a)
Gọi oxit \(A\) là \(Fe_aO_b\).
Cho \(A\) tác dụng với \(HCl\).
\(F{e_a}{O_b} + 2bHCl\xrightarrow{{}}F{e_a}C{l_{2b}} + b{H_2}O\)
Ta có:
\({n_{HCl}} = 0,8.1 = 0,8{\text{ mol}}\)
\( \to {n_{F{e_a}{O_b}}} = \frac{{{n_{HCl}}}}{{2b}} = \frac{{0,8}}{{2b}} = \frac{{0,4}}{b}\)
\( \to {M_{F{e_a}{O_b}}} = 56a + 16b = \frac{{23,2}}{{\frac{{0,4}}{b}}} = 58b\)
\( \to 56a = 42b \to \frac{a}{b} = \frac{{42}}{{56}} = \frac{3}{4}\)
Vậy oxit là \(Fe_3O_4\).
b)
Gọi oxit \(B\) là \(Fe_xO_y\).
Khử oxit trên
\(F{e_x}{O_y} + y{H_2}\xrightarrow{{{t^o}}}xFe + y{H_2}O\)
Ta có:
\({n_{{H_2}O}} = \frac{{10,8}}{{18}} = 0,6{\text{ mol}}\)
\( \to {n_{F{e_x}{O_y}}} = \frac{{{n_{{H_2}}}}}{y} = \frac{{0,6}}{y}\)
\( \to {M_{F{e_x}{O_y}}} = 56x + 16y = \frac{{32}}{{\frac{{0,6}}{y}}} = \frac{{160y}}{3}\)
\( \to 56x = \frac{{112y}}{3} \to x:y = \frac{{112}}{3}:56 = 2:3\)
Vậy oxit \(B\) là \(Fe_2O_3\)
