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

Question

The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35 g of 0.0122 M KMnO4 are required to titrate 1.072 g of a household H202 solution. Except for answers to (a) and (e), use E notation. For example 0.00245 is 2.45E-3. Do not include units, just enter numerical values 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 H202 in sample. d) Calculate the number of grams of H202 in the sample. e) Calculate the % m/m H2O2 in the household H2O2 solution.

Explanation / Answer

Solution:

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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