2) 

\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

\(3C{l_2} + 6KOH\xrightarrow{{}}KCl{O_3} + 5KCl + 3{H_2}O\)

\(2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2}\)

\(KCl + {H_2}S{O_{4{\text{ đặc}}}}\xrightarrow{{}}KHS{O_4} + HCl\)

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

\(C{l_2} + Ca{(OH)_2}\xrightarrow{{}}CaOC{l_2} + {H_2}O\)

3)

Phản ứng xảy ra:

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

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

\(\to {n_{{H_2}}} = {n_{Fe}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {{\text{m}}_{F{e_2}{O_3}}} = 8 - 5,6 = 2,4{\text{ gam}}\)

\(\to \% {m_{Fe}} = \frac{{5,6}}{8} = 70\%  \to \% {n_{F{e_2}{O_3}}} = 30\% \)

Ta có: \({n_{F{e_2}{O_3}}} = \frac{{2,4}}{{160}} = 0,015{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{Fe}} + 6{n_{F{e_2}{O_3}}} = 0,1.2 + 0,015.6 = 0,29{\text{ mol}} \to {{\text{m}}_{HCl}} = 10,585{\text{ gam}}\)

\(\to {m_{dd{\text{ HCl}}}} = \frac{{10,585}}{{10\% }} = 105,85{\text{ gam}} \to {{\text{V}}_{dd{\text{HCl}}}} = \frac{{105,85}}{{1,1}} = 96,23ml = 0,09623{\text{ lít}}\)

Sau phản ứng dung dịch chứa:

\({n_{FeC{l_2}}} = {n_{Fe}} = 0,1{\text{ mol}} \to {{\text{C}}_{M{\text{ FeC}}{{\text{l}}_2}}} = \frac{{0,1}}{{0,09623}} = 1,04M;{\text{ }}{{\text{n}}_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,02{\text{ mol}} \to {{\text{C}}_{M{\text{ FeC}}{{\text{l}}_3}}} = \frac{{0,02}}{{0,09623}} = 0,208M\)