Gọi \(X_3\) có dạng \(Fe_xO_y\)
\( \to x:y = \frac{{\% {m_{Fe}}}}{{56}}:\frac{{\% {m_O}}}{{16}} = \frac{{72,4\% }}{{56}}:\frac{{27,6\% }}{{16}} = 3:4\)
Vậy \(X_3\) là \(Fe_3O_4\)
Suy ra \(X_1\) là \(FeO\); \(X_2\) là \(Fe_2O_3\); \(X_3\) là \(Fe_3O_4\)
Các phản ứng xảy ra:
(1) \(FeO + {H_2}\xrightarrow{{{t^o}}}Fe + {H_2}O\)
(2) \(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
(3) \(F{e_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3Fe + 4{H_2}O\)
(4) \(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
(5) \(FeS{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + FeC{l_2}\)
(6) \(FeC{l_2} + 2NaOH\xrightarrow{{}}Fe{(OH)_2} + 2NaCl\)
(7) \(Fe{(OH)_2}\xrightarrow{{{t^o}}}FeO + {H_2}O\)
(8) \(2Fe + 3C{l_2}\xrightarrow{{{t^o}}}2FeC{l_3}\)
(9) \(FeC{l_3} + 3NaOH\xrightarrow{{}}Fe{(OH)_3} + 3NaCl\)
(10) \(2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O\)