\(n_{khí}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH:
\(Fe_2O_3+3H_2\xrightarrow[t^o]{}2Fe+3H_2O\left(1\right)\\ Fe_2O_3+3CO\xrightarrow[t^o]{}2Fe+3CO_2\left(2\right)\)
Theo PTHH \(\left(1\right)\left(2\right):n_{Fe_2O_3}=\dfrac{1}{3}n_{khí}=\dfrac{1}{3}.0,6=0,2\left(mol\right)\)
=> m = 0,2.160 = 32 (g)