Đáp án:

\(\begin{array}{l}
a)\\
{m_{Al}} = 2,7g\\
{m_{Fe}} = 8,4g\\
b)\\
CTHH:F{e_3}{O_4}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 11,1(1)\\
\dfrac{3}{2}a + b = 0,3(2)\\
 \text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,15\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
{m_{Fe}} = n \times M = 0,15 \times 56 = 8,4g\\
b)\\
{M_x}{O_y} + y{H_2} \to xM + y{H_2}O\\
{n_{{M_x}{O_y}}} = \dfrac{{{n_{{H_2}}}}}{y} = \dfrac{{0,3}}{y}(mol)\\
{M_{{M_x}{O_y}}} = \dfrac{m}{n} = 58y\\
 \Rightarrow x{M_M} + y{M_O} = 58y\\
 \Rightarrow {M_M} = 21 \times \dfrac{{2y}}{x}\\
\dfrac{{2y}}{x} = \dfrac{8}{3}\\
 \Rightarrow {M_M} = 21 \times \dfrac{8}{3} = 56dvC\\
 \Rightarrow M:(Fe)\\
CTHH:F{e_3}{O_4}
\end{array}\) 

a,

Gọi x, y là số mol $Al$, $Fe$

$\Rightarrow 27x+56y=11,1$               (1)

$2Al+6HCl\to 2AlCl_3+3H_2$

$Fe+2HCl\to FeCl_2+H_2$

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$\Rightarrow 1,5x+y=0,3$                   (2)

$(1)(2)\Rightarrow x=0,1; y=0,15$

$m_{Al}=0,1.27=2,7g$

$m_{Fe}=0,15.56=8,4g$

c,

Gọi CTHH oxit là $M_xO_y$

$M_xO_y+ yH_2\to xM+yH_2O$

$\Rightarrow n_{M_xO_y}=\dfrac{0,3}{y}(mol)$

$\Rightarrow M_{M_xO_y}=\dfrac{17,4y}{0,3}=58y=Mx+16y$

$\Leftrightarrow M=\dfrac{42y}{x}$

$\Rightarrow x=3; y=4; M=56(Fe)$

Vậy oxit là $Fe_3O_4$