$n_{HCl} = \dfrac{54,75}{36,5} = 1,5(mol)$
$n_{KOH} = \dfrac{56}{56} = 1(mol)$
$KOH + HCl \to KCl + H_2O$
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
Theo PTHH :
$n_{HCl} = n_{KOH} + 2n_{Ba(OH)_2}$
$\Rightarrow 1,5 = 1 + 2n_{Ba(OH)_2}$
$\Rightarrow n_{Ba(OH)_2} = 0,25(mol)$
$m_{dd\ Ba(OH)_2\ đã\ dùng} = \dfrac{0,25.171}{25\%} = 171(gam)$