Here is the exact question:An unknown amount of potassium oxalate, K2C2O4, was dissolved in 80.0 ml of 2.0 M sulfuric acid and then titrated with 26.5 ml of 0.0284 M KMnO4. Calculate the amount of K2C2O4 in grams used in the titration.
My plan is to first determine the number of moles in K2C2O4 from K2C2O4 + H2SO4 ?? And then do the other part. Or should I put the whole thing together as in:
K2C2O4 + H2SO4 + KMnO4 --> ????
2007-03-13 02:13:26 · 3 answers · asked by elizabeth_is_a_fiesty_gal 2 in Science & Mathematics ➔ Chemistry
why doesn't the number of moles of H2SO4 play in the answer?
2007-03-13 03:34:16 · update #1
The balanced equation is:
5K2C2O4 +8 H2SO4 + 2KMnO4 --> 2 MnSO4 + 10CO2 + 8H2O + 6K2SO4
Determine the number of moles from the concentration x volume of KMnO4 (0.0265L x 0.0284Mol/L) = 7.526E-4 mole
7.526E-4 mol of KMnO4 x 5mol K2C2O4/(2mole KMnO4) =1.882E-3 mole K2C2O4 or 1.882 x 10^-3 Mols
mass = moles x Molar mass of K2C2O4
(note you have just 3 sig figs in the original problem)
The sulfuric acid was added to provide an acidic environment for oxidation/reduction to occur. KMnO4 is a strong oxidizer.
Hope this helps ;o}
2007-03-13 02:53:53 · answer #1 · answered by docrider28 4 · 0⤊ 1⤋
K2C2O4 + H2SO4 --> H2C2O4 + K2SO4
2007-03-13 09:52:17 · answer #2 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋
Yeah, I do think it is alright to link them together like that. But there aren't any chemical formula of the products, how do you want to balance them? Or... there isn't any products for this experiment at all?
2007-03-13 09:52:15 · answer #3 · answered by Cherry Shortcake 3 · 0⤊ 0⤋