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
Edson, let's knock this out. Step 1: Calculate the molar mass of sulphuric acid (H_2SO_4). Given atomic masses: H = 1.0, S = 32.0, O = 16.0. Molar mass of H_2SO_4 = (2 × 1.0) + 32.0 + (4 × 16.0) = 2.0 + 32.0 + 64.0 = 98.0 g/mol Step 2: Calculate the molar concentration of the sulphuric acid solution. Given concentration of H_2SO_4 = 4.9 g/dm^3. Molar concentration of H_2SO_4 = Mass concentrationMolar mass = 4.9 g/dm^398.0 g/mol = 0.05 mol/dm^3 Step 3: Calculate the moles of H_2SO_4 used in the neutralization. Volume of H_2SO_4 used = 12 cm^3 = (12)/(1000) dm^3 = 0.012 dm^3. Moles of H_2SO_4 = Molar concentration × Volume = 0.05 mol/dm^3 × 0.012 dm^3 = 0.0006 mol Step 4: Determine the moles of MOH that reacted using the stoichiometric ratio. The balanced chemical equation is: 2MOH_aq + H_2SO_4 → M_2SO_4 + 2H_2O. From the equation, 2 moles of MOH react with 1 mole of H_2SO_4. Moles of MOH = 2 × Moles of H_2SO_4 = 2 × 0.0006 mol = 0.0012 mol Step 5: Calculate the molar concentration of the MOH solution. Volume of MOH solution used = 6 cm^3 = (6)/(1000) dm^3 = 0.006 dm^3. Molar concentration of MOH = Moles of MOHVolume of MOH solution = 0.0012 mol0.006 dm^3 = 0.2 mol/dm^3 Step 6: Calculate the molar mass of MOH. The initial MOH solution was prepared by dissolving 11.2 g of MOH in enough solvent to make one litre (1 dm^3) of solution. Since the concentration is 0.2 mol/dm^3, one litre of solution contains 0.2 moles of MOH. Molar mass of MOH = Mass of MOHMoles of MOH = 11.2 g0.2 mol = 56 g/mol Step 7: Calculate the relative atomic mass of M. The formula of the hydroxide is MOH. Molar mass of MOH = Relative atomic mass of M + Relative atomic mass of O + Relative atomic mass of H. 56 = M + 16.0 + 1.0 56 = M + 17.0 M = 56 - 17.0 M = 39.0 The relative atomic mass of M is 39.0. Send me the next one 📸