Đáp án:
\({{\text{m}}_{Fe}} = 39,2{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{FeO}} = \frac{{14,4}}{{56 + 16}} = 0,2{\text{ mol;}}{{\text{n}}_{F{e_2}{O_3}}} = \frac{{16}}{{56.2 + 16.3}} = 0,1{\text{ mol;}}{{\text{n}}_{F{e_3}{O_4}}} = \frac{{23,2}}{{56.3 + 16.4}} = 0,1{\text{ mol}}\)
Phản ứng xảy ra:
\(FeO + {H_2}\xrightarrow{{{t^o}}}Fe + {H_2}O\)
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{}}2Fe + 3{H_2}O\)
\(F{e_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3Fe + 4{H_2}O\)
\( \to {n_{{H_2}}} = {n_{FeO}} + 3{n_{F{e_2}{O_3}}} + 4{n_{F{e_3}{O_4}}} = 0,2 + 0,1.3 + 0,1.4 = 0,9{\text{ mol}} \to {{\text{V}}_{{H_2}O}} = 0,9.22,4 = 20,16{\text{ lít}}\)
\({n_{Fe}} = {n_{FeO}} + 2{n_{F{e_2}{O_3}}} + 3{n_{F{e_3}{O_4}}} = 0,2 + 0,1.2 + 0,1.3 = 0,7{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,7.56 = 39,2{\text{ gam}}\)
