a) Gọi $n_{Al\ pư} = a(mol)$
$2Al + 3CuSO_4 \to Al_2(SO_4)_3 + 3Cu$
Theo PTHH : $n_{Cu} = \dfrac{3}{2}n_{Al} = 1,5a(mol)$
Ta có : $m_{Cu} - m_{Al} = 1,38 \Rightarrow 1,5a.64 - 27a = 1,38$
$\Rightarrow a = 0,02$
b) $n_{CuSO_4} = \dfrac{3}{2}n_{Al} = 0,03(mol)$
$C_{M_{CuSO_4}} = \dfrac{0,03}{0,1} = 0,3M$