Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\)
PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)
Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)