$n_{BaO}=\dfrac{15,3}{153}=0,1 mol$
$BaO+H_2O\to Ba(OH)_2$
$\Rightarrow n_{Ba(OH)_2}=0,1 mol$
$m_A=500+15,3=515,3g$
Trong $1/2A$ có:
$m_{dd}=515,3:2=257,65g$
$n_{Ba(OH)_2}=0,1:2=0,05 mol$
$n_{CO_2}=\dfrac{1,68}{22,4}=0,075 mol$
$\dfrac{2n_{Ba(OH)_2}}{n_{CO_2}}=1,3\Rightarrow$ tạo 2 muối: $BaCO_3$ (x mol), $Ba(HCO_3)_2$ (y mol)
$Ba(OH)_2+CO_2\to BaCO_3+H_2O$
$Ba(OH)_2+2CO_2\to Ba(HCO_3)_2$
$\Rightarrow x+y=0,05; x+2y=0,075$
$\Rightarrow x=y=0,025$
$m_{dd\text{spứ}}=0,075.44+257,65-0,025.197=256,025g$
$\to C\%_{Ba(HCO_3)_2}=\dfrac{0,025.259.100}{256,025}=2,53\%$