Đáp án:
\(V = 2,24{\text{ lít}}\)
\({m_{Fe}} = 3,73{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol}} = {n_{{H_2}}}\)
Ta có:
\(V = {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
Dẫn khí \(H_2\) qua \(Fe_2O_3\)
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Ta có:
\({n_{F{e_2}{O_3}}} = \frac{{16}}{{56.2 + 16.3}} = 0,1{\text{ mol > }}\frac{1}{3}{n_{{H_2}}}\)
\( \to {n_{Fe}} = \frac{2}{3}{n_{{H_2}}} = \frac{{0,2}}{3}{\text{ mol}}\)
\( \to {m_{Fe}} = \frac{{0,2}}{3}.56 = 3,73{\text{ gam}}\)