Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 53,85\% \\
\% {m_{Mg}} = 46,15\% \\
b)\\
{C_\% }MgC{l_2} = 6,13\% \\
{C_\% }FeC{l_2} = 4,1\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
56a + 24b = 10,4
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{10,4}} \times 100\% = 53,85\% \\
\% {m_{Mg}} = 100 - 53,85 = 46,15\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{7,3\% }} = 300g\\
{m_{{\rm{dd}}spu}} = 10,4 + 300 - 0,3 \times 2 = 309,8g\\
{C_\% }MgC{l_2} = \dfrac{{0,2 \times 95}}{{309,8}} \times 100\% = 6,13\% \\
{C_\% }FeC{l_2} = \dfrac{{0,1 \times 127}}{{309,8}} \times 100\% = 4,1\%
\end{array}\)
