Câu b bạn làm sai ý 1 do quên $HCl$ phản ứng)
b,
$n_{HCl\text{pứ}}=3x+2y=1,3(mol)$
$n_{HCl\text{dư}}=n_{NaOH}=0,2(mol)$
$\Rightarrow n_{HCl\text{bđ}}=1,5(mol)$
$m_{HCl}=1,5.36,5=54,75g$
$\to m_{dd HCl}=54,75:36,5\%=150g$
$m_{dd A}=m_{Al}+m_{Mg}+m_{dd HCl}-m_{H_2}=12+150-0,65.2=160,7g$
$n_{AlCl_3}=n_{Al}=x=0,4(mol)$
$\to C\%_{AlCl_3}=\dfrac{0,4.133,5.100}{160,7}=33,2\%$
$n_{MgCl_2}=n_{Mg}=y=0,05(mol)$
$\to C\%_{MgCl_2}=\dfrac{0,05.95.100}{160,7}=2,96\%$