Question

a sample of iron ore (m=0,330g)consisting of a mixture of FeO and Fe2O3 was dissolved in dilute sulfuric acid and titrated with a potassium permanganate solution containing 0.0200g of KMnO4 per litre, and required 20,3 mL for complete reaction. The second sample of iron ore (m=0,370g)was dissolved in dilute sulfuric acid solution and fe3+ is reduced to fe2+ then titrated with the same potassium permanganate solution. 42,6mL was required for the second titration. Calculate the mass of each iron oxide in the original sample.
im not sure about chemical reactions because Fe2o3 is not reacting with KMno4

Answer 1

Ans. Part A: Concentration of KMnO₄ solution

Number of moles of KMnO₄ solution in 0.0200 g sample = mass/ molar mass

                        = 0.0200 g / 158.09 g mol^-1 = 1.2651 x 10^-4 moles

Molarity = number of moles of solute / liter of solution. Since, 0.020 g (= 1.2651 x 10^-4 moles) dissolved in 1 L,

Molarity of KMnO₄ solution = 1.2651 x 10^-4 M                    

Part B: Ore Sample 1: 10 FeO + 2 KMnO4 + 3 H2SO4---> 5 Fe2O3 + 2 MnSO4 + K2SO4 + 3 H2O

#.Fe2O3 does not react with KMnO4

Moles of KMnO4 required for complete reaction = Molarity x volume (in L) of KMnSO4 solution

                                    = 1.2651 x 10^-4 M x 0.0203 L = 2.568157 x 10^-6 moles

Since 2 moles KMnO4 reacts with 10 moles FeO in the above stoichiometry, Or, 1 mol KMnO4 reacts with 5 moles FeO. Thus, the number of moles of FeO in the sample is equal to 5 times the number of moles of KMnO4

            So, moles of FeO = 5 x 2.568157 x 10^-6 moles = 1.2840 x 10^-5 moles

Mass of FeO = moles of FeO x molar mass

                        = 1.2840 x 10^-5 moles x 71.844 g mol^-1= 0.00092253 gram

Mass of Fe2O3 (originally present in the ore) = total mass of ore- mass of FeO

                        = 0.330 g - 0.00092253 g = 0.32907747 g

Part C: Ore Sample 2: It’s assumed that the reactions are similar to previous sample.

Moles of KMnO4 required for complete reaction = Molarity x volume (in L) of KMnSO4 solution

                                    = 1.2651 x 10^-4 M x 0.0426 L = 5.389335 x 10^-6 moles

            Now, moles of FeO = 5 x 5.389335 x 10^-6 moles = 2.69466 x 10^-5 moles

Mass of FeO = moles of FeO x molar mass

                        = 2.69466 x 10^-5 moles x 71.844 g mol^-1= 0.0019359 gram

Mass of Fe2O3 (originally present in the ore) = total mass of ore- mass of FeO

                        = 0.370 g - 0.0019359 g = 0.3680641g