Question

2 MnO4 (aq) + 5 H2O2(aq) + 6 H(aq) → 5 O2(aq) + 2 Mn2+ (aq) + 8 H20(1) completely oxidize 25.0 mL of a H2O2 solution, calculate the molarity of the H20, solution

Answer 1

Answer:

Answers (6) a moy cag.) + 5H2O2 Caqi) + Ghilarg.) - 502 (94-) + 2M, + (99.) + 8H06) When an element is onidised, it's oxidation state increases and echen an element is reduced, it's onidation State decreases, for, men For mnou mant Herme, Foo mnogo let's onidation state of Mnan :: nto (-27x450) A r =- => mz+844 = +7 and for my 2t = oxidation state = +2. . Oxidation state of mnou changes form 47 to ta, there it is reduced. in mnout is reduced.] and H2O, a ondation state of o=a. . For H₂O₂, let's " oxidation state of o=a o 2x1 + 2xx=0 2x=-2 »n=-1 and for elemerrteel Oy = Oxidation - 0 state ofo = : Onidation state of H2Oe is increased a to o . HzOn is omiclised from

(6) from above balanced reaction - Number of moles of mnou Number of moles of H20₂ 5 cherre, 2 and 5 are coefficients of moon and the and we know that, Number of moles = concentration X Volume, O 2 . i [mmor] XV. Choo.] XV2 - 0 S . 5 where V,= 28.8mL , [mnou = 0.028m Va= 25.0mL , CH20) = 2 .. [H202] = 3 x [mna] XV = (.02] = 3 sm x 0.02872 x 28.8mL » [Maba] = 0.08 m :: Molasity of Helly = 0.08m 5X 0.028m x 28.8mL 25mL