Permanganate ion oxidizes ferrous ion quantitatively in acid solution at room temperature and the reaction can be used to determine the amount of ironin a sample  A) Write a balanced net ionic...

Permanganate ion oxidizes ferrous ion quantitatively in acid solution at room temperature and the reaction can be used to determine the amount of iron

in a sample 

A) Write a balanced net ionic equation for the reaction between MnO4^- and Fe ^2+ in acid solution 

B) A sample of iron ore weighing 0.7545 g is dissolved in acid and all the iron is completely converted to Fe ^2+ ions. This solution is titrated with 0.0400M KMnO4, and 31.96 mL of the permanganate solution are required to reach the end point in the titration.

i)How many millimoles of Fe^2+ were titrated?

ii)What was the percentage of iron in the iron ore? 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Here There is a little correction.

In iron ore we have `Fe^(2+)` and when we add `MnO_4^-` it will oxidize to Fe^(3+).

A)

The reaction takes place in an acidic medium.

`5Fe^(2+)+MnO_4^-+8H^+ -----gt 5Fe^(3+)+Mn^(2+)+4H_2O`

B)

i)

Amount of `MnO_4^-` moles cosumed = 0.04/1000*31.96

                                                    = 1.2784*10^(-3)mol

According to the balance ionic equation `MnO_4^(-):Fe^(2+)` = 1:5

Therefore `Fe^(2+)` moles in sample = `1.2784*10^(-3)*5`

                                                =  `6.392*10^(-3) mol`

So in iron ore there were 6.392 milimol of `Fe^(2+)`

 

ii)

The form of iron in iron ore is FeO.

Weight of one mole of FeO approximately = (56+16) = 72 g/mol

Number of moles in irom Ore = 0.7545/72 mol

                                          =0.01 mol

                                          = 10milimol

 

In Pure iron ore we should have 10milimol of `Fe^(2+)` when we take 0.7545g of iron ore.

But the titration shows that we have only 6.392milimol

So % of iron in iron ore           = 6.392/10*100%

                                                = 63.92%

Sources:

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