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 — let's solve these questions. 13. i) Determine the formula of the hydrated salt. Step 1: Calculate the moles of each component in a 100 g sample. Assume a 100 g sample of the hydrated salt. Mass of Iron (Fe) = 20.2 g Mass of Sulphur (S) = 11.5 g Mass of Oxygen (O) = 23.0 g Mass of Water (H_2O) = 45.3 g Moles of Fe = 20.2 g56 g/mol = 0.3607 mol Moles of S = 11.5 g32 g/mol = 0.3594 mol Moles of O = 23.0 g16 g/mol = 1.4375 mol Moles of H_2O = 45.3 g18 g/mol = 2.5167 mol Step 2: Find the simplest whole-number mole ratio by dividing by the smallest number of moles (0.3594 mol). For Fe: (0.3607)/(0.3594) ≈ 1 For S: (0.3594)/(0.3594) = 1 For O: (1.4375)/(0.3594) ≈ 4 For H_2O: (2.5167)/(0.3594) ≈ 7 The ratio of Fe : S : O : H_2O is 1:1:4:7. This indicates the anhydrous part is FeSO_4 and there are 7 molecules of water of crystallization. The formula of the hydrated salt is FeSO_4 · 7H_2O. 13. ii) Calculate the concentration of the resulting salt solution in moles per litre. Step 1: Calculate the moles of the hydrated salt. Given mass of hydrated salt = 6.95 g Given molecular mass of the salt = 278 g/mol Moles of salt = MassMolecular mass = 6.95 g278 g/mol = 0.025 mol Step 2: Calculate the concentration in moles per litre. Given total volume = 250 cm^3 = 0.250 dm^3 Concentration = Moles of saltVolume of solution in dm^3 = 0.025 mol0.250 dm^3 = 0.1 mol/dm^3 The concentration of the resulting salt solution is 0.1 M. 14. (i) Calculate the mass in grams of lead (II) iodide formed. Step 1: Calculate the moles of iodide ions (I^-). Given volume of iodide ions = 300 cm^3 = 0.300 dm^3 Given molarity of iodide ions = 0.1 M Moles of I^- = Molarity × Volume = 0.1 mol/dm^3 × 0.300 dm^3 = 0.030 mol Step 2: Use stoichiometry to find the moles of lead (II) iodide (PbI_2) formed. The balanced equation is: Pb^2+ (aq) + 2I^- (aq) → PbI_2 (s) From the equation, 2 moles of I^- react to produce 1 mole of PbI_2. Moles of PbI_2 = (1)/(2) × Moles of I^- = (1)/(2) × 0.030 mol = 0.015 mol Step 3: Calculate the molar mass of lead (II) iodide. Atomic mass of Pb = 207 Atomic mass of I = 127 Molar mass of PbI_2 = 207 + (2 × 127) = 207 + 254 = 461 g/mol Step 4: Calculate the mass of lead (II) iodide formed. Mass of PbI_2 = Moles × Molar mass = 0.015 mol × 461 g/mol = 6.915 g The mass of lead (II) iodide formed is 6.915 g. 14. (ii) Identify the colour of the product formed in (i). The product formed is lead (II) iodide (PbI_2). The colour of lead (II) iodide is yellow. Got more? Send 'em!