Sục V lít CO2 (đktc) vào bình đựng 1 lít dung dịch Ba(OH)2 0,2M...

Theo đầu bài ta có:
$\begin{aligned}
& {{n}_{Ba{{(OH)}_{2}}}}=0,2(mol)>{{n}_{BaC{{O}_{3}}}}=0,1(mol) \\
& C{{O}_{2}}+\underbrace{Ba{{(OH)}_{2}}}_{0,2(mol)}\left\{ \begin{aligned}
& \xrightarrow{(1)}\left\{ \begin{aligned}
& BaC{{O}_{3}}:0,1(mol) \\
& Ba{{(OH)}_{2(\text{d)}}}:0,1(mol) \\
\end{aligned} \right. \\
& \xrightarrow{(2)}\left\{ \begin{aligned}
& BaC{{O}_{3}}:0,1(mol) \\
& Ba{{(HC{{O}_{3}})}_{2}}:0,1(mol) \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned}$
TH1: CO2​ + Ba(OH)2​ $\to $ BaCO3​ $\downarrow $ + H2​O (1)
0,1 $\to $ 0,1
Từ (1) $\to {{n}_{C{{O}_{2}}}}={{n}_{BaC{{O}_{3}}}}=0,1(mol)\to {{V}_{C{{O}_{2}}}}=2,24(l)$
TH2: CO2​ + Ba(OH)2​ $\to $ BaCO3​ $\downarrow $ + H2​O (2)
0,1 $\leftarrow $ 0,1 $\leftarrow $ 0,1
2CO2​ + Ba(OH)2​ $\to $ Ba(HCO3​)2​ (3)
0,2 $\leftarrow $ 0,1 $\to $ 0,1
Từ (2), (3) $\to \sum{nC{{O}_{2}}}=0,3(mol)\to {{V}_{C{{O}_{2}}}}=6,72(l)$
Vậy, từ (1), (2), (3) $\to \left[ \begin{aligned}
& {{V}_{C{{O}_{2}}}}=2,24(l) \\
& {{V}_{C{{O}_{2}}}}=6,72(l) \\
\end{aligned} \right.$