Đáp án:
\(\begin{array}{l}
2.\\
a.Fe + 2HCl \to FeC{l_2} + {H_2}\\
b.\\
\% {m_{Fe}} = 56\% \\
\% {m_{Cu}} = 44\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
2.\\
a.Fe + 2HCl \to FeC{l_2} + {H_2}\\
b.\\
{n_{{H_2}}} = 0,3mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,3mol\\
\to {m_{Fe}} = 16,8g\\
\to \% {m_{Fe}} = \dfrac{{16,8}}{{30}} \times 100\% = 56\% \\
\to \% {m_{Cu}} = 100\% - 56\% = 44\%
\end{array}\)
