Gọi n Na và Ba=x -->ta có
2Na+2H2O--->2NaOH+H2
x-----------------x----------0,5x
Ba+2H2O--->Ba(OH)2+H2
x------------------x------------x
NaOH+HCl--->NaCl+H2O
x----------x
Ba(OH)2+2HCl---->BaCl2+2H2O
x-----------2x
n H2=1,344/22,4=0,06(mol)
-->0,5x+x=0,06
-->1,5x=0,06-->x=0,04(mol)
Tổng n HCl=3x=0,12(mol)
V HCl=0,12/2=0,06(l)
\(n_{H_2}=0,06\left(mol\right)\\ BT\text{ }e\Rightarrow n_{Na}+2n_{Ba}=2n_{H_2}\\ BTĐT\Rightarrow n_{OH^-}=n_{Na^+}+2n_{Ba^{2+}}\\ \Rightarrow n_{OH^-}=2n_{H_2}=0,12\left(mol\right)\\\)
H+ + OH- ---> H2O
0,12____0,12
\(\Rightarrow n_{HCl}=n_{H^+}=0,12\left(mol\right)\\ \Rightarrow V=0,06\left(l\right)\)