a sample of iron ore (m=0,330g)consisting of a mixture of FeO and Fe2O3 was diss
ID: 1058058 • Letter: A
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
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
Explanation / Answer
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
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