a,
$n_{CuSO_4}=\dfrac{512.25\%}{160}=0,8(mol)$
Đặt $n_{Al\text{pứ}}=x(mol)$
$2Al+3CuSO_4\to Al_2(SO_4)_3+3Cu$
$\to n_{Cu}=1,5x(mol)$
Ta có: $16,2-27x+64.1,5x=32,85$
$\to x=0,2413$
Vậy $m_{Al\text{pứ}}=27x=6,5151g; m_{Cu}=64.1,5x=23,1648g$
b,
Dung dịch $A$ có:
$\begin{cases} Al_2(SO_4)_3: 0,12065(mol)\\ CuSO_4: 0,8-0,2413.1,5=0,43805(mol)\end{cases}$
$m_{dd\text{spu}}=6,5151+512-23,1643=495,3508g$
Vậy $C\%_{Al_2(SO_4)_3}=\dfrac{0,12065.342.100}{495,3508}=8,33\%; C\%_{CuSO_4}=\dfrac{0,43805.160.100}{495,3508}=14,15\%$