If 32.6mL of 0.0118M permanganate solution are used to titrate a sample of oxala

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The balanced reaction of the titration of permanganate withoxalate is - 2 MnO4-(aq) + 5C2O42-(aq) + 16H+(aq) ---> 10CO2(g) + 2 Mn2+(aq) + 8H2O(l)Therefore number of moles of permanganate present in thesolution = 0.0118 mol/L * 0.0326 L                                                                                                      = 0.0003846 mole Thus number of moles of oxalate is required for reaction =0.0003846 mol*(5 mol C2O42-/ 2 MnO4-)                                                                                      = 0.0009617 mol                                                                                      = 0.9617 mmol

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