Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 16,6\\
\dfrac{3}{2}a + b = 0,5
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,2\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1mol\\
{V_{HCl}} = \dfrac{1}{{0,5}} = 2l = 2000ml\\
c)\\
{m_{{\rm{dd}}HCl}} = 2000 \times 1,05 = 2100g\\
{m_{{\rm{dd}}spu}} = 16,6 + 2100 - 0,5 \times 2 = 2115,6g\\
{n_{AlC{l_2}}} = {n_{Al}} = 0,2mol\\
C{\% _{AlC{l_3}}} = \dfrac{{0,2 \times 133,5}}{{2115,6}} \times 100\%  = 1,262\% \\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
C{\% _{FeC{l_2}}} = \dfrac{{0,2 \times 127}}{{2115,6}} \times 100\%  = 1,2\% 
\end{array}\)

Đáp án:

Giải thích các bước giải:

a)$n_{H_2}=\frac{11,2}{22,4}=0,5(mol)$

Gọi $\left \{ {{n_{Al}=a(mol)} \atop {n_{Fe}=b(mol)}} \right.$ 

$⇒27a+56b=16,6$

$⇔56b=16,6-27a$

$⇔b=\frac{16,6-27a}{56}$

$2Al+6HCl→2AlCl_3+3H_2↑$

$a:3a:a:1,5a$

$Fe+2HCl→FeCl_2+H_2↑$

$b:2b:b:b$

$∑n_{H_2}=1,5a+b=0,5$

$⇒1,5a+\frac{16,6-27a}{56}=0,5$

$⇒a=0,2$

$⇒b=\frac{16,6-27.0,2}{56}=0,2$

$⇒\left \{ {{m_{Al}=0,2.27=5,4(g)} \atop {m_{Fe}=0,2.56=11,2(g)}} \right.$ 

b)$V_{dd HCl}=\frac{0,2.3+0,2.2}{0,5}=2(l)$

c)$C_{ddHCl}=\frac{1.36,5}{2000.1,05}=0,017$%

Xin hay nhất!!!