JPJC 2026 Mole & Stoichiometry Assignment

2026 JC1 H2 Chemistry (9476) 1 JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) Topical Assignment Topic 2: The Mole Concept & Stoichiometry Name : ______________________________________ Penalty −1 / NA 10 Class : 26S ___ • Missing 3 s.f. in final answer • Missing/wrong units • Working not shown Date : ____________ Duration : 20 minutes Answer all questions in the space provided. You are provided with a Data-Booklet. 1 (a) 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. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. [1] (b) (i) 1.0 cm3 of ‘20-volume’ hydrogen peroxide produces 20.0 cm 3 of oxygen gas at room temperature and pressure, as shown in equation 3. equation 3 2H2O2(aq) → 2H2O(l) + O2(g) Use this information to calculate the concentration, in mol dm −3, of ’20 -volume’ hydrogen peroxide. [2]

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 cm 3 of ’20-volume’ hydrogen peroxide reacts completely at room temperature and pressure, according to equation 2. [2] [Total: 5] 2 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 cm 3 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. 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–(aq) + I2(aq) → 2I–(aq) + S4O62–(aq) (a) Construct an ionic equation between I– ions and acidified MnO4– ions to liberate iodine, I2. [1]

2026 JC1 H2 Chemistry (9476) 3 (b) Calculate the amount of iodine liberated by the remaining MnO4–. Hence, calculate the amount of remaining MnO4– ions after reaction with NO2– ions. [2] (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. [2] [Total: 5]