Complete the following table for this titration. Data Table P3: Titration of hyd

Question

Complete the following table for this titration. Data Table P3: Titration of hydrogen peroxide with potassium permanganate. concentration of MnO4

Explanation / Answer

. . . .MnO4(-) + 8 H(+) + 5 e(-) -----> Mn(2+) + 4 H2O . . . . . . . . ] x 2 . . . .H2O2 . . . . . . . . . . . . . . ..------> O2 + 2 H(+) + 2 e(-) . . . . . ] x 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 MnO4(-) + 5 H2O2 + 6 H(+) ------> 2 Mn(2+) + 5 O2 + 8 H2O mmol of MnO4(-) ? - - - - - - - - - - - - - C1 x V1 = 0,561 x 16,12 = 9,04 mmol of MnO4(-) mmol of H2O2 ? - - - - - - - - - - - . . . according to the equation : n(H2O2) = 5/2 n(MnO4(-)) C2 x V2 = 5/2 x C1 x V1 C2 x V2 = 5/2 x n(MnO4(-)) C2 x V2 = 5/2 x 9,04 C2 x V2 = 22,61 mmol of H2O2 mass of H2O2 ? - - - - - - - - - - - m(H2O2) = n(H2O2) x M(H2O2) m(H2O2) = 22,61•10^-3 x 34 m(H2O2) = 0,769 g of H2O2 mass % of H2O2 in the original sample ? - - - - - - - - - - - - - - - - - - - - - - - - - - - - %m(H2O2) = 100 x m(H2O2) / m(solution of H2O2) %m(H2O2) = 100 x 0,769 / 17,88 %m(H2O2) = 4,30 % molarity of H2O2 in the original sample ? - - - - - - - - - - - - - - - - - - - - - - - - - - - . . . according to the equation : n(H2O2) = 5/2 n(MnO4(-)) C2 x V2 = 5/2 x C1 x V1 C2 = 5/2 x C1 x V1 / V2 C2 = 5/2 x 0,561 x 16,12 / 17,80 C2 = 1,27 mol/L

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