20.141 The percent iron in iron ore can be determined by di ing the ore in acid,
ID: 502728 • Letter: 2
Question
20.141 The percent iron in iron ore can be determined by di ing the ore in acid, then reducing the ron to ISSolN- Fe2+ and finally titrating the Fe with aqueous KMnou. The reac- tion products are Fe and Mn (a) Write a balanced net ionic equation for the titration reaction. (b) Use the Eo Values in Appendix D to calculate AGo kilojoules) and the equilibrium constant for the reaction (c) Draw a crystal field energy-level diagram for the reactants and products. MnO [Fe(H20) and [Mn(H2O)6] and predict the num ber of unpaired electrons for each. (d) Does the paramagnetism of the solution increase or decrease as the reaction proceeds? Explain (e) What is the mass Fe in the iron ore if titration of the Fe2 from a 1.265 g sample of ore requires 34.83 mLof 0.051 32 M KMno4 to reach the equivalence point?Explanation / Answer
Volume of KMnO4 used = 34.83 mL = 0.03483 L
Concentration of KMnO4 = 0.05132 M
No of moles of KMnO4 = Concentration * Volume = 0.05132 * 0.03483 = 0.00179 moles
As the reaction between KMnO4 with Fe2+
MnO4- + 5 Fe2+ + 8H+ ------------> 5 Fe3+ + Mn2+ + 4H2O
thus according to equation
1 mole of KMnO4 reacts with 5 moles of Fe(II)
so 0.00179 moles moles will react with = 5 *0.00179 = 0.00895 moles of Fe(II)
no of moles of Fe(II) present in sample = 0.00895 moles
the mass of iron in the sample is = no of moles * atomic mass of Fe = 0.00895 * 56 = 0.5012 grams
mass percent of Fe in sample = mass of iron in sample *100 / mass of sample = 0.5012 *100/ 1.265 =39.62 % Fe
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