\(n_{Ba\left(OH\right)_2}=0,5.0,2=0,1\left(mol\right)\)
PTHH: Ba(OH)2 + 2HCl -----> BaCl2 + 2H2O
Mol: 0,1 0,2
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100\%}{15\%}=\dfrac{146}{3}\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{\dfrac{146}{3}}{1,25}=\dfrac{584}{15}=38,9\left(ml\right)\)