a,
Mỗi phần có $7,22:2=3,61g$ A.
Mỗi phần có $x$ mol $Fe$ và $y$ mol $M$ (hoá trị $n$)
$n_{H_2}=\dfrac{2,128}{22,4}=0,095(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$2R+2nHCl\to 2RCl_n+ nH_2$
$\Rightarrow x+0,5ny=0,095$ $(1)$
$n_{NO}=\dfrac{1,792}{22,4}=0,08(mol)$
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$3M+4nHNO_3\to 3M(NO_3)_n+nNO+2nH_2O$
$\Rightarrow x+\dfrac{1}{3}.ny=0,08$ $(2)$
$(1)(2)\Rightarrow x=0,05; ny=0,09$
$\Leftrightarrow y=\dfrac{0,09}{n}$
Ta có $0,05.56+\dfrac{0,09M}{n}=3,61$
$\Leftrightarrow M=9n$
$n=3\Rightarrow M=27(Al)$
b,
$\%m_{Fe}=\dfrac{0,05.56.100}{3,61}=77,56\%$
$\%m_{Al}=22,44\%$