What volume of 0.15 M KMnO4 would be required to react with 1g of formic acid?

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This topic contains 1 reply, has 0 voices, and was last updated by  Ethon 12th May 2018 at 8:56 am.

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    Ethon
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    The unbalanced equation provided is MnO4- + HCO2H –> Mn2+ + CO2.

    After balancing it, I get 10H2O + 5HCO2H + 2MnO4- –> 2Mn2+ + 5CO2. Could someone please check this equation? 

    Furthermore, then I get 1g HCO2H * (1 mol HCO2H / 46g/mol) = 2.17 * 10^-2 mol of HCO2H. 

    Then I get 2.17 * 10^-2 mol of HCO2H * (2 mol MnO4-/5 mol  5HCO2H) = 8.4 * 10^-3 mol

    Then I divide the moles by the molarity, 8.4 * 10^-3 mol of MnO4-/0.15M to get 5.6*10^-2 L of MnO4-. Is this right? Could someone please, please check all my calculations? Thank you SO much. 

    UPDATE: I really can’t seem to get the redox reaction!

    I have 
    2[5e-+8H3O+ + MnO4- –> Mn2+ +12H2O] 
    and 
    5[2H2O + HCO2H –> CO2 + 2H3O+ +2e-] 
    which gives me 
    6H3O+ + 5HCO2H + 2MnO4- –> 2Mn2+ + 5CO2 + 14H2O.

    but that’s 24 hydrogens on the left side and 28 hydrogens on the left side! Why is this?!

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