\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe(NO3)3 + 3KOH --> Fe(OH)3\(\downarrow\) + 3KNO3
________0,1<-------------------0,1
2Fe(OH)3 --to--> Fe2O3 + 3H2O
_0,1<-------------0,05
=> nFe(NO3)3 = 0,1(mol)
=> \(C_M=\dfrac{0,1}{0,2}=0,5M\)