Đáp án:

\(\begin{array}{l}
a)\\
\% Zn = 51,7\% \\
\% Fe = 48,3\% \\
b)\\
{C_{{M_{HCl}}}} = 0,6l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
hh:Zn(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,5\\
65a + 56b = 30,16
\end{array} \right.\\
 \Rightarrow a = 0,24mol;b = 0,26mol\\
{m_{Zn}} = 0,24 \times 65 = 15,6g\\
\% Zn = \dfrac{{15,6}}{{30,16}} \times 100\%  = 51,7\% \\
\% Fe = 100 - 51,7 = 48,3\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1mol\\
{n_{HC{l_{tt}}}} = \dfrac{{1 \times 120}}{{100}} = 1,2mol\\
{C_{{M_{HCl}}}} = \dfrac{{1,2}}{2} = 0,6l
\end{array}\)

Cho $\begin{cases} Zn: x(mol)\\ Fe: y(mol)\\\end{cases}$

Ta có: `65x+56y=30,16g(1)`

`n_{H_2}=\frac{11,2}{22,4}=0,5(mol)`

Phương trình:

`Zn+2HCl\to ZnCl_2+H_2`

`Fe+2HCl\to FeCl_2+H_2`

Ta nhận thấy: `n_{H_2}=x+y=0,5(mol)(2)`

Từ `(1),(2)` suy ra $\begin{cases} x=0,24(mol)\\y=0,26(mol)\\\end{cases}$

`a)` `%m_{Zn}=\frac{65.0,24.100%}{30,16}\approx 51,72%`

`%m_{Fe}=\frac{56.0,26.100%}{30,16}\approx 48,27%`

`b)` Ta nhận thấy: `2n_{H_2}=n_{HCl}=1(mol)`

Do dùng dư `20%`

`=> m_{HCl \text{ bđ}}=\frac{1.120}{100}=1,2(mol)`

`=> V_{HCl}=\frac{n}{V}=\frac{1,2}{2}=0,6(l)`