23,4 gam chứ 23,3 thì hơi lẻ em hi
\(4H_2+Fe_3O_4\rightarrow\left(t^o\right)3Fe+4H_2O\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ Vì:\dfrac{0,1}{1}>\dfrac{0,2}{4}\Rightarrow H_2hết,Fe_3O_4dư\\ n_{Fe_3O_4\left(p.ứ\right)}=\dfrac{0,2}{4}=0,05\left(mol\right)\\ n_{Fe_3O_4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ m_{Fe_3O_4\left(dư\right)}=0,05.232=11,6\left(g\right)\\ n_{H_2O}=n_{H_2}=0,2\left(mol\right)\\ m_{H_2O}=0,2.18=3,6\left(g\right)\)