Em tham khảo nha :
\(\begin{array}{l}
1)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{HCl}} = 0,6 \times 1 = 0,6mol\\
{n_{AlC{l_3}}} = \dfrac{{{n_{HCl}}}}{3} = 0,2mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
2)\\
NaOH\,20\% :\\
{m_{NaOH}} = \dfrac{{200 \times 20}}{{100}} = 40g\\
NaOH\,10\% \\
{m_{NaOH}} = \dfrac{{400 \times 10}}{{100}} = 40g\\
C{\% _{NaOH}} = \dfrac{{40 + 40}}{{200 + 400}} \times 100\% = 13,33\%
\end{array}\)