$a)n_{CO_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$n_{Ba(OH)_2}=\dfrac{285.12\%}{171}=0,2(mol)$

$\Rightarrow n_{Ba^{2+}}=0,2(mol);n_{OH^-}=2n_{Ba(OH)_2}=0,4(mol)$

$T=\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0,4}{0,3}=1,33$

$\Rightarrow $ Tạo 2 muối $CO_3^{2-};HCO_3^-$

$CO_2+2OH^-\to CO_3^{2-}+H_2O$

$CO_2+OH^-\to HCO_3^-$

Đặt $(n_{BaCO_3};n_{Ba(HCO_3)_2})=(x;y)(mol)$

$\Rightarrow n_{CO_2}=x+y=0,3$

$n_{OH^-}=2x+y=0,4$

Giải hệ ta được: $x=0,1;y=0,2$

So sánh $n_{Ba^{2+}}>n_{CO_3^{2-}}\Rightarrow Ba^{2+}$ dư so vs $CO_3^{2-}$

$\Rightarrow m_{BaCO_3}=0,1.197=19,7(g)$

$\Rightarrow m_{Ba(HCO_3)_2}=0,2.259=51,8(g)$

Lập tỉ lệ: $\dfrac{n_{Ba(OH)_2}}{1}>\dfrac{n_{CO_2}}{2}\Rightarrow Ba(OH)_2$ dư

$b)$ Vì $Ba(OH)_2$ dư nên $A$ gồm $Ba(OH)_2,Ba(HCO_3)_2$

$m_{CO_2}=0,3.44=13,2(g);n_{Ba(OH)_2(dư)}=0,2-0,15=0,05(mol)$

$\Rightarrow m_{Ba(OH)_2}=0,05.171=8,55(g)$

$\Rightarrow C\%_{Ba(OH)_2}=\dfrac{8,55}{13,2+285}.100\%\approx 2,87\%$

$\Rightarrow C\%_{Ba(HCO_3)_2}=\dfrac{38,35}{13,2+285}.100\%\approx 12,86\%$