Em tham khảo nha:
\(\begin{array}{l}
3)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{HCl}} = 2{n_{Mg}} = 0,2\,mol\\
{C_M}HCl = \dfrac{{0,2}}{{0,5}} = 0,4M\\
4)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3\,mol\\
{C_M}HCl = \dfrac{{0,3}}{{0,5}} = 0,6\,M\\
5)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 8,3\\
a + 1,5b = 0,25
\end{array} \right.\\
\Rightarrow a = b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{8,3}} \times 100\% = 67,47\% \\
\% {m_{Al}} = 100 - 67,47 = 32,53\%
\end{array}\)