If a student weighs out 1.000g of oxalate unknown and it takes 24.6mL of permang

Question

If a student weighs out 1.000g of oxalate unknown and it takes 24.6mL of permanganate solution to titrate it, how much oxalate unknown should the student weigh out in order to use 35.0mL of permanganate solution?

Explanation / Answer

The balanced reaction of the titration of permanganate withoxalate is - 2 MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) ---> 10CO2(g) + 2 Mn2+(aq) + 8H2O(l)We take the oxalate ion is formed from the Na2C2O4,then Therefore number of moles of oxalate present in thesolution = 1.0 g / 134.0 g/mol                                                                                            = 0.007462 mole Therefore the number of moles of permonganate required =0.007462 mol* ( 2 mol MnO4- / 5 moleC2O42-)                                                                                        = 0.002984 mole To neutralize this amount of oxalate requires 24.6 mL of thepermangane, hence - The concentration of the permanganate = 0.002984 mol / 0.0246L                                                             = 0.1213 M Thus the amount of oxalate is to be used to neutralize 35.0mL of the permanganate =                                                 ( 0.1213 mol/L*0.035 L)(5 C2O42- /2 MnO4-) (134.0 g /1 molC2O42-)                                             = 1.42 g

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