Ta có: \(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)
PTHH: \(CuSO_4+BaCl_2--->BaSO_4\downarrow+CuCl_2\)
Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\)
Vậy CuSO4 dư.
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)
=> \(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)