Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 36,48\% \\
\% {m_{Zn}} = 63,52\% \\
b)\\
{V_{{\rm{dd}}HCl}} = 0,5l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
56a + 65b = 30,7\\
a + b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,3\\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{30,7}} \times 100\% = 36,48\% \\
\% {m_{Zn}} = 100 - 36,48 = 63,52\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{1}{2} = 0,5l
\end{array}\)