Asked by Megan
If 20.00mL of a solution that contains oxalate ion requires 10.00mL of 0.020M MnO4- solution in a titration, what was the molarity of the oxalate solution?
5C2O42- + 2MnO4- + 12H+ --> 2Mn2+ + 10CO2 + 8 H2O
5C2O42- + 2MnO4- + 12H+ --> 2Mn2+ + 10CO2 + 8 H2O
Answers
Answered by
DrBob222
Your equation isn't balanced. I think the 12 H^+ should be 16H^+. (H doesn't balance now and neither is the charge on the left the same as the charge on the right).
moles MnO4^- = M x L.
moles C2O4^-2 = moles MnO4^- x (5/2)
M of C2O4^- = moles/0.020
moles MnO4^- = M x L.
moles C2O4^-2 = moles MnO4^- x (5/2)
M of C2O4^- = moles/0.020
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