Ba(OH)2 + 2HCl \(\rightarrow\)BaCl2 + 2H2O
nHCl=\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
nBa(OH)2=0,25.0,8=0,2(mol)
Vì 0,125<0,2 nên Ba(OH)2 dư 0,075 mol
Theo PTHH ta có:
\(\dfrac{1}{2}\)nHCl=nBaCl2=0,125(mol)
CM dd BaCl2 =\(\dfrac{0,125}{0,25}=0,5M\)
CM dd Ba(OH)2=\(\dfrac{0,075}{0,25}=0,3M\)