\(a)\\ n_{Al} = \frac{10,8}{27} = 0,4(mol)\\ 2Al + 3CuSO_4\to Al_2(SO_4)_3 + 3Cu\\ TPT: n_{CuSO_4} = \frac{3}{2} n_{Al} = \frac{3}{2} . 0 , 4=0,6(mol)\\ \Rightarrow C\%\ CuSO_4 = \frac{0,6 . 160}{200} . 100\%=48\%\\ b)\\ TPT:n_{Al_2(SO_4)_3}=\frac{1}{2} n_{Al} = \frac{1}{2} . 0,4=0,2(mol)\\ \Rightarrow C\%\ CuSO_4 = \frac{0,2 . 342}{200 + 10,8} \approx32,45\%\)