`a)`

`2Al + 6HCl → 2AlCl_3 + 3H_2`

`n_{HCl} = {375}/{1000} . 0,2= 0,075(mol)`

`n_{H_2} = {0,672}/{22,4} = 0,03(mol)`

Theo pt: 

`n_{Al} = 2/3n_{H_2}`

`= {0,03 . 2}/3 = 0,02(mol)`

`n_{AlCl_3} = n_{Al} = 0,02(mol)`

`n_{HCl} ≠ 2n_{H_2}`

`→ HCl` dư 

`n_{HCl} (text{dư}) = 0,075 - 2n_{H_2}`

`= 0,075 - 0,03 . 2 = 0,015(mol)`

`C_{M_{AlCl_3}} = {0,02}/{{375}/{1000}} = 0,053(M)`

`C_{M_{HCl}} = {0,015}/{{375}/{1000}} = 0,04(M)`

`c)`

`n_{Al} = {0,02.  3,54}/{1,18} = 0,06(mol)`

`n_S = {3,54 - 0,06 .27}/{32} = 0,06(mol)`

`2Al + 3S`$\xrightarrow{t°}$`Al_2S_3`

`{n_{Al}}/2 > {n_S}/3`

`→ Al` dư 

`n_{Al} (text{dư}) = 0,06 - 3/2n_S`

`= 0,06 - {0,06. 2}/3 = 0,02(mol)`

Bảo toàn khối lượng: 

`m_C = m_A`

`%m_{Al} = {0,02.27}/{3,54} .100 = 15,25%`

`%m_{Al_2S_3} = 100 - 15,25 = 84,75%`

\(\begin{array}{l}
a)\\
375ml=0,375l\\
2Al+6HCl\to 2AlCl_3+3H_2\\
n_{H_2}=\frac{0,672}{22,4}=0,03(mol)\\
Theo\,PT:\,n_{HCl(du)}=0,375.0,2-0,03.2=0,015(mol)\\
Theo\,PT:\,n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,02(mol)\\
\to \begin{cases} C_{M\,HCl(du)}=\frac{0,015}{0,375}=0,04M\\ C_{M\,AlCl_3}=\frac{0,02}{0,375}\approx 0,053M \end{cases}\\
b)\\
Theo\,PT:\,n_{Al}=n_{AlCl_3}=0,02(mol)\\
\to n_S=\frac{1,18-0,02.27}{32}=0,02(mol)\\
2Al+3S\xrightarrow{t^o}Al_2S_3\\
Vi\,\frac{n_{Al}}{2}>\frac{n_S}{3}\to Al\,du\\
Theo\,PT:\,n_{Al_2S_3}=\frac{1}{3}n_S=\frac{1}{150}(mol)\\
n_{Al(du)}=0,02-\frac{2}{3}n_S=\frac{1}{150}(mol)\\
\to \begin{cases} m_{Al_2S_3}=\frac{1}{150}.150=1(g)\\ m_{Al(du)}=\frac{1}{150}.27=0,18(g) \end{cases}
\end{array}\)