Đáp án đúng: D
Giải chi tiết:\(\begin{gathered} \begin{array}{*{20}{l}} {{n_{Al}}{\text{ }} = {\text{ }}0,4{\text{ }}mol} \\ {{n_{F{e_3}{O_4}}}{\text{ }} = {\text{ }}0,15{\text{ }}mol} \end{array} \hfill \\ Gia\,su\,H = x \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8Al\,\,\,\,\,\,\,\, + 3F{e_3}{O_4}\xrightarrow{{{t^o}}}4A{l_2}{O_3} + 9Fe \hfill \\ Bd:\,\,\,\,\,\,\,\,\,0,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15 \hfill \\ Pu:\,\,\,\,\,\,\,\,\,\,0,4x\,\,\,\,\,\,\,\,\,\,\,0,15x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2x\,\,\,\,\,\,\,\,0,45x \hfill \\ Sau:0,4 - 0,4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,45x \hfill \\ {n_{{H_2}}} = 1,5{n_{Al}} + {n_{Fe}} \to 0,48 = 1,5.(0,4 - 0,4x) + 0,45x \hfill \\ \to x = 0,8 \hfill \\ \to H = 80\% \hfill \\ \end{gathered} \)
Đáp án D