\(2Na +2H_2O \to 2NaOH + H_2\\ n_{H_2} = \dfrac{1}{2}n_{Na} = \dfrac{1}{2}.\dfrac{11,5}{23}= 0,25(mol)\\ n_{H_2\ thu\ được} = 0,25 - 0,25.20\% = 0,2(mol)\\ n_{Fe_3O_4} = \dfrac{17,4}{232} =0,075(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ n_{H_2} = 0,2 < 4n_{Fe_3O_4} =0,075.4 = 0,3\Rightarrow Fe_3O_4\ dư\\ n_{Fe} = \dfrac{3}{4}n_{H_2} = \dfrac{3}{4}.0,2 = 0,15(mol)\\ m_{Fe} = 0,15.56 = 8,4(gam)\)