Ba(OH)2 + 2HCl → BaCl2 + 2H2O
\(n_{Ba\left(OH\right)_2}=0,05\times0,5=0,025\left(mol\right)\)
\(n_{HCl}=0,15\times0,1=0,015\left(mol\right)\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Ba\left(OH\right)_2}=\dfrac{5}{3}n_{HCl}\)
Vì \(\dfrac{5}{3}>\dfrac{1}{2}\) ⇒ Ba(OH)2 dư
Dung dịch A gồm: Ba(OH)2 dư và BaCl2
Theo PT: \(n_{Ba\left(OH\right)_2}pư=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,015=0,0075\left(mol\right)\)
\(\Rightarrow n_{Ba\left(OH\right)_2}dư=0,025-0,0075=0,0175\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}dư=\dfrac{0,0175}{0,2}=0,0875\left(M\right)\)
Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,015=0,0075\left(mol\right)\)
\(\Rightarrow C_{M_{BaCl_2}}=\dfrac{0,0075}{0,2}=0,0375\left(M\right)\)
