Ta có: \(n_{Ba\left(OH\right)_2}=0,01\left(mol\right)\)
PT: \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)
_____0,02____0,01 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,02}{0,25}=0,08\left(l\right)=80\left(ml\right)\)
Ta có: n B a ( O H ) 2 = 0 , 01 ( m o l ) PT: 2 H C l + B a ( O H ) 2 → B a C l 2 + 2 H 2 O _____0,02____0,01 (mol) ⇒ V d d H C l = 0 , 02 0 , 25 = 0 , 08 ( l ) = 80 ( m l )