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A sample of 1.55 g of iron ore is dissolved in an acid solution in which the iron is converted into fe2 . the solution formed is then titrated with kmno4 which oxidises fe2 to fe3 while the mno4- ions are reduced to mn2 ions. 92.95 ml of 0.020 m kmno4 is required for the titration to reach the equivalence point. a) write the balanced equation for the titration. b) calculate the percentage of iron in the sample.​

Respuesta :

(a) The reaction is redox in nature.

(b) The percentage of Fe2+ is 34.3%.

The overall balanced redox reaction equation is:

[tex]MnO4^{-} + 8H^{+} + 5Fe^{2} + -------- > Mn^{2} + 4H2O + 5Fe^{3}[/tex]

Number of permanganate moles = 0.020 M * 92.95/1000 L = 0.0019 moles

According to the reaction equation:

5 moles of  Fe^{2}+ (permanganate) reacts with 1 mole of permanganate

x moles of Fe^{2}+ (permanganate) reacts with 0.0019 moles of permanganate

x = 5 * 0.0019 /1 = 0.0095 moles of Fe^{2}+

Mass of Fe^{2}+ = 0.0095 moles of Fe^{2}+ * 56 g/mol = 0.532 g

Iron percentage = 0.532 g/1.55 g * 100 = 34.3%

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