Đáp án:
\(\begin{array}{l}
5)\\
2.\\
{m_{AlC{l_3}}} = 26,7g\\
3.\\
{C_M}HCl = 3M\\
6)\\
2.\\
{m_{FeC{l_2}}} = 76,2g\\
3.\\
{C_M}HCl = 6M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5)\\
1.\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
2.\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{AlC{l_3}}} = \dfrac{{0,3 \times 2}}{3} = 0,2\,mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
3.\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{C_M}HCl = \dfrac{{0,6}}{{0,2}} = 3M\\
6)\\
1.\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2.\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
{n_{FeC{l_2}}} = {n_{{H_2}}} = 0,6\,mol\\
{m_{FeC{l_2}}} = 0,6 \times 127 = 76,2g\\
3.\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6 \times 2 = 1,2\,mol\\
{C_M}HCl = \dfrac{{1,2}}{{0,2}} = 6M
\end{array}\)