Phương trình hóa học:

\(R + {H_2}S{O_4}\xrightarrow{{}}RS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \\\to {{\text{n}}_R} = {n_{{H_2}}} = 0,15{\text{ mol}}\)

\( \to {M_R} = \dfrac{{8,4}}{{0,15}} = 56 \\\to R:F{\text{e}}\)(sắt)

Hoà tan hỗn hợp \(X\)

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

\(F{{\text{e}}_2}{O_3} + 6HCl\xrightarrow{{}}2F{\text{e}}C{l_3} + 3{H_2}O\)

\(F{\text{e}}C{l_2} + 3{\text{A}}gN{O_3}\xrightarrow{{}}2{\text{A}}gCl + Ag + F{\text{e}}{(N{O_3})_3}\)

\(F{\text{e}}C{l_3} + 3{\text{A}}gN{O_3}\xrightarrow{{}}F{\text{e}}{(N{O_3})_3} + 3{\text{A}}gCl\)

\({n_{F{\text{e}}}} = \dfrac{{11,2}}{{56}} = 0,2{\text{ mol = }}\\{{\text{n}}_{F{\text{e}}C{l_2}}};{n_{F{{\text{e}}_2}{O_3}}} = \dfrac{{4,8}}{{56.2 + 16.3}} = 0,03{\text{ mol}} \to {{\text{n}}_{F{\text{e}}C{l_3}}} = 2{n_{F{{\text{e}}_2}{O_3}}} = 0,06{\text{ mol}}\)

\( \to {n_{HCl}} = 2{n_{F{\text{e}}C{l_2}}} + 3{n_{F{\text{e}}C{l_3}}} = 0,2.2 + 0,06.3 = 0,58{\text{ mol}}\)

\( \to V = {V_{HCl}} = \dfrac{{0,58}}{2} = 0,29{\text{ lít = 290 ml}}\)

\({n_{Ag}} = {n_{F{\text{e}}C{l_2}}} = 0,2{\text{ mol;}}\\{{\text{n}}_{AgCl}} = {n_{HCl}} = 0,58{\text{ mol}}\)

\( \to m = 0,2.108 + 0,58.143,5 = 104,83{\text{ gam}}\)

Hoà tan \(Fe_3O_4\) và \(H_2SO_4\) loãng dư

\(F{{\text{e}}_3}{O_4} + 4{H_2}S{O_4}\xrightarrow{{}}F{\text{e}}S{O_4} + F{{\text{e}}_2}{(S{O_4})_3} + 4{H_2}O\)

Dung dịch chứa \(H_2SO_4;FeSO_4;Fe_2(SO_4)_3\)

Các chất có phản ứng là \(K_2Cr_2O_7;K_2S;NaOH;Fe\)

+\(6F{\text{e}} + {K_2}C{{\text{r}}_2}{O_7} + 7{H_2}S{O_4}\xrightarrow{{}}3F{{\text{e}}_2}{(S{O_4})_3} + {K_2}S{O_4} + C{{\text{r}}_2}{(S{O_4})_3} + 7{H_2}O\)

+

\({K_2}S + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + {H_2}S\)

\(F{\text{e}}S{O_4} + {K_2}S\xrightarrow{{}}F{\text{e}}S + {K_2}S{O_4}\)

\(F{{\text{e}}_2}{(S{O_4})_3} + 3{K_2}S\xrightarrow{{}}2F{\text{e}}S + S + 3{K_2}S{O_4}\)

+

\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)

\(F{\text{e}}S{O_4} + 2NaOH\xrightarrow{{}}F{\text{e}}{(OH)_2} + N{a_2}S{O_4}\)

\(F{{\text{e}}_2}{(S{O_4})_3} + 6NaOH\xrightarrow{{}}2F{\text{e}}{(OH)_3} + 3N{a_2}S{O_4}\)

+

\(F{\text{e}} + {H_2}S{O_4}\xrightarrow{{}}F{\text{e}}S{O_4} + {H_2}\)

\(F{{\text{e}}_2}{(S{O_4})_3} + F{\text{e}}\xrightarrow{{}}3F{\text{e}}S{O_4}\)