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
Welcome back — been a while! Let's pick up where you left off. The tables for mass and burette readings are empty, so I will assume some values to demonstrate the calculation steps. Assumed Data: Mass of sodium oxalate (Na_2C_2O_4) = 1.250 g Volume of sodium oxalate solution prepared = 250 cm³ = 0.250 dm³ Volume of solution G (KMnO_4) reacting with 25.0 cm³ of sodium oxalate (Titre 1) = 20.00 cm³ Volume of solution G (KMnO_4) reacting with 25.0 cm³ of solution H (NaNO_2) (Titre 2) = 22.50 cm³ 1 (a) Calculate the concentration (mol dm⁻³) of your sodium oxalate solution (Na_2C_2O_4, C = 12.0, O = 16.0). Step 1: Calculate the molar mass of Na_2C_2O_4. M(Na_2C_2O_4) = (2 × 23.0) + (2 × 12.0) + (4 × 16.0) M(Na_2C_2O_4) = 46.0 + 24.0 + 64.0 = 134.0 g mol^-1 Step 2: Calculate the moles of sodium oxalate. Moles = MassMolar mass = 1.250 g134.0 g mol^-1 Moles = 0.009328 mol Step 3: Calculate the concentration of sodium oxalate solution. Concentration = MolesVolume (dm^3) = 0.009328 mol0.250 dm^3 Concentration = 0.0373 mol dm^-3 1 (b) The concentration (mol dm⁻³) of the potassium permanganate in solution G. The balanced equation for the reaction between KMnO_4 and Na_2C_2O_4 is: 2MnO_4^-(aq) + 5C_2O_4^2-(aq) + 16H^+(aq) → 2Mn^2+(aq) + 10CO_2(g) + 8H_2O(l) The mole ratio of MnO_4^- to C_2O_4^2- is 2:5. Step 1: Calculate the moles of C_2O_4^2- in 25.0 cm³ of solution. Moles of C_2O_4^2- = Concentration × Volume (dm^3) Moles of C_2O_4^2- = 0.0373 mol dm^-3 × (25.0)/(1000) dm^3 Moles of C_2O_4^2- = 0.0009325 mol Step 2: Calculate the moles of MnO_4^- that reacted. Moles of MnO_4^- = Moles of C_2O_4^2- × (2)/(5) Moles of MnO_4^- = 0.0009325 mol × (2)/(5) = 0.000373 mol Step 3: Calculate the concentration of KMnO_4 in solution G. Concentration of KMnO_4 = Moles of MnO_4^-Volume of KMnO_4 (dm^3) Concentration of KMnO_4 = 0.000373 mol20.00/1000 dm^3 Concentration of KMnO_4 = 0.01865 mol dm^-3 2 Find the concentration (mol dm⁻³) of the sodium nitrite in solution H. The balanced equation for the reaction between KMnO_4 and NaNO_2 is given: 2MnO_4^-(aq) + 5NO_2^-(aq) + 6H^+(aq) → 2Mn^2+(aq) + 5NO_3^-(aq) + 3H_2O(l) The mole ratio of MnO_4^- to NO_2^- is 2:5. Step 1: Calculate the moles of MnO_4^- that reacted with solution H. Moles of MnO_4^- = Concentration of KMnO_4 × Volume of KMnO_4 (dm^3) Moles of MnO_4^- = 0.01865 mol dm^-3 × (22.50)/(1000) dm^3 Moles of MnO_4^- = 0.0004196 mol Step 2: Calculate the moles of NO_2^- that reacted. Moles of NO_2^- = Moles of MnO_4^- × (5)/(2) Moles of NO_2^- = 0.0004196 mol × (5)/(2) = 0.001049 mol Step 3: Calculate the concentration of NaNO_2 in solution H. Concentration of NaNO_2 = Moles of NO_2^-Volume of NaNO_2 (dm^3) Concentration of NaNO_2 = 0.001049 mol25.0/1000 dm^3 Concentration of NaNO_2 = 0.04196 mol dm^-3 3 Find the mass of sodium nitrite actually present in 1 dm³ of solution H (O = 16.0). Step 1: Calculate the molar mass of NaNO_2. M(NaNO_2) = 23.0 + 14.0 + (2 × 16.0) M(NaNO_2) = 23.0 + 14.0 + 32.0 = 69.0 g mol^-1 Step 2: Calculate the mass of NaNO_2 in 1 dm³ of solution H. Mass = Concentration × Molar mass × Volume Mass = 0.04196 mol dm^-3 × 69.0 g mol^-1 × 1 dm^3 Mass = 2.895 g 4 Calculate the percentage purity of the sodium nitrite used. To calculate the percentage purity, the initial mass of the impure sodium nitrite used to prepare solution H is required. This information is not provided in the problem. If, for example, 3.00 g of impure sodium nitrite was dissolved to make 1 dm³ of solution H, then the calculation would be: Percentage purity = Mass of pure NaNO_2 in 1 dm^3Mass of impure NaNO_2 used to make 1 dm^3 × 100\% Percentage purity = 2.895 g3.00 g × 100\% = 96.5\% (This is an example calculation assuming an initial impure mass of 3.00 g.) What's next? Send 'em! 📸