a,
$n_{Ba}=\dfrac{13,7}{137}=0,1(mol)$
$n_{H_2SO_4}=\dfrac{100.2,94\%}{98}=0,03(mol)$
$Ba+H_2SO_4\to BaSO_4+H_2$
$\to n_{Ba\text{pứ}}=n_{H_2SO_4}=n_{BaSO_4}=0,03(mol)$
Còn $0,1-0,03=0,07$ mol $Ba$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$\to n_{Ba(OH)_2}=0,07(mol)$
Ta có: $n_{H_2}=n_{BaSO_4}+n_{Ba(OH)_2}=0,1(mol)$
$\to V=0,1.22,4=2,24l$
$m_{BaSO_4}=0,03.233=6,99g$
b,
$m_{\text{dd spu}}=13,7+100-6,99-0,1.2=106,51g$
$\to C\%_{Ba(OH)_2}=\dfrac{0,07.171.100}{106,51}=11,24\%$
