Đáp án:

b) Oxi dư 27,733 gam

c) \({{\text{m}}_{F{e_3}{O_4}}} = 15,467{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)

\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol; }}{{\text{n}}_{{O_2}}} = \frac{{22,4}}{{22,4}} = 1{\text{ mol > }}\frac{2}{3}{n_{Fe}} \to {O_2}\) dư.

\({n_{{O_2}{\text{ phản ứng}}}} = \frac{2}{3}{n_{Fe}} = \frac{{0,4}}{3}{\text{ mol}} \to {{\text{n}}_{{O_2}{\text{ dư}}}}1 - \frac{{0,4}}{3} = \frac{{13}}{{15}}{\text{ mol}} \to {{\text{m}}_{{O_2}{\text{ dư}}}} = \frac{{13}}{{15}}.32 = 27,733{\text{ }}gam\)

\({n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = \frac{{0,2}}{3}{\text{ mol}} \to {{\text{m}}_{F{e_3}{O_4}}} = \frac{{0,2}}{3}.(56.3 + 16.4) = 15,467{\text{ gam}}\)