\(n_{Fe_3O_4} = \dfrac{23,2}{232} = 0,1(mol)\\ n_{H_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ n_{Fe_3O_4} = 0,1 < \dfrac{n_{H_2}}{4} = 0,125 \to H_2\ dư\\ n_{Fe} = 3n_{Fe_3O_4} = 0,3(mol)\\ m_{Fe} = 0,3.56 = 16,8(gam)\)
PTHH: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\) \(\Rightarrow\) H2 còn dư, Fe3O4 p/ứ hết
\(\Rightarrow n_{Fe}=0,3\left(mol\right)\) \(\Rightarrow m_{Fe}=0,3\cdot56=16,8\left(g\right)\)
