Question

CHE 101 EXPERIMENT 10 PRE-LAB; OXIDATION - REDUCTION TITRATION Name Gisselle. Caballo Lab Day Timetaday op Date II2929 5 Points (3 Points not completed before lab begins) Consider the following molecular equation for the reaction wolved in today's experiment: (Place a and b at the end of the Procedure for Part 2) KMnO4 + Na2C204 + H2504 - MnSO4 + CO2 + Na2SO4 + H2O a. Write a balanced, unmultiplied ionic half-reaction for the oxidation: b. Write a balanced, unmultiplied ionic half-reaction for the reduction: c. Write the balanced net-ionic reaction: d. Write the balanced molecular reaction: 2. How many mL of a 1.356 N KMnO4 solution are needed to completely react with (titrate) 2.500 grams of Na2C204 in acid solution. Show all work and units.

Answer 1

Solution ① Given reaction, KMnO4 + NazCOu + H₂SO4 Musou + CO₂ + Na, South2o a Balanced, Unmultiplied @ agonic half-reaction for the oxidation. cou 200₂ + 26- O Balanced , un multiplied, ionic hal-reachion for reduction Mnou +84 + 50 M2+ + 4H2O © Balanced Net ionic reaction. 10 co₂+ 2M2 + +8H₂O 5 sou + 2 MnOu +64t Balance molecular reaction. 2 KMn Oy + 5 Naz C₂O4 + 8 4₂504 - Kq sou +10 co₂t 9 nSO4 + 5 Na₂SO4 +8H2O

solution 2 is Balance reaction Kasou + 10 co₂+ 2M2.504 +5 Nasou 2 KMnO4 + 5 Na₂C₂O4 + 8 H₂SO4 + 8H2O we see that 5 mole. of Bla₂GO4 2 mole, of KMnO4 = now y = 5 x. No. of moles of k No. of moles of Naz Cou = 5 x. ples of Na, - 3 2.500g -0.0186m 734 g 1mor No. of moles. Of NqC ou = molarity of KMnOy = normality of Km nou no facfor 2 1.356 m =0.2712 M let required volume of KMnou z xml No of moles. f KMnou in (WmL = x x 0.2712 mot novo from @ we get 0.0186 = 5x Xx0,2912 0.0186 = 3/3 1ooo x = 27.4 ~2715 ml