Đáp án:

\( {m_{F{e_2}{O_3}}} = 11,2{\text{ gam}}\)

Giải thích các bước giải:

Phương trình hóa học:

\(Fe + 2HCl \to FeCl_2+ {H_2}\)

\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)
Ta có:

\({n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,05.56 = 2,8{\text{ gam}} \to {{\text{m}}_{F{e_2}{O_3}}} = 7,2{\text{ gam}}\)

\( \to {n_{F{e_2}{O_3}}} = \dfrac{{7,2}}{{56.2 + 16.3}} = 0,045{\text{ mol}}\)

\( \to {n_{FeC{l_2}}} = {n_{Fe}} = 0,05{\text{ mol;}}{{\text{n}}_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,09{\text{ mol}}\)

\(FeC{l_2} + 2NaOH\xrightarrow{{}}Fe{(OH)_2} + 2NaCl\)

\(FeC{l_3} + 3NaOH\xrightarrow{{}}Fe{(OH)_3} + 3NaCl\)

\(2Fe{(OH)_2} + \dfrac{1}{2}{O_2}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 2{H_2}O\)

\(2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O\)

\( \to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{FeC{l_2}}} + \dfrac{1}{2}{n_{FeC{l_3}}} = 0,07{\text{ mol}}\)

\( \to {m_{F{e_2}{O_3}}} = 0,07.160 = 11,2{\text{ gam}}\)