Gọi n Na2O = a(mol) => n BaO = 2a(mol)
=> 62a + 153.2a = 36,8
=> a = 0,1
$Na_2O + H_2O \to 2NaOH$
$BaO + H_2O \to Ba(OH)_2$
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$NaOH + HCl \to NaCl + H_2O$
n HCl = n NaOH + 2n Ba(OH)2 = 2n Na2O + 2n BaO = 0,1.2 + 0,1.2.2 = 0,6(mol)
m dd HCl = 0,6.36,5/20% = 109,5(gam)