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Topic: balanced equation/ titration  (Read 3795 times)

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faye

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balanced equation/ titration
« on: July 08, 2004, 11:14:11 AM »
If the question is : A 50.0 mL solution of H2C2O4 requires 24.35 mL of .125M NaOH to neutralize both protons. It takes 15.83mL of KMnO4 (product Mn2+) to react with all the oxalic acid in another 5.0 mL sample of oxalic acid (product CO2). A 2.500g sample containing iron(II)chloride is dissolved in acid and titrated with 31.87 mL of the KMnO4 solution.  How do you writie out the balanced equation for this reaction??? THere are three of them aren't there? And  how do you find the concentration of permanganate solution?

Offline Donaldson Tan

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Re:balanced equation/ titration
« Reply #1 on: July 08, 2004, 01:50:06 PM »
(1) H2C2O4 + 2NaOH => Na2C2O4 + H2O
This'll help you derrive the concentration of the ethandioc (oxalic) acid.

(2) 2[MnO4]- + 6H+ + 5H2C2O4 -> 10CO2 +2[Mn]2+ + 8H2O
This'll help u derrive the no. of moles of permangate in 15.83mL of solution.

(3) [MnO4]- + 5[Fe]2+ + 8H+ => [Mn]2+ + 5[Fe]3+ + 4H2O
This equation should help you confirm your answer derived from the results in (2)
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