\(n_{HCl}=0,25\left(mol\right);n_{H_2SO_4}=0,125\left(mol\right)\)
=> \(n_{H^+\left(bđ\right)}=0,5\left(mol\right)\)
\(n_{H_2}=\dfrac{4,368}{22,4}=0,195\Rightarrow n_{H^+\left(pứ\right)}=2n_{H_2}=0,39\left(mol\right)\)
Vì \(n_{H^+\left(bđ\right)}>n_{H^+\left(pứ\right)}\)=> Axit còn dư
\(n_{H^+\left(dư\right)}=0,5-0,39=0,11\left(mol\right)\)
Trung hòa axit dư :
H+ + OH- ---> H2O
=> \(n_{OH^-}=n_{H^+}=0,11\left(mol\right)\)
Gọi V là thể tích của hỗn hợp bazo trung hòa axit dư
=> 0,02V + 0,01.2.V =0,11
=> V=2,75 (l)
=> \(n_{NaOH}=0,055\left(mol\right);n_{Ba\left(OH\right)_2}=0,0275\left(mol\right)\)
\(m_{muối}=m_{KL\left(Mg,Al,Na,Ba\right)}+m_{Cl^-+SO_4^{2-}}=29,7775\left(g\right)\)
