Đáp án:

$↓$

Giải thích các bước giải:

$a,$

Đặt $n_{Fe}=x(mol),\ n_{Al}=y(mol)$

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

$Fe+2HCl→FeCl_2+H_2$

$2Al+6HCl→2AlCl_3+3H_2$

Theo PT: $n_{H_2}=x+1,5y=0,4(mol)$ $(1)$

$m_{h^2}=56x+27y=11(g)\ (2)$

Từ $(1)(2)⇒\ x=0,1;\ y=0,2$

$→\%Fe=\dfrac{0,1×56}{11}×100\%=50,9\%$

$→\%Al=100-50,9=49,1\%$

$b,$

$FeCl_2+2NaOH→Fe(OH)_2+2NaCl$

$AlCl_3+3NaOH→Al(OH)_3+3NaCl$

$Al(OH)_3+NaOH→NaAlO_2+2H_2O$

Theo phương trình: $n_{Fe(OH)_2}=n_{FeCl_2}=n_{Fe}=0,1(mol)$

$→m_Y=m_{Fe(OH)_2}=0,1×90=9(g)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
\frac{3}{2}a + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
\% Al = \dfrac{{5,4}}{{11}} \times 100\%  = 49,1\% \\
\% Fe = 100 - 49,1 = 50,9\% \\
b)\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
Y:Fe{(OH)_2}\\
{n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{Fe{{(OH)}_2}}} = 0,1 \times 90 = 9g
\end{array}\)