\(a,n_{FeO}=\dfrac{6,12}{72}=0,085\left(mol\right)\\
pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
0,085 0,085
\(m_{Fe}=0,085.56=4,76g\)
\(b,n_{Fe_2O_3}=\dfrac{20,8}{160}=0,13\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,13 0,26
\(m_{Fe}=0,26.56=14,56g\)
\(c,n_{Fe_3O_4}=\dfrac{51,04}{232}=0,22\left(mol\right)\\
pthh:Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
0,22 0,66
\(m_{Fe}=0,66.56=36,96g\)