a)
$n_{HCl} = \dfrac{150.14,6\%}{36,5} = 0,6(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = 0,3(mol)$
$n_{Al} = n_{AlCl_3} = \dfrac{1}{3}n_{HCl} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b)
$m_{dd} = 5,4 + 150 - 0,3.2 = 154,8(gam)$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{154,8}.100\% = 17,25\%$