a,
$2Al+6HCl\to 2AlCl_3+3H_2$
b,
$m_{Cu}=4g$
$\Rightarrow m_{Al}=14,8-4=10,8g$
$\Rightarrow n_{Al}=\dfrac{10,8}{27}=0,4(mol)$
Theo PTHH, $n_{AlCl_3}=n_{Al}=0,4(mol)$
$\Rightarrow m_{AlCl_3}=0,4.133,5=53,4g$
c,
Theo PTHH, $n_{HCl}=3n_{Al}=1,2(mol)$
$\Rightarrow C\%_{HCl}=\dfrac{1,2.36,5.100}{300}=14,6\%$