Home › Forums › Stimulants › What volume of 0.15 M KMnO4 would be required to react with 1g of formic acid?
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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!
2[5e-+8H3O+ + MnO4- –> Mn2+ +12H2O]
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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