Đáp án:

\(\begin{array}{l}
a)\\
\% Al = 34,6\% \\
\% A{l_2}{O_3} = 65,4\% \\
b)\\
{m_{{\rm{dd}}HCl}} = 300g\\
c)\\
C{\% _{AlC{l_3}}} = 8,68\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
{m_{A{l_2}{O_3}}} = 7,8 - 2,7 = 5,1g\\
\% Al = \dfrac{{2,7}}{{7,8}} \times 100\%  = 34,6\% \\
\% A{l_2}{O_3} = 100 - 34,6 = 65,4\% \\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{m}{M} = \dfrac{{5,1}}{{102}} = 0,05mol\\
{n_{HCl}} = 3{n_{Al}} + 6{n_{A{l_2}{O_3}}} = 3 \times 0,1 + 6 \times 0,05 = 0,6mol\\
{m_{HCl}} = n \times M = 0,6 \times 36,5 = 21,9g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{21,9 \times 100}}{{7,3}} = 300g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} + 2{n_{A{l_2}{O_3}}} = 0,2mol\\
{m_{AlC{l_3}}} = n \times M = 0,2 \times 133,5 = 26,7g\\
{m_{{\rm{dd}}spu}} = 7,8 + 300 - 0,15 \times 2 = 307,5g\\
C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{307,5}} \times 100\%  = 8,68\% 
\end{array}\)