Đáp án:

\( {{\text{V}}_{{H_2}}} = 8,96{\text{ lít}}\)

\({{\text{C}}_{M{\text{ HCl}}}}  = 4M\)

\( m = 25,6{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Gọi số mol \(Zn;Fe\) lần lượt là \(x;y\)

\( \to 65{\text{x}} + 56y = 24,2{\text{ gam}}\)

Ta có:

\({n_{ZnC{l_2}}} = {n_{Zn}} = x;{n_{F{\text{e}}C{l_2}}} = {n_{F{\text{e}}}} = y\)

\( \to 136{\text{x}} + 127y = 52,6{\text{ gam}}\)

Giải được:

\(x=y=0,2\)

\( \to {n_{{H_2}}} = {n_{F{\text{e}}}} + {n_{Zn}} = 0,4{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,4.22,4 = 8,96{\text{ lít}}\)

\({n_{HCl}} = 2{n_{{H_2}}} = 0,8{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,8}}{{0,2}} = 4M\)

Khử oxit

\(CuO + {H_2}\xrightarrow{{{t^o}}}CuO + {H_2}O\)

\(F{{\text{e}}_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2F{\text{e}} + 3{H_2}O\)

Ta có:

\({m_{CuO}} = {m_{F{{\text{e}}_2}{O_3}}} = 0,5m\)

\({n_{CuO}} = \frac{{0,5m}}{{80}} = \frac{m}{{160}};{n_{F{{\text{e}}_2}{O_3}}} = \frac{{0,5m}}{{56.2 + 16.3}} = \frac{m}{{320}}\)

\( \to {n_{{H_2}}} = \frac{m}{{160}} + \frac{m}{{320}}.3 = \frac{{5m}}{{320}} = 0,4 \to m = 25,6{\text{ gam}}\)