This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
here are the answers to your questions: 3. a) Two reasons why calcium oxide is used in the set up: • To dry the chlorine gas. • To absorb any unreacted acidic gases (like excess chlorine or hydrogen chloride). b) The chlorine gas was dried by passing it through a wash bottle containing concentrated sulphuric acid. c) i) Substance A is aluminium chloride (AlCl_3). ii) The equation for the reaction that took place in the combustion tube is: 2Al(s) + 3Cl_2(g) 2AlCl_3(s) iii) Calculate the volume of chlorine gas used: Molar mass of Al = 27.0 g/mol Molar gas volume = 24 dm^3/mol Step 1: Calculate the moles of aluminium. Moles of Al = Mass of AlMolar mass of Al = 0.8 g27.0 g/mol ≈ 0.02963 mol Step 2: Determine the moles of chlorine gas required from the stoichiometry. From the balanced equation, 2 moles of Al react with 3 moles of Cl_2. Moles of Cl_2 = Moles of Al × (3)/(2) = 0.02963 mol × 1.5 ≈ 0.044445 mol Step 3: Calculate the volume of chlorine gas. Volume of Cl_2 = Moles of Cl_2 × Molar gas volume = 0.044445 mol × 24 dm^3/mol ≈ 1.067 dm^3 6. a) Write an equation for the following: i) Chlorine gas reacting with iron: 2Fe(s) + 3Cl_2(g) 2FeCl_3(s) ii) Chlorine gas reacting with hot concentrated sodium hydroxide solution: 3Cl_2(g) + 6NaOH(aq) hot, conc. 5NaCl(aq) + NaClO_3(aq) + 3H_2O(l) b) Determine the empirical formula of the oxide: Mass of oxide = 1.83 g Mass of oxygen (O) = 1.12 g Mass of chlorine (Cl) = 1.83 g - 1.12 g = 0.71 g Molar mass of O = 16.0 g/mol Molar mass of Cl = 35.5 g/mol Step 1: Calculate the moles of each element. Moles of O = 1.12 g16.0 g/mol = 0.07 mol Moles of Cl = 0.71 g35.5 g/mol = 0.02 mol Step 2: Find the simplest mole ratio by dividing by the smallest number of moles. For Cl: (0.02)/(0.02) = 1 For O: (0.07)/(0.02) = 3.5 Step 3: Multiply by a factor to get whole numbers. Multiply by 2: Cl: 1 × 2 = 2 O: 3.5 × 2 = 7 The empirical formula is Cl_2O_7. c) Other than the manufacture of weed killers, two other uses of chlorine are: • Water treatment (disinfection). • Manufacture of plastics (e.g., PVC). 7. a) i) The formula of another reagent that can be reacted with concentrated hydrochloric acid to produce chlorine gas is MnO_2 (manganese(IV) oxide). ii) The chlorine gas would be dried in the laboratory by passing it through a wash bottle containing concentrated sulphuric acid. 8. a) The conditions under which sodium hydroxide reacts with chlorine gas to form sodium hypochlorite are cold and dilute sodium hydroxide solution. b) The uses of sodium hypochlorite include: • As a bleaching agent (e.g., in household bleach). • As a disinfectant or antiseptic. Last free one today — make it count tomorrow, or type /upgrade for unlimited.