$PTHH : \\Fe+2HCl\to FeCl_2+H_2(1) \\2Al+6HCl\to 2AlCl_3+3H_2(2) \\n_{H_2}=\dfrac{5,6}{22,4}=0,25mol⇒m_{H_2}=0,25.2=0,5g \\Gọi\ n_{Fe}=a(mol);n_{Al}=b(mol) \\Ta\ có : \\m_{Fe}+m_{Al}=8,3(g) \\⇒56a+27b=8,3(I) \\∑n_{H_2}=n_{Fe}+1,5.n_{Al}=0,25mol \\⇒a+1,5b=0,25(II) \\Từ\ (I)\ và\ (II):$

$⇒\left\{\begin{matrix} 56a+27b=8,3 & \\  a+1,5b=0,25 &  \end{matrix}\right.⇔\left\{\begin{matrix} a=0,1 & \\  b=0,1 &  \end{matrix}\right.$

$⇒\%m_{Al}=\dfrac{0,1.27}{8,3}.100\%=32,53\% \\\%m_{Fe}=100\%-32,53\%=67,47\%$

b.Dung dịch A là FeCl2 ; AlCl3 và HCl dư

$Theo\ pt\ (1) : \\n_{FeCl_2}=n_{Fe}=0,1mol \\⇒m_{FeCl_2}=0,1.127=12,7g \\n_{HCl(1)\ pư}=2.n_{Fe}=2.0,1=0,2mol \\Theo\ pt\ (2) : \\n_{AlCl_3}=n_{Al}=0,1mol \\⇒m_{AlCl_3}=0,1.133,5=13,35g \\n_{HCl(2)\ pư}=3.n_{Al}=3.0,1=0,3mol \\⇒∑n_{HCl\ pư}=0,2+0,3=0,5mol$

vì dd HCl dùng dư 20% so với lượng phản ứng

$⇒n_{HCl\ dư}=0,5.20\%=0,1mol \\⇒m_{HCl\ dư}=0,1.36,5=3,65g \\⇒n_{HCl\ bđ}=0,5+0,1=0,6mol \\⇒m_{HCl\ bđ}=0,6.36,5=21,9g \\⇒m_{dd\ HCl}=\dfrac{21,9}{7,3\%}=300g \\m_{dd\ spu}=8,3+300-0,5=307,8g \\⇒C\%_{FeCl_2}=\dfrac{12,7}{307,8}.100\%=4,13\% \\C\%_{AlCl_3}=\dfrac{13,35}{307,8}.100\%=4,34\% \\C\%_{HCl\ dư}=\dfrac{3,65}{300}.100\%=1,22\%$