\(a)Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ b)n_{H_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ n_{Fe} = \dfrac{3}{4}n_{H_2} = 0,3(mol)\\ n_{Fe_3O_4\ pư} = \dfrac{1}{4}n_{H_2} = 0,1(mol)\\ \Rightarrow m_{chất\ rắn\ sau\ phản\ ứng} = 0,3.56 + (34,8 -0,1.232)=28,4(gam)\\ c) \%m_{Fe_3O_4\ bị\ khử} = \dfrac{0,1.232}{34,8}.100\% = 66,67\%\)