\(m_{Ba\left(OH\right)_2}=\dfrac{100\cdot17,1\%}{100\%}=17,1\left(g\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\\ PTHH:Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+H_2O\\ n_{BaCl_2}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\ \Rightarrow m_{muối.sau.p/ứ}=m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\)