Question

Answer 1

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

Ta có: \(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,1+0,12.2=0,34\left(mol\right)\)

\(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=2x+3x.2=8x\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

Theo PT: \(n_{OH^-}=n_{H^+}\Rightarrow0,34=8x\Rightarrow x=0,0425\left(mol\right)\)

Có: \(n_{H_2O}=n_{OH^-}=0,34\left(mol\right)\)

Theo ĐLBT KL, có: m_NaOH + m_Ba(OH)2 + m_HCl + m_H2SO4 = m _muối + m_H2O

⇒ m _muối = 0,1.40 + 0,12.171 + 2.0,0425.36,5 + 3.0,0425.98 - 0,34.18 = 33,9975 (g)