chem

posted by tomi

The following equation describes the oxidation of an acidified solution of sodium oxalate (Na2C2O4) by potassium permanganate(KMnO4)

5C2O4 + 2MnO4 + 16H2SO4 --> 10CO2 + 2Mn + 16HSO4 + 8H20

What volume of 0.127M sodium oxalate in sulphuric acid will react with 3.57g of potassium permanganate?

  1. DrBob222

    Convert 3.57 g KMnO4 to moles. mols = grams/molar mass.
    Using the coefficients in the balanced equation, convert mols KMnO4 to mols Na2C2O4.
    molarity Na2C2O4 = mols/L. YOu know mols from the above calculation, you know M, calculate volume.
    Post your work if you get stuck.

  2. MSingh

    0.071L

  3. Dr. E

    M KMnO4 = 158.0339g/mol

    1 mol
    3.57g X -----------------
    158.0339g KMnO4
    5 mol C2O4-
    = 0.02259009 mol KMnO4 X -------------
    2 mol KMnO4

    1 L
    =0.056475225 mol C2O4-X ---------------
    0.127 mol C2O4-

    = 0.445 L

  4. Dr. E

    Woops sorry about the spacing. I didn't realize it would come out like that. Hope you can read it.

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