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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I will solve the questions based on the provided images, interpreting the blurry text as accurately as possible. --- Question 2: (Titration Problem) Given: • Mixture of NaOH and NaCl = 2.0 g • Total volume of solution = 250 cm^3 • Volume of aliquot = 25.0 cm^3 • Concentration of H_2SO_4 = 0.1 M • Volume of H_2SO_4 used = 20.0 cm^3 b) a) Write the equation for the reaction. The reaction is between sodium hydroxide (NaOH) and sulfuric acid (H_2SO_4). H_2SO_4(aq) + 2NaOH(aq) → Na_2SO_4(aq) + 2H_2O(l) b) b) Calculate the number of moles of sulphuric (VI) acid used. Step 1: Convert the volume of H_2SO_4 to dm^3. 20.0 cm^3 = 20.0 × 10^-3 dm^3 = 0.020 dm^3 Step 2: Calculate the moles of H_2SO_4 using its concentration and volume. Moles of H_2SO_4 = Concentration × Volume Moles of H_2SO_4 = 0.1 mol/dm^3 × 0.020 dm^3 = 0.002 mol The number of moles of sulphuric (VI) acid used is 0.002 mol. b) c) Calculate the number of moles of the substance that reacted with sulphuric (VI) acid. The substance that reacted with sulfuric acid is NaOH. From the balanced equation, 1 mole of H_2SO_4 reacts with 2 moles of NaOH. Moles of NaOH in 25.0 cm^3 aliquot = 2 × Moles of H_2SO_4 Moles of NaOH = 2 × 0.002 mol = 0.004 mol The number of moles of NaOH that reacted is 0.004 mol. b) d) Calculate the number of moles of the substance that would react with sulphuric (VI) acid in one litre of solution. The 0.004 mol of NaOH was present in a 25.0 cm^3 aliquot. The original solution had a total volume of 250 cm^3. Step 1: Calculate the total moles of NaOH in the 250 cm^3 solution. Moles of NaOH in 250 cm^3 = 0.004 mol × 250 cm^325.0 cm^3 = 0.004 mol × 10 = 0.040 mol Step 2: Calculate the moles of NaOH in one litre (1000 cm^3) of solution. Moles of NaOH in 1000 cm^3 = 0.040 mol × 1000 cm^3250 cm^3 = 0.040 mol × 4 = 0.160 mol The number of moles of NaOH in one litre of solution is 0.160 mol. b) e) Determine the mass of the unreacted substance in the mixture of solution and hence calculate its percentage in the mixture. The mixture contains NaOH (reacted substance) and NaCl (unreacted substance). Step 1: Calculate the mass of NaOH in the 250 cm^3 solution. Molar mass of NaOH = 23.0 (Na) + 16.0 (O) + 1.0 (H) = 40.0 g/mol. Mass of NaOH = Moles × Molar mass = 0.040 mol × 40.0 g/mol = 1.6 g. Step 2: Calculate the mass of the unreacted substance (NaCl). Total mass of mixture = 2.0 g. Mass of NaCl = Total mass of mixture - Mass of NaOH Mass of NaCl = 2.0 g - 1.6 g = 0.4 g Step 3: Calculate the percentage of the unreacted substance (NaCl) in the mixture. Percentage of NaCl = Mass of NaClTotal mass of mixture × 100\% Percentage of NaCl = 0.4 g2.0 g × 100\% = 20\% The mass of the unreacted substance (NaCl) is 0.4 g. Its percentage in the mixture is 20\%. --- Question 3: (Combustion Analysis of a Hydrocarbon) Given: • Mass of $