Đáp án:

 $22A$

$23B$

$24B$

Giải thích các bước giải:

\(\begin{array}{l}
22)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Cu + C{l_2} \to CuC{l_2}\\
hh:Fe(a\,mol),Cu(b\,mol)\\
{n_{C{l_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
\left\{ \begin{array}{l}
56a + 64b = 12\\
1,5a + b = 0,25
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{12}} \times 100\%  = 46,67\% \\
23)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{KMn{O_4}}} = \dfrac{{20,54}}{{158}} = 0,13\,mol\\
{n_{C{l_2}}} = 0,13 \times \frac{5}{2} = 0,325\,mol\\
{V_{C{l_2}}} = 0,325 \times 22,4 = 7,28l\\
24)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{KMn{O_4}}} = \dfrac{{12,64}}{{158}} = 0,08\,mol\\
{n_{C{l_2}}} = 0,08 \times \frac{5}{2} = 0,2\,mol\\
H = 60\%  \Rightarrow {V_{C{l_2}}} = 0,2 \times 60\%  \times 22,4 = 2,688l
\end{array}\)