\(n_{Ba\left(OH\right)2}=0,04\left(mol\right)\)
\(\text{+ PTHH: Ba +2 H 2 O→ B a ( O H ) 2 + H 2 ↑}\)
...................0,04.....................0,04...........................(mol)
\(\text{⇒ m= 0.04x137=5.48 g}\)
Ba+2H2O---->Ba(OH)2+H2
n Ba(OH)2=0,2.0,2=0,04(mol)
Theo pthh
n Ba=n Ba(OH)2=0,04(mol)
m Ba=m=0,04.137=5,48(g)