\(n_{Ba\left(OH\right)_2}=\dfrac{150.10\%}{171}=\dfrac{15}{171}=\dfrac{5}{57}\left(mol\right)\\Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ n_{HCl}=2.\dfrac{5}{57}=\dfrac{10}{57}\left(mol\right)\\ m_{HCl}=\dfrac{10.36,5}{57}=\dfrac{365}{57}\\ m_{ddHCl}=\dfrac{365.100}{15.57}=\dfrac{36500}{1387}\left(g\right)\)