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    Analysis of Hydrogen Peroxide Solutions

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    Molarity of KMnO4: .2 M
    Volume of H2O2: 10 mL

    Titration of KMnO4 and new H2O2
    Trial 1(Rough) Trial 2(Fine) Trial 3(Fine)
    Initial volume 12.00mL 12.00mL 12.00mL
    Final volume 13.98mL 12.10mL 12.05mL

    Titration of KMnO4 and old H2O2 (Fine Titrations Only)
    Trial 1(Rough) Trial 2(Fine) Trial 3(Fine)
    Initial volume 12.00mL 12.00mL 12.00mL
    Final volume 12.99mL 12.10mL 12.05mL

    Calculations/Interpretations: Show all math performed (give the formula, show your setup, and give the result. Use proper significant figures and include proper labels. Also, answer ALL questions asked in the lab.

    Trial 2 Trial 3 Average
    Volume of KMnO4 (new H2O2):
    Volume of KMnO4 (old H2O2):

    Moles KMnO4 (new H2O2):
    Moles KMnO4 (old H2O2):

    Balanced equation:
    2 KMnO4 + 5 H2O2 + 3 H2SO4  K2SO4 + 2 MnSO4 + 8H2O + 5 O2

    Using the average values from trials 2 and 3 and the balanced equation, calculate the following:
    Moles H2O2 (new):
    Moles H2O2 (old):

    Molarity H2O2 (new):
    Molarity H2O2 (old):

    © BrainMass Inc. brainmass.com October 10, 2019, 12:12 am ad1c9bdddf
    https://brainmass.com/chemistry/stoichiometry/analysis-hydrogen-peroxide-solutions-282921

    Solution Preview

    Trial 2
    Volume of KMnO4 (new H2O2) = 12.10mL - 12.00mL = 0.10mL
    Volume of KMnO4 (old H2O2) = 12.10mL - 12.00mL = 0.10mL
    Moles KMnO4 (new H2O2) = (0.2mol/L)*(0.10mL)*(1L/1000mL) = 2x10^-5 mol
    Moles KMnO4 (old H2O2) = (0.2mol/L)*(0.10mL)*(1L/1000mL) = 2x10^-5 mol

    Trial 3
    Volume of KMnO4 (new H2O2) = 12.05mL - 12.00mL = 0.05mL
    Volume of KMnO4 ...

    Solution Summary

    An analysis of hydrogen peroxide solutions

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