How do you calculate moles to neutralise oxalic acid?

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SUMMARY

The discussion focuses on calculating the moles of potassium manganate (KMnO4) required to oxidize oxalic acid (H2C2O4) using the reaction equation 2KMnO4 + 5H2C2O4 + 16H+ → 2Mn2+ + 10CO2 + H2O. The correct stoichiometric ratio indicates that 2 moles of KMnO4 are needed for every 5 moles of oxalic acid. Given the concentration of oxalic acid at 0.05 M and its amount at 5x10^-4 moles, the moles of KMnO4 required can be calculated using the formula n = c x V, leading to a required volume of 0.01085 M solution for the reaction.

PREREQUISITES
  • Understanding of stoichiometry in chemical reactions
  • Familiarity with molarity and concentration calculations
  • Knowledge of oxidation-reduction reactions
  • Proficiency in using the equations n = c x V and n = m/Ar
NEXT STEPS
  • Study the stoichiometric coefficients in redox reactions
  • Learn how to balance chemical equations involving KMnO4 and organic acids
  • Explore the concept of molarity and its applications in titration
  • Investigate the role of potassium manganate as an oxidizing agent in organic chemistry
USEFUL FOR

Chemistry students, educators, and laboratory technicians involved in analytical chemistry and redox reaction calculations will benefit from this discussion.

Daniel2244
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Homework Statement


calculate the number of moles of potassium manganate required to neutralise oxalic acid.
2KMn)4- + 5H2C2O42-+ 16H+ --> 2Mn2+ + 10CO2 + H2O

Homework Equations


n = c x V
n= m/Ar

The Attempt at a Solution


Moles of oxalic acid = 5x10-4
Volumeol of KMnO4 = 0.01085dm-3
Concentration of oxalic acid = 0.05 therefore concentration of manganate is the same (might be wrong).

moles manganate= c x V = 0.05 X 5x10-4= 5.425x10-4.
However, I think this is wrong and that you're meant to use the mole ratio difference to get the moles, but I'm not sure how.
 
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Just follow the stoichiometry. How many moles of KMnO4 are needed to oxidize (not neutralize!) 5×10-4 moles of oxalic acid? In what volume of 0.01085 M solution will you find this amount of permanganate?

There are several problem with your reaction equation, KMnO4/(COOH)2 ratio is still OK though.
 

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