Answer to Question #67585 in General Chemistry for Gage

Question

If a 0.1593 gram sample of iron metal is oxidized to from iron (II) in aqueous solution, the iron (II) requires 28.54 mL of KMnO4(aq) solution to reach the endpoint. Determine the molarity of the KMNO4(aq).

Answer 1

Chemical equation:
10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2 (SO4)3 + 2MnSO4 + K2SO4 + 8H2O
At the equivalence point we have:
n(Fe) = C(KMnO4)·V(KMnO4)
The number of moles of Fe equals:
n(Fe) = m(Fe)/M(Fe) = 0.1593/55.85 = 0.0029 (moles)
Thus, molarity of the KMnO4 equals:
C(KMnO4) = (n(Fe))/(V(KMnO4))
So, plug in values to find
C(KMnO4) = 0.0029/0.02854 = 0.102 (mol/L)

Answer: C(KMnO4)= 0.102 (mol/L).