a, \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
Ba(OH)2: bari hydroxit
H2: hydro
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Ba}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Ba}=0,2.137=27,4\left(g\right)\)
\(\Rightarrow m_{BaO}=42,7-27,4=15,3\left(g\right)\)