Asked by Meso
Suppose 22.93mL of KMnO4 solution used in the question above (KMnO4 solution in above question had concentration of 0.02036M)are needed to oxidize Fe(2+) to Fe(3+) in a 0.4927g pill. What is the mass percent of FeSO4*7H2O (molar mass 278.03g/mol) in the pill?
I know I am supposed to use the KMnO4 to find mols of Fe(II). But after doing that I am unsure as to how i should use Fe(II) to find the mass of FeSO4*7H2O
I know I am supposed to use the KMnO4 to find mols of Fe(II). But after doing that I am unsure as to how i should use Fe(II) to find the mass of FeSO4*7H2O
Answers
Answered by
bobpursley
moles Fe+2 is the same as the moles of FeSO4.7H2O. mass of that is molmass * numbermoles
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