\(pthh:4H_2+Fe_3O_4\overset{t^o}{--->}3Fe+4H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\end{matrix}\right.\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\)
Vậy H2 dư.
Theo pt: \(n_{Fe}=3.0,1=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
Theo pt: \(n_{H_{2_{PỨ}}}=4.0,1=0,4\left(mol\right)\)
\(\Rightarrow m_{H_{2_{dư}}}=\left(0,5-0,4\right).2=0,2\left(g\right)\)