$A+NaOH$ dư $\to $ kết tủa
$\to A$ có $Ba(HCO_3)_2$
$n_{BaCO_3(2)}=\dfrac{9,85}{197}=0,05(mol)$
Bảo toàn $Ba$: $n_{Ba(HCO_3)_2}=n_{BaCO_3(2)}=0,05(mol)$
$n_{Ba(OH)_2}=0,5.0,3=0,15(mol)$
Bảo toàn $Ba$:
$n_{BaCO_3(1)}=n_{Ba(OH)_2}-n_{Ba(HCO_3)_2}=0,1(mol)$
$\to x=0,1.197=19,7g$
Bảo toàn $C$:
$n_{CO_2}=n_{BaCO_3(1)}+2n_{Ba(HCO_3)_2}=0,2(mol)$
$\to V=0,2.22,4=4,48l$