Đáp án đúng: D
27,95%; 72,05%.
sau pư đầu tiên còn dư Al sau đó Al td vs NaOH đuợc khí H2 thoát ra$\displaystyle \begin{array}{l}{Al+NaOH+}{{{H}}_{{2}}}{O}\xrightarrow{{}}{ NaAl}{{{O}}_{{2}}}{ + }\frac{3}{2}{{{H}}_{{2 }\!\!~\!\!{ }}}\\{{{n}}_{{{{H}}_{{2}}}}}{=0,3}\Rightarrow {{{n}}_{{Al}}}{=0,2 }\!\!~\!\!{ }\end{array}$nếu cho hỗn hợp sau pư 1 tác dụng HCl thì chỉ có Al(dư) và Fe tác dụng HCl tạo khí H2 $\displaystyle \begin{array}{l}{Al +3HCl}\xrightarrow{{}}{ AlC}{{{l}}_{{3}}}{+}\frac{{3}}{{2}}{{{H}}_{{2}}}\\{0,2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{ 0,3 }\!\!~\!\!{ }\\{Fe +2HCl }\xrightarrow{{}}{ FeC}{{{l}}_{{2}}}{+ }{{{H}}_{{2}}}\\{ }\!\!~\!\!{ a }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a }\!\!~\!\!{ }\end{array}$
$\displaystyle {{{n}}_{{{{H}}_{{2}}}}}$ tạo thành sau 2 pư = 26,88/22,4=1,2 a=1,2-0,3=0,9 $\displaystyle \begin{array}{l}{3F}{{{e}}_{{3}}}{{{O}}_{{4}}}{+8Al }\xrightarrow{{}}{ 9Fe +4A}{{{l}}_{{2}}}{{{O}}_{{3}}}{ }\!\!~\!\!{ }\\\,\,\,\,\,{0,3 }\,\,\,\,\,\,{0,8 }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{0,9 }\!\!~\!\!{ }\\{{{n}}_{{Al}}}{=0,8+0,2=1}\\{{{n}}_{{F}{{{e}}_{{3}}}{{{O}}_{{4}}}}}{= 0,3}\end{array}$
