`-` `n_(H_2) = (5,6)/(22,4) = 0,25 (mol)`

a)  PTHH :  `2Al + 6HCl → 2AlCl_3 + 3H_2↑`

`-` Theo PTHH : `n_(Al) = (0,25 × 2)/3 = 1/6 (mol)`

`→` `m_(Al) = 1/6 × 27 = 4,5 (gam)`

b) `-` Theo PTHH : `n_(HCl) = 2n_(H_2) = 0,25 × 2 = 0,5 (mol)`

`→` `m_(HCl) = 0,5 × 36,5 = 18,25 (gam)`

`→` `C%_(ddHCl) = (18,25)/120 × 100% = 15,2%`

c) `-` `m_(dd) = 4,5 + 120 - 0,25 × 2 = 124 (gam)`

`-` Theo PTHH : `n_(AlCl_3) = n_(Al) = 1/6 (mol)`

`→` `m_(AlCl_3) = 1/6 × 133,5 = 22,25 (gam)`

`→` `C%_(AlCl_3) = (22,25)/124 × 100% = 17,94%`

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = \frac{1}{6}mol\\
{m_{Al}} = \dfrac{1}{6} \times 27 = 4,5g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5mol\\
{m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
C{\% _{HCl}} = \dfrac{{18,25}}{{120}} \times 100\%  = 15,2\% \\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = \frac{1}{6}mol\\
{m_{AlC{l_3}}} = \dfrac{1}{6} \times 133,5 = 22,25g\\
{m_{ddspu}} = 120 + 4,5 - 0,25 \times 2 = 124g\\
C{\% _{AlC{l_3}}} = \dfrac{{22,25}}{{124}} \times 100\%  = 17,94\% 
\end{array}\)