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?
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+).
The reaction takes place in an acidic medium.
`5Fe^(2+)+MnO_4^-+8H^+ -----gt 5Fe^(3+)+Mn^(2+)+4H_2O`
Amount of `MnO_4^-` moles cosumed = 0.04/1000*31.96
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+)`
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
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%