Đáp án đúng: C
0,80%.
$\begin{array}{l}\text{C}{{\text{O}}_{\text{2}}}\,\text{+}\,\text{2KOH}\,\,\xrightarrow{{}}\,\,{{\text{K}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}\\{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\frac{\text{0,44}}{\text{44}}\,\,\text{=}\,\text{0,01}\,\,\text{mol}\,\,\Rightarrow \,\,{{\text{n}}_{\text{C}}}\,\text{=}\,{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\text{0,01}\,\,\text{mol}\end{array}$
${\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}\text{C}\,\text{=}\,\,\frac{\text{0,01}\text{.12}}{\text{15}}\,\text{.}\,\text{100}{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}\,\,\text{=}\,\text{0,8}{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}$