The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35g of 0.0122 M KMnO4

Question

The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35g of 0.0122 M KMnO4 are required to titrate 1.072g of a household H2O2 solution. (a) calculate the mL of MnO4 added to reach the endpoint. (b) calculate moles of MnO4 added to reach the endpoint. (c) calculate the number of moles of H2O2 in sample. (d) calculate the number of grams of H2O2 in the sample. (e) calculate the % m/m H2O2 in the household H2O2 solution.

Explanation / Answer

2KMnO4 +3 H2O2 ---> 2KOH + 2MnO2 + 3O2 + 2H2O, a) 26.35 gm = (26.35/1.037 ) = 25.41 ml , b) M = 0.0122 , moles of KMnO4 = 0.0122 x25.41 /1000 = 0.00031, c) moles of H2O2 = (3/2 x KMnO4 moles) = (3/2 x0.00031) = 0.000465, d) mass of H2O2 = (0.000465 x34.1) = 0.015856 grams , e) mass% = (0.015856/1.072) x100 = 1.48 %

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