Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 22,31\% \\
\% {m_{Zn}} = 77,69\% \\
b)\\
{m_{FeC{l_2}}} = 12,7g\\
{m_{ZnC{l_2}}} = 40,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,4\\
56a + 65b = 25,1
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,3\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{25,1}} \times 100\% = 22,31\% \\
\% {m_{Zn}} = 100 - 22,31 = 77,69\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,3\,mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{m_{ZnC{l_2}}} = 0,3 \times 136 = 40,8g
\end{array}\)
