Đáp án:
\(\begin{array}{l}
a)\\
\% Fe = 53,85\% \\
\% Mg = 46,15\% \\
b)\\
{C_{{M_{HCl}}}} = 1,2M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe(a\,mol);Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
56a + 24b = 10,4
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{10,4}} \times 100\% = 53,85\% \\
\% Mg = 100 - 53,85 = 46,15\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,6}}{{0,5}} = 1,2M
\end{array}\)