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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Here is the solution to the problem. Question 7: A solution of ammonia in water with a density of 0.93 g/cm^3 contained 18% by mass of ammonia. What volume of this solution is required to form 30g of ammonium tetraoxosulphate(VI) based on the reaction: 2NH_3 + H_2SO_4 → (NH_4)_2SO_4. Step 1: Calculate the molar mass of ammonium tetraoxosulphate(VI), (NH_4)_2SO_4. Molar mass of (NH_4)_2SO_4 = 2(N) + 8(H) + 1(S) + 4(O) = 2(14.01 g/mol) + 8(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 28.02 + 8.08 + 32.07 + 64.00 = 132.17 g/mol Step 2: Calculate the moles of (NH_4)_2SO_4 to be formed. Moles = MassMolar mass Moles of (NH_4)_2SO_4 = 30 g132.17 g/mol = 0.22698 mol Step 3: Use stoichiometry from the balanced equation to find the moles of NH_3 required. The balanced equation is: 2NH_3 + H_2SO_4 → (NH_4)_2SO_4. From the equation, 2 moles of NH_3 react to form 1 mole of (NH_4)_2SO_4. Moles of NH_3 = 0.22698 mol (NH_4)_2SO_4 × 2 mol NH_31 mol (NH_4)_2SO_4 = 0.45396 mol NH_3 Step 4: Calculate the mass of NH_3 required. Molar mass of NH_3 = 1(N) + 3(H) = 14.01 g/mol + 3(1.01 g/mol) = 14.01 + 3.03 = 17.04 g/mol Mass of NH_3 = Moles × Molar mass = 0.45396 mol × 17.04 g/mol = 7.735 g Step 5: Calculate the mass of the ammonia solution required. The solution contains 18% by mass of ammonia. Mass of solution = Mass of pure NH_3Fractional abundance of NH_3 = 7.735 g0.18 = 42.972 g Step 6: Calculate the volume of the ammonia solution required. The density of the ammonia solution is 0.93 g/cm^3, which is equivalent to 0.93 g/mL. Volume of solution = Mass of solutionDensity of solution = 42.972 g0.93 g/mL = 46.206 mL Rounding to three significant figures: Volume of solution = 46.2 mL The volume of the ammonia solution required is 46.2 mL.