Em tham khảo nha :

\(\begin{array}{l}
{n_{Al}} = \dfrac{{8,1}}{{27}} = 0,3mol\\
{n_{HCl}} = \dfrac{{150 \times 49\% }}{{36,5}} = 2mol\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
\dfrac{{0,3}}{2} < \dfrac{2}{6} \Rightarrow HCl\text{ dư}\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,3mol\\
{m_{AlC{l_3}}} = 0,3 \times 133,5 = 40,05g\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,45mol\\
{V_{{H_2}}} = 0,45 \times 22,4 = 10,08l\\
C{\% _{AlC{l_3}}} = \dfrac{{40,05}}{{150 + 8,1 - 0,45 \times 2}} \times 100\%  = 25,5\% 
\end{array}\)

`n_{Al}=(8,1)/(27)=0,3(mol)`

`m_{HCl}=150.49%=73,5(g)`

`⇒n_{HCl}=(73,5)/(36,5)≈2,01(mol)`

`PTHH:

`2Al+6HCl→2AlCl_3+3H_2`

0,3       2.01   (mol)

Lập tỉ lệ : 

`(n_{Al})/2 =(0,3)/2 < (n_{HCl})/6 = (2,01)/6`

`⇒` sau p/ứ Al hết , HCL dư

`n_{AlCl_3}=n_{Al}=0,3(mol)`

`⇒m=m_{AlCl_3}=0,3.133,5=40,05(g)`

`n_{H_2}= 3/2 n_{Al}= 3/2 . 0,3= 0,2(mol)`

`⇒V_{H_2}=0,2.22,4=4,48(lít)`

`n_{HCl dư}=2,01-0,3.2=1,41(mol)`

Ta có khối lượng dung dịch sau p/ứ :

`m_{dd sau pứ}=m_{Al}+m_{HCl bđ}-m_{H_2}=8,1+150-0,2.2=157,7(g)`

⇒C%$_{AlCl_3}$=`(40,05)/(157,7).100`%≈25,4%

C%$_{HCl dư}$=`(1,41.36,5)/(157,7).100`%≈32,63%