Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{HCl}} = 0,6 \times 2 = 1,2mol\\
\dfrac{{0,2}}{2} < \dfrac{{1,2}}{6} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{HC{l_d}}} = 1,2 - 3 \times 0,2 = 0,6mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2mol\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,2}}{{0,6}} = \dfrac{1}{3}M\\
{C_{{M_{HCl}}}} = \dfrac{{0,6}}{{0,6}} = 1M
\end{array}\)
