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