\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ 4CO+Fe_3O_4\rightarrow\left(t^o\right)3Fe+4CO_2\\ V\text{ì}:\dfrac{0,1}{1}>\dfrac{0,2}{4}\Rightarrow Fe_3O_4d\text{ư}\\ \Rightarrow n_{Fe}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ n_{Fe_3O_4\left(d\text{ư}\right)}=0,1-\dfrac{0,2}{4}=0,05\left(mol\right)\\ m_{r\text{ắn}}=m_{Fe_3O_4\left(d\text{ư}\right)}+m_{Fe}=0,05.232+0,15.56=20\left(g\right)\)