Đáp án:

\({m_{Al}} = 13,5{\text{ gam}};{{\text{m}}_{F{{\text{e}}_3}{O_4}}}  = 34,8{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(8{\text{A}}l + 3F{{\text{e}}_3}{O_4}\xrightarrow{{{t^o}}}4{\text{A}}{l_2}{O_3} + 9F{\text{e}}\)

\(2NaOH + 2{\text{A}}l + 2{H_2}O\xrightarrow{{}}2NaAl{O_2} + 3{H_2}\)

\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{Al{\text{ dư}}}} = \frac{2}{3}{n_{{H_2}}} = 0,1{\text{ mol}}\)

Gọi số mol \(Al\) phản ứng là \(x\)

\( \to {m_{F{{\text{e}}_3}{O_4}}} = \frac{3}{8}{n_{Al{\text{ phản ứng}}}} = \frac{3}{8}x = 0,375{\text{x}}\)

\( \to 27{\text{x}} + 0,375{\text{x}}.(56.3 + 16.4) + 0,1.27 = 48,3{\text{ gam}}\)

\( \to x=0,4\)

\( \to {m_{Al}} = 0,4.27 + 0,1.27 = 13,5{\text{ gam}} \to {{\text{m}}_{F{{\text{e}}_3}{O_4}}} = 48,3 - 13,5 = 34,8{\text{ gam}}\)