a,
$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$\to n_{Ba}=n_{H_2}=0,25(mol)$
$\to m_{Ba}=0,25.137=34,25g$
Ta có: $\dfrac{m_{Ba}}{m_{BaO}}=\dfrac{137}{153}$
$\to m_{BaO}=\dfrac{34,25.153}{137}=38,25g$
Vậy $m=m_X=38,25+34,25=72,5g$
b,
$n_{BaO}=\dfrac{38,25}{153}=0,25(mol)$
$BaO+H_2O\to Ba(OH)_2$
$\to n_{Ba(OH)_2}=n_{Ba}+n_{BaO}=0,5(mol)$
Vậy $C\%_{Ba(OH)_2}=\dfrac{0,5.171.100}{400}=21,375\%$