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

Question

The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35 g of 0.0122M KMnO4 are required to titrate 1.072 of a household H2O2 solution. Except for answers to (a) and (e), use scientific notation. Do not include units, just enter numerical values.

a) calculate the ml of MnO4- added to reach the endpoint.

b) calculate the moles of MnO4- added to reach the endpoint.

c) calculate the number of moles of H2O2 in the 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 + 5H2O2 + 3H2SO4 <---------> 2MnSO4 + K2SO4 + 5O2 + 8H2O

vol of KMnO4 = 26.35/1.037 = 25.41 ml , M = 0.0122 ,

a) vol of KMnO4 = 25.4 1

b) moles of KMnO4 = 0.0122 x ( 25.41/1000) = 0.00031 = 31 E-5

c) moles of H2O2 = (5/2) x 0.00031 = 0.000775 = 775 E-6

d) H2O2 in grams = 0.000775 x 34 =0.02635 = 2635 E-5

e) m/m% = 0.02635 x100/1.072 = 2.46 = 246 E-2

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