Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{ZnO}} = 77,14\% \\
\% {m_{FeO}} = 22,86\% \\
b)\\
{m_{F{e_2}{O_3}}} = 8g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
ZnO + {H_2} \xrightarrow{t^0} Zn + {H_2}O\\
FeO + {H_2} \xrightarrow{t^0} Fe + {H_2}O\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:ZnO(a\,mol),FeO(b\,mol)\\
\left\{ \begin{array}{l}
81a + 72b = 31,5\\
a + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,3;b = 0,1\\
\% {m_{ZnO}} = \dfrac{{0,3 \times 81}}{{31,5}} \times 100\%  = 77,14\% \\
\% {m_{FeO}} = 100 - 77,14 = 22,86\% \\
b)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
ZnC{l_2} + 2KOH \to Zn{(OH)_2} + 2KCl\\
Zn{(OH)_2} + 2KOH \to {K_2}Zn{O_2} + 2{H_2}O\\
FeC{l_2} + 2KOH \to Fe{(OH)_2} + 2KCl\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \xrightarrow{t^0} 4Fe{(OH)_3}\\
2Fe{(OH)_3} \xrightarrow{t^0} F{e_2}{O_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{m_{F{e_2}{O_3}}} = 0,05 \times 160 = 8g
\end{array}\)