes
13	(a) In the reaction A = B the concentration of A falls from 0.80 M to 0.20 M in 40 s. Find the average rate of consumption of A.	Answer: rate = d[A] / dt = (0.80 – 0.20) / 40 = 0.015 M per second.
14	(b) A first order reaction has a half-life of 25 minutes. Find the rate constant and the time for the concentration to fall from 0.40 M to 0.050 M.	Answer: k = ln2 / t-half = 0.693 / 25 = 0.028 min^-1. Going from 0.40 to 0.050 needs 3 half-lives, so the time is 3 x 25 = 75 minutes.
15	(c) At a certain temperature the equilibrium mixture of H2 + I2 = 2HI has [H2] = 0.20 M, [I2] = 0.20 M and [HI] = 1.60 M. Find Kc.	Answer: Kc = [HI]^2 / ([H2][I2]) = 2.56 / 0.040 = 64.00.
16	(d) At another temperature Kc = 4.00. Starting from [H2] = [I2] = 1.00 M and no HI, find the equilibrium [HI], and state the sign of dH with one reason.	Answer: let 2x mol dm-3 of HI form. Then (2x)^2 / (1.00 – x)^2 = 4.00, giving 2x / (1.00 – x) = 2 and x = 0.50, so [HI] = 1.00 M. Since Kc falls from 64.0 to 4.0 as temperature rises, the forward reaction is exothermic and dH is negative.

Part 5: Redox and electrochemistry (16 marks)

Take the Faraday constant as 96 500 C/mol.

#	Question	Answer and marking notes
17	(a) Find the oxidation number of chromium in K2Cr2O7 and of manganese in MnO4-.	Answer: in K2Cr2O7 potassium is +1 and oxygen -2, so 2 + 2x – 14 = 0 and x = +6. In MnO4-, 4 x (-2) + x = -1 gives x = +7.
18	(b) A current of 2.00 A passes through copper(II) sulphate solution for 30 minutes. Find the charge passed and the mass of copper deposited at the cathode (Ar = 63.5).	Answer: Q = I x t = 3,600 C, so moles of electrons = 0.037. Since Cu2+ + 2e- = Cu, the mass of copper is 0.037 / 2 x 63.5 = 1.18 g.
19	(c) Given E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = -0.76 V, find the emf of the zinc-copper cell and write the anode half equation.	Answer: E-cell = 0.34 – (-0.76) = 1.10 V. The anode half equation is Zn = Zn2+ + 2e-.
20	(d) 25.00 cm3 of iron(II) sulphate solution needs 0.020 M potassium manganate(VII) solution, the titre being 24.00 cm3. Given MnO4- is reduced to Mn2+ and Fe2+ is oxidised to Fe3+, find the concentration of iron(II) ions.	Answer: moles of MnO4- = 0.000. From MnO4- + 8H+ + 5Fe2+ = Mn2+ + 4H2O + 5Fe3+, moles of Fe2+ = 5 x 0.000 = 0.002, so the concentration is 0.096 M.

Part 6: Organic chemistry and polymers (15 marks)

Relative atomic masses: H 1.0, C 12.0, O 16.0, Cl 35.5, Br 80.0.

#	Question	Answer and marking notes
21	(a) An alkane contains 5 carbon atoms. Give its molecular formula, relative molecular mass and the number of structural isomers.	Answer: the alkane formula CnH2n+2 gives C5H12 with relative molecular mass 72. Pentane has 3 structural isomers: pentane, 2-methylbutane and 2,2-dimethylpropane.
22	(b) 28.00 g of ethene reacts completely with bromine by addition. Find the moles of ethene and the mass of bromine needed.	Answer: moles of ethene = 28.00 / 28.0 = 1.00 mol. C2H4 + Br2 = C2H4Br2, so the mass of bromine is 1.00 x 160.0 = 160 g.
23	(c) 46.00 g of ethanol and 60.00 g of ethanoic acid are esterified with acid catalysis, giving 66.00 g of ethyl ethanoate. Find the theoretical yield and the percentage yield.	Answer: moles of ethanol and of acid are both 1.00, so the theoretical yield is 1.00 x 88.0 = 88.00 g. Percentage yield = 66.00 / 88.00 x 100% = 75.00%.
24	(d) The monomer of PVC is chloroethene, C2H3Cl, with relative molecular mass 62.50. Find the relative molecular mass of a sample with degree of polymerisation 1,600.	Answer: relative molecular mass = 62.50 x 1,600 = 100,000.

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