Question

Consider a different titration for this exercise. Potassium permanganate (KMn04) is the titrant and hydrogen peroxide (H202) the analyte according to the following balanced chemical equation. 2 MnO45 H202 + 6 H+ »2 Mn2+ 5 02 + 8 H20 (a) What is the stoichiometry of MnO4 to H202? 8:3 5:2 2:1 1:1 1:2 O2:5 3:8 (b) Complete the following table for this titration. Data Table P3: Titration of hydrogen peroxide with potassium permanganate concentration of MnOa 0.554 M volume H202 solution 17.71 mL mass H202 solution 17.79 g volume of MnO solution 16.11 mL 4.0 mmol of MnO4 mmol mmol of H202 40J mmol 4.0 mass of H202 g mass % of H202 in the original sample 40J molarity of H202 in the original sample 40

Answer 1

Q 2:5 b); Moles of Mmoh - 0.554 mol x 16.11 mL = 8.925 x103 moi = 8.925 mmol (* 1 mol = 10*2 mmol) i mmol of Mnog = 18.925 mmol. i mmol of Minda 2 mmol of Ho-5 * 8.925_- mmol of H₂O → mmol of H2O 5x8.925 - 22.312 mer 0.759 mmol of H2O2 = 22. 312) mmol iu) mass of H2O2 = 22.312 x16mot 34 g = 0.759] , i) mass % of Hz Q2 in the original sample=77 = 4.266) % V) Molarity of 1722 iu tie original sample 22.312 x10 mol, 1000 ML = 1,260 mol/L 17.11 mL - 11.260) M. 17.79 %100% IL