\(\begin{array}{l}
a)\\
Dat\,n_{Al}=x(mol);n_{Fe}=y(mol)\\
\to 27x+56y=5,5(1)\\
n_{H_2}=\frac{4,48}{22,4}=0,2(mol)\\
2Al+6HCl\to 2AlCl_3+3H_2\\
Fe+2HCl\to FeCl_2+H_2\\
Theo\,PT:\,1,5x+y=0,2(2)\\
(1)(2)\to x=0,1(mol);y=0,05(mol)\\
\to \%m_{Al}=\frac{0,1.27}{5,5}.100\%\approx 49,09\%\\
\to \%m_{Fe}=100-49,09=50,91\%\\
b)\\
Theo\,PT:\,n_{HCl}=2n_{H_2}=0,4(mol)\\
n_{AlCl_3}=x=0,1(mol);n_{FeCl_2}=y=0,05(mol)\\
m_{dd\,HCl}=\frac{0,4.36,5}{14,6\%}=100(g)\\
m_{dd\,spu}=5,5+100-0,2.2=105,1(g)\\
C\%_{FeCl_2}=\frac{0,05.127}{105,1}.100\%\approx 6,04\%\\
C\%_{AlCl_3}=\frac{0,1.133,5}{105,1}.100\%\approx 12,\%
\end{array}\)

Câu \(c)\) đề thiếu.

`#Ly`

$a)$

`n_(H_2` `=` `(V)/(22,4)` `=` `(4,48)/(22,4)=0,2` $(mol)$

PTHH:

$(1):$ $2Al$ $+$ $6HCl$ `->` $2AlCl_3$ $+$ $3H_2↑$

$(2):$ $Fe$ $+$ $2HCl$ `->` $FeCl_2$ $+$ $H_2↑$

Gọi $x,y$ là số $mol$ $Al,Fe$ pư

Theo PT $(1)$, `n_(H_2` `=` `(3)/(2)n_(Al` `=` $1,5x$ $(mol)$

Theo PT $(2)$, `n_(H_2` `=` `n_(Fe` `=` `y` $(mol)$

`∑``n_(H_2` `=` `1,5+y=0,2` $(mol)$

`m_(hhKL` `=` `27x+56y=5,5` $(g)$

Ta có hệ PT:

`1,5+y=0,2`

`27x+56y=5,5`

$⇒$ $\begin{cases} x=0,1\\y=0,05 \end{cases}$

`m_(Al` `=` `n.M=0,1.27=2,7` $(g)$

`%m_(Al` `=` `(m_{Al})/(m_{hhKL}).100%` `=` `(2,7)/(5,5).100%≈49,09%`

`%m_(Fe` `=` `100-49,09=50,91%`

$b)$

Từ $(1)$ và $(2)$ `=>` `n_(HCl` `=` `2n_(H_2` `=` `0,4` $(mol)$

`m_(HCl` `=` `n.M=0,4.36,5=14,6` $(g)$

`m_(ddHCl` `=` `(m_{HCl}.100)/(C%)` `=` `(14,6.100)/(14,6)=100` $(g)$

`m_(dd(spư)` `=` `5,5+100-0,2.2=105,1` $(g)$

Theo PT $(1)$, `n_(AlCl_3` `=` `n_(Al` `=` `0,1` $(mol)$

`m_(AlCl_3` `=` `n.M=0,1.133,5=13,35` $(g)$

Theo PT $(2)$, `n_(FeCl_2` `=` `n_(Fe` `=` `0,05` $(mol)$

`m_(FeCl_2` `=` `n.M=0,05.127=6,35` $(g)$

`C%_(AlCl_3` `=` `(m_{AlCl_3})/(m_{dd(spư)}).100%` `=` `(13,35)/(105,1).100%≈12,7%`

`C%_(FeCl_2` `=` `(m_{FeCl_2})/(m_{dd(spư)}).100%` `=` `(6,35)/(105,1).100%≈6,04%`

$c)$ Chưa đủ dữ kiện