Ba(OH)2 + CuSO4 → Cu(OH)2↓ + BaSO4↓
\(n_{Ba\left(OH\right)_2}=0,1.4=0,4\left(mol\right)\)
\(n_{CuSO_4}=n_{BaSO_4}=n_{Cu\left(OH\right)_2}=n_{Ba\left(OH\right)_2}=0,4\left(mol\right)\)
=>\(x=CM_{CuSO_4}=\dfrac{0,4}{0,1}=4M\)
\(m=0,4.98+0,4.233=132,4\left(g\right)\)