50g of steel wire was dissolved in excessive of diluted sulphuric acid and the solution was made up to 250ml in a standard flask. 25ml of this solution was pipetted into a conical flask and needed 25.45ml of 0.02M KMnO4 for complete oxidation.
calculate the percentage of iron in the sample.

Question

50g of steel wire was dissolved in excessive of diluted sulphuric acid and the solution was made up to 250ml in a standard flask. 25ml of this solution was pipetted into a conical flask and needed 25.45ml of 0.02M KMnO4 for complete oxidation.
calculate the percentage of iron in the sample.

anonymous · Answer

apa??

anonymous · Answer

Your first equation:
Fe+H2SO4->FeSO4+ H2
When you add KMnO4,
n factor for KMnO4=5(Mn7+ to Mn2+) and for Fe it is 1(Fe2+ to Fe3+)
so by law of equivalents you have
5*0.02*25.45=x*1*25. (x is molarity of FeSO4).
x=2.545/25=0.1018
No of moles in 250ml or 1/4 L= 0.1018*1/4=0.02545.
Moles of FeSO4=moles of Fe.
Moles of Fe in the wire=0.02545.
Weight=moles*molecular weight=0.02545*55=1.4g.
Now find precentage of iron. (I have a feeling i made a mistake in calculations- i always do. The method is correct).