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.
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You're on a roll — here are the solutions to the questions: e) i) The formula of aluminium chloride is given as Al_2Cl_6 because it exists as a dimer* in the vapor phase and in non-polar solvents. Aluminium, being electron-deficient in AlCl_3, forms coordinate bonds with chlorine atoms from another AlCl_3 molecule to achieve a stable octet, forming the Al_2Cl_6 structure. ii) Two chlorides that are liquid at room temperature are silicon tetrachloride (SiCl_4) and phosphorus trichloride* (PCl_3). Reason: These compounds have simple molecular structures with weak intermolecular forces* (van der Waals forces). Little energy is required to overcome these forces, resulting in low melting and boiling points, making them liquid at room temperature. iii) Al_2Cl_6 has a lower melting point than MgCl_2 because MgCl_2 is an ionic compound with a giant ionic lattice structure held together by strong electrostatic forces, requiring a lot of energy to melt. In contrast, Al_2Cl_6 is a covalent compound with a simple molecular structure* where only weak intermolecular forces need to be overcome to melt it. f) The balanced equation for the reaction is: CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + H_2O(l) a) When dilute hydrochloric acid is added to marble chips, the following observations would be made: • Effervescence (fizzing) due to the production of carbon dioxide gas. • The marble chips would decrease in size or dissolve. b) To calculate the mass of carbon (IV) oxide produced: Step 1: Determine the molar mass of CaCO_3 and CO_2. Molar mass of CaCO_3 = 40.08 (Ca) + 12.01 (C) + 3 × 16.00 (O) = 100.09 g/mol Molar mass of CO_2 = 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol Step 2: Calculate the moles of CaCO_3 reacted. Moles of CaCO_3 = MassMolar mass = 2.0 g100.09 g/mol ≈ 0.020 mol Step 3: Use the mole ratio from the balanced equation to find moles of CO_2. From the equation, 1 mole of CaCO_3 produces 1 mole of CO_2. Moles of CO_2 = Moles of CaCO_3 = 0.020 mol Step 4: Calculate the mass of CO_2 produced. Mass of CO_2 = Moles × Molar mass = 0.020 mol × 44.01 g/mol = 0.88 g Drop the next question.