a)Ba+ 2H2O--->Ba(OH)2+H2
0,2-----------------0,2---------0,2
b) n\(_{Ba}=\frac{27,4}{137}=0,2\left(mol\right)\)
V\(_{H2}=0,2.22,4=4,48\left(l\right)\)
c) m\(_{H2}=0,4\left(g\right)\)
m\(_{Ba\left(OH\right)2}=0,2.172=34,2\left(g\right)\)
Gọi mH2O=a
Suy ra
\(\frac{34,2}{a+27,4-0,4}.100\%=8\%\)
=> \(\frac{34,2}{a+27}=0,08\)
=> 34,2=0,08a+2,16
=> 0,08a=32,04(g)
=>a=400,5(g)
Vậy m H2O=400,5(g)