Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 27,03\% \\
\% {m_{Ag}} = 72,97\% \\
b){m_{C{l_2}}} = 3,905g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,025mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,025mol\\
\to {m_{Fe}} = 1,4g\\
a)\\
\% {m_{Fe}} = \dfrac{{1,4}}{{5,18}} \times 100\% = 27,03\% \\
\% {m_{Ag}} = 100\% - 27,03\% = 72,97\% \\
b)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2Ag + C{l_2} \to 2AgCl\\
{n_{Ag}} = 0,035mol\\
\to {n_{C{l_2}}} = \dfrac{3}{2}{n_{Fe}} + \dfrac{1}{2}{n_{Ag}} = 0,055mol\\
\to {m_{C{l_2}}} = 3,905g
\end{array}\)