\(n_{H_2}=\dfrac{4}{2}=2\left(mol\right)\\ 4H_2+Fe_3O_4\rightarrow\left(t^o\right)3Fe+4H_2O\\ n_{Fe}=\dfrac{3}{4}.2=1,5\left(mol\right)\\ \Rightarrow m_{Fe}=1,5.56=84\left(g\right)\\ \Rightarrow ChọnB\)
n H2=2 mol
Fe3O4+4H2-to>3Fe+4H2O
0,5---------2-------1,5
=>mFe=1,5.56=84g
=>B