a,
Gọi x, y là số mol $Al$, $Al_2O_3$
$\Rightarrow 27x+102y=7,8$ (1)
$2Al+6HCl\to 2AlCl_3+3H_2$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15 mol$
$\Rightarrow 1,5x=0,15$ (2)
(1)(2)$\Rightarrow x=0,1; y=0,05$
$\%m_{Al}=\dfrac{0,1.27.100}{7,8}=34,6\%$
$\%m_{Al_2O_3}=65,4\%$
b,
$n_{HCl}=3x+6y=0,6(mol)$
$\to V_{HCl}=0,6:3=0,2(l)$