Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
a)\\
{n_{Zn}} = 0,1mol\\
{m_{HCl}} = \dfrac{{200 \times 14,6\% }}{{100\% }} = 29,2g\\
\to {n_{HCl}} = 0,8mol\\
\to {n_{HCl}} > {n_{Fe}} \to {n_{HCl}}dư\\
\to {n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
\to {V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
\to {m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{HCl(pt)}} = 2{n_{Fe}} = 0,2mol\\
\to {n_{HCl(dư)}} = 0,6mol\\
\to {m_{HCl(dư)}} = 0,6 \times 36,5 = 21,9g\\
\to {m_{ddFeC{l_2}}} = {m_{Fe}} + {m_{ddHCl}} - {m_{{H_2}}} = 5,6 + 200 - 0,2 = 205,4g\\
\to C{\% _{ddZnC{l_2}}} = \dfrac{{12,7}}{{205,4}} \times 100\% = 6,18\% \\
\to C{\% _{HCl(dư)}} = \dfrac{{21,9}}{{205,4}} \times 100\% = 10,66\%
\end{array}\)
