Đáp án đúng: C
28%.
$\begin{array}{l}\text{C}{{\text{O}}_{\text{2}}}\,\xrightarrow{\text{NaOH}}\,\left\{ \begin{array}{l}\text{CO}_{\text{3}}^{\text{2-}}\\\text{HCO}_{\text{3}}^{\text{-}}\end{array} \right.\xrightarrow{\text{BaC}{{\text{l}}_{\text{2}}}}\,\left\langle \begin{array}{l}\text{BaC}{{\text{O}}_{\text{3}}}\downarrow \text{:}\,\text{2,955}\,\text{gam}\\\text{HCO}_{\text{3}}^{\text{-}}\,\xrightarrow{\text{Ba(OH}{{\text{)}}_{\text{2}}}}\,\text{BaC}{{\text{O}}_{\text{3}}}\,\downarrow \,\text{11,82}\,\text{gam}\end{array} \right.\\\Rightarrow {{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,{{\text{n}}_{\text{BaC}{{\text{O}}_{\text{3}}}}}\,\text{=}\,\frac{\text{2,955}\,\text{+}\,\text{11,82}}{\text{197}}\,\text{=}\,\text{0,075}\,\text{mol}\\\Rightarrow \,{{\text{V}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\text{1,68}\,\text{lit}\,\Rightarrow \,{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}{{\text{V}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\text{28}{\scriptstyle{}^{o}\!\!\diagup\!\!{}_{o}\;}\end{array}$
