JPJC 2026 Mole & Stoichiometry Assignment Answer

2026 JC1 H2 Chemistry (9476) 1 [Turn Over JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) Topical Assignment Topic 2: The Mole Concept & Stoichiometry (Suggested Answers) 1 (a) [N2019 P2 Q2(a)(b) – modified] Hydrogen peroxide can behave as an oxidising agent or a reducing agent depending on the reactant added to it. When solid silver(I) oxide is added to aqueous hydrogen peroxide, the reaction shown in equation 2 takes place. equation 2 Ag2O(s) + H2O2(aq) → 2Ag(s) + O2(g) + H2O(l) Consider the changes in oxidation numbers that occur during this reaction, state whether hydrogen peroxide is behaving as an oxidising agent or a reducing agent in this reaction. Note: As a reducing agent, H2O2 is oxidised to O2. As an oxidising agent, H2O2 is reduced to H2O. Since the oxidation state of Ag decreases from +1 in Ag 2O to 0 in Ag, it implies that Ag2O is reduced by H 2O2 and hence, H2O2 is a reducing agent. [1] (preferred ans.) Or, Since the oxidation state of O increases from −1 in H2O2 to 0 in O2, it implies that H2O2 is oxidised by Ag 2O and hence H2O2 is a reducing agent. [1] [1] (b) (i) 1.0 cm3 of ‘20-volume’ hydrogen peroxide produces 20.0 cm3 of oxygen gas at room temperature and pressure, as shown in equation 3. equation 3 2H2O2(aq) → 2H2O(l) + O2(g) 1.0 cm3 20.0 cm3 Use this information to calculate the concentration, in mol dm−3, of ’20-volume’ hydrogen peroxide. n(O2) produced at r.t.p. = 20.0 24 000 = 8.333  10−4 mol [1] n(H2O2) in 1.0 cm3 = 2  (8.333 × 10−4) = 1.667  10−3 mol [H2O2] = 0.001667  1.0 1000 = 1.667 = 1.67 mol dm−3 [1] 3 s.f. + units [2] (+1) (-1) (0) (0) [R] [O] H2O2 : O2 2 : 1 1 mol of any gas occupies 24 dm3 at r.t.p.

2026 JC1 H2 Chemistry (9476) 2 (ii) Use the value obtained in (b)(i) to calculate the minimum mass of silver(I) oxide which must be used to ensure that 10.0 cm3 of ’20-volume’ hydrogen peroxide reacts completely at room temperature and pressure, according to equation 2. n(H2O2) in 10.0 cm3 = 1.667  1000 10 (OR, 10  1.667  10−3) = 0.01667 mol = n(Ag2O) mass of Ag2O required = 0.01667  231.8 = 3.86 g [1] 3 s.f. + units [1] [Total: 4] 2 [N2020 P3 Q1(f) - modified] Acidified KMnO4 can be used to determine amounts of reducing agents. Sodium nitrite, NaNO 2, is a reducing agent used as a preservative for meat. When NaNO 2 dissolves in water, aqueous nitrite ions, NO2– aq), form. A 1.00 g sample of preserved meat is shaken in deionised water to dissolve the sodium nitrite. A 50.00 cm3 sample of 0.00500 mol dm −3 acidified MnO4–(aq) ions, an excess, is added to the solution. The reaction occurs as shown by the equation 4. All the nitrite ions are oxidised. equation 4 2MnO4–(aq) + 6H+(aq) + 5NO2–(aq) → 2Mn2+(aq) + 3H2O(l) + 5NO3–(aq) An excess of iodide solution is then added to react with the remaining MnO 4– to liberate iodine. { I- + leftover MnO4- → I2  I2 formed is then reacted with S2O32-} The iodine solution formed is titrated against 0.0150 mol dm –3 Na2S2O3. The end-point is reached when 22.40 cm3 of 0.0150 mol dm–3 Na2S2O3 has been added. 2S2O32– + I2 → 2I– + S4O62– (a) Construct an ionic equation between I– ions and acidified MnO4– ions to liberate iodine, I2. { I- + MnO4- + H+ → I2} Note: In acidic medium, MnO4- is reduced to Mn2+. [O] 2I− → I2 + 2e− (5) [R]: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (2)  2MnO4- + 16H+ + 10I- → 2Mn2+ + 8H2O + 5I2 (ans) [1] [1] (b) Calculate the amount of iodine liberated by the remaining MnO4–. Hence, calculate the amount of remaining MnO4– ions after reaction with NO2– ions. n(S2O32-) reacted =  22.400.0150 1000 = 3.36  10-4 mol n(I2) liberated from remaining MnO4- = 1 2  3.36  10-4 = 1.68  10-4 mol [1] n(MnO4-) remained after reaction with NO2- = 2 5  1.68  10-4 = 6.72  10-5 mol [1] 3 s.f. + units [2] S2O32- : I2 2 : 1 MnO4- : I2 2 : 5 Ag2O : H2O2 1 : 1

2026 JC1 H2 Chemistry (9476) 3 [Turn Over (c) Use the value obtained in (b) and equation 4 to calculate the percentage by mass of NaNO2 in the 1.00 g sample of preserved meat. n(MnO4-) reacted with NO2- =   50.000.00500 1000 - 6.72  10-5 = 1.828  10-4 mol (ecf) n(NO2-) reacted = 5 2  1.828  10-4 = 4.57  10-4 mol [1] = n(NaNO2) in 1.00 g sample of preserved meat Mass of NaNO2 in 1.00 g sample = 4.57  10-4  69.0 = 0.03153 g (ecf) % by mass of NaNO2 =  0.03153 1001.00 = 3.15 % [1] 3 s.f. + units [2] [Total: 5] MnO4- : NO2- 2 : 5 NaNO2 : NO2- 1 : 1