Đáp án:
\(\begin{array}{l}
\% {m_{Fe}} = 14,9\% \\
\% {m_{F{e_2}{O_3}}} = 85,1\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}(1)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O(2)\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}(3)
\end{array}\)
\(\begin{array}{l}
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Fe}} = 5,6g\\
{n_{Fe(3)}} = 0,4mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe(3)}} = 0,2mol\\
\to {m_{F{e_2}{O_3}}} = 32g\\
\to a = {m_{Fe}} + {m_{F{e_2}{O_3}}} = 37,6g
\end{array}\)
\(\begin{array}{l}
\to \% {m_{Fe}} = \dfrac{{5,6}}{{37,6}} \times 100\% = 14,9\% \\
\to \% {m_{F{e_2}{O_3}}} = \frac{{32}}{{37,6}} \times 100\% = 85,1\%
\end{array}\)
