\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(MOL\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,05 0,1
\(m_{Fe\left(lt\right)}=0,1.56=5,6\left(g\right)\\
m_{Fe\left(tt\right)}=\dfrac{5,6.90}{100}=5,04\left(g\right)\)