a)
$KOH + HNO_3 \to KNO_3 + H_2O$
$Ba(OH)_2 + 2HNO_3 \to Ba(NO_3)_2 + 2H_2O$
b) $n_{HNO_3} = \dfrac{189}{63} = 3(mol) ;n_{KOH} = \dfrac{112}{56} = 2(mol)$
Ta có :
$n_{HNO_3} = n_{KOH} + 2n_{Ba(OH)_2}$
$\Rightarrow n_{Ba(OH)_2} = \dfrac{3 - 2}{2} = 0,5(mol)$
$\Rightarrow m_{dd\ Ba(OH)_2} = \dfrac{0,5.171}{25\%} = 342(gam)$