Đáp án + Giải thích các bước giải:

Gọi $x(mol) = n_{Al}, y(mol) = n_{Fe}$

$n_{Cl_2} = \dfrac{V}{22,4} = \dfrac{10,08}{22,4} = 0,45(mol)$

 PTHH:
$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2 \uparrow$
$Fe  + 2HCl \longrightarrow FeCl_2 + H_2 \uparrow$

$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$

$n_{H_2} = \dfrac{V}{22,4} = \dfrac{8,96}{22,4} = 0,4(mol)$

Theo PTHH:
$n_{H_2} = \dfrac{3}{2}n_{Al} + n_{Fe}=\dfrac{3}{2}x + y = 0,4(1)$

$n_{Cl_2} = \dfrac{3}{2}n_{Al} + \dfrac{3}{2}n_{Fe} = \dfrac{3}{2}x + \dfrac{3}{2}y = 0,45(2)$
Từ $(1)$ và $(2)$, ta có:
$\begin {cases} \dfrac{3}{2}x + y = 0,4 \\ \dfrac{3}{2}x + \dfrac{3}{2}y = 0,45 \end {cases}$
$\Leftrightarrow \begin {cases} x = 0,2 \\ y = 0,1 \end {cases}$

$\rightarrow m_{Al} = n . M = 0,2 . 27 = 5,4(g)$
$m_{Fe} = n . M = 0,1 . 56 = 5,6(g)$

$\rightarrow \%_{Al} = \dfrac{5,4}{5,4 + 5,6} . 100\% \approx 49,09\%$

$\%_{Fe} = 100\% - 49,09\% = 50,91\%$

@$\text{Ly}$

$n_{H_2}$ $=$ $\dfrac{8,96}{22,4}$ $=$ $0,4$ $mol$

$n_{Cl_2}$ $=$ $\dfrac{10,08}{22,4}$ $=$ $0,45$ $mol$

Gọi $x,y$ lần lượt là số $mol$ của $Al,Fe$ phản ứng $(x,y>0)$

PTHH:

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

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

$(3):$ $2Al$ $+$ $3Cl_2$ $\xrightarrow[]{t^o}$ $2AlCl_3$

$(4):$ $2Fe$ $+$ $3Cl_2$ $\xrightarrow[]{t^o}$ $2FeCl_3$

Theo PT $(1)$, $n_{H_2}$ $=$ $1,5n_{Al}$ $=$ $1,5x$ $mol$

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

Theo PT $(3)$, $n_{Cl_2}$ $=$ $1,5n_{Al}$ $=$ $1,5x$ $mol$

Theo PT $(4)$, $n_{Cl_2}$ $=$ $1,5n_{Fe}$ $=$ $1,5y$ $mol$

Ta có:

$n_{H_2(bđ)}$ $=$ $n_{H_2(1)}$ $+$ $n_{H_2(2)}$ $=$ $1,5x+y=0,4$ $mol$

$n_{Cl_2(bđ)}$ $=$ $n_{Cl_2(3)}$ $+$ $n_{Cl_2(4)}$ $=$ $1,5x+1,5y=0,45$ $mol$

Ta có hệ PT sau:

$1,5x+y=0,4$

$1,5x+1,5y=0,45$

$⇒$ $\begin{cases} x=0,2\\y=0,1 \end{cases}$

$m_{Al}$ $=$ $0,2.27=5,4$ $(g)$

$m_{Fe}$ $=$ $56.0,1=5,6$ $(g)$

$m_{hh}$ $=$ $5,4+5,6=11$ $(g)$

`%m_(Al` $=$ $\dfrac{5,4}{11}$`.100%` $≈$ $49,09$`%`

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