${{n}_{BaO}}=\frac{{{m}_{BaO}}}{{{M}_{BaO}}}=\frac{22,95}{153}=0,15\left( mol \right)$

$BaO+{{H}_{2}}O\to Ba{{(OH)}_{2}}$

0,15             $\to $  0,15            (mol)

Theo phương trình :

${{n}_{Ba{{\left( OH \right)}_{2}}}}={{n}_{BaO}}=0,15$ (mol) 

 $\Rightarrow {{m}_{Ba{{\left( OH \right)}_{2}}}}={{n}_{Ba{{\left( OH \right)}_{2}}}}.{{M}_{Ba{{\left( OH \right)}_{2}}}}=0,15.171=25,65\left( g \right)$

$\Rightarrow $ C% $_{Ba{{\left( OH \right)}_{2}}}=\frac{{{m}_{Ba{{\left( OH \right)}_{2}}}}}{{{m}_{\text{dd}}}}.100=\frac{25,65}{500}.100=5,13$ (%)

$\Rightarrow $Chọn đáp án B