$n_{H2}$=$\frac{17,92}{22,4}$=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%