$n_{H_2}=17,92/22,4=0,8mol$

$PTHH :$

$2Al+6HCl\to 2AlCl_3+3H_2↑(1)$

$Fe+2HCl\to FeCl_2+H_2↑(2)$

\(a.Gọi\ n_{Al}=a;n_{Fe}=b(a,b>0)\\Ta\ có :\\m_{hh}=27a+56b=22g\\ n_{H_2}=1,5+b=0,8mol\\ Ta\ có\ hpt:\ \left\{\begin{matrix} 27a+56b=22& \\ 1,5a+b=0,8 & \end{matrix}\right.⇔\left\{\begin{matrix} a=0,4 & \\ b=0,2 & \end{matrix}\right.\\ ⇒\%m_{Al}=\dfrac{0,4.27}{22}.100\%=49,1\%\\ \$m_{Fe}=100\%-49,1\%=50,9\%\\ b.Theo\ pt (1)\ và\ (2)\ :\\ n_{HCl}=3a+2b=1,6mol\\ ⇒m_{HCl}=1,6.36,5=58,4g\\ ⇒m_{dd\ HCl}=\dfrac{58,4}{7,3\%}=800g\\c.Theo\ pt\ (1)\ và\ (2)\ :\\ n_{AlCl_3}=n_{Al}=0,4mol\\ ⇒m_{AlCl_3}=0,4.133,5=53,4g\\ n_{FeCl_2}=n_{Fe}=0,2mol\\⇒m_{FeCl_2}=0,2.127=25,4g\\ m_{dd\ spu}=22+800-0,8.2=820,4g\\ C\%_{AlCl_3}=\dfrac{53,4}{820,4}.100\%=6,5\%\\ C\%_{FeCl_2}=\dfrac{25,4}{820,4}.100\%=3,09\%\)

$n_{H2}$=$\frac{17,92}{22,4}$=0,8 (mol)

a,

2Al + 6HCl -> 2AlCl3 + 3H2

 2a       6a           2a         3a

Fe + 2HCl -> FeCl2 + H2

 b       2b           b         b

$\left \{ {{54a+56b=22} \atop {3a+b=0,8}} \right.$

$\left \{ {{x=0,2} \atop {y=0,2}} \right.$

⇒$m_{Al}$=0,2.54=10,8 (g) 

    $m_{Fe}$=22-10,8=11,2 (g)

b, $mct_{HCl}$=36,5.(6.0,2+2.0,2)=58,4 (g)

$mdd_{HCl}$=$\frac{58,4.100}{7,3}$=800 (g)

c, $m_{AlCl3}$=0,2.2.133,5=53,4 (g)

    $m_{FeCl2}$=0,2.127=25,4 (g)

$mdd_{spu}$=800+22=822 (g)

$Cphầntrăm_{AlCl3}$=$\frac{53,4.100}{822}$≈6,49635%

$Cphầntrăm_{FeCl2}$=$\frac{25,4.100}{822}$≈3,09%