Đáp án đúng: D
29,7.
$\begin{array}{l}{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\text{0}\text{,3}\,\text{mol;}\,{{\text{n}}_{\text{O}{{\text{H}}^{\text{-}}}}}\,\text{=}\,\text{0}\text{,5}\,\text{mol}\\\left\{ \begin{array}{l}{{\text{n}}_{\text{CO}_{\text{3}}^{\text{2-}}}}\,\text{+}\,{{\text{n}}_{\text{HCO}_{\text{3}}^{\text{-}}}}\,\text{=}\,{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\,\text{=}\,\text{0}\text{,3}\,\text{mol}\\\text{2}{{\text{n}}_{\text{CO}_{\text{3}}^{\text{2-}}}}\,\text{+}\,{{\text{n}}_{\text{HCO}_{\text{3}}^{\text{-}}}}\,\text{=}\,{{\text{n}}_{\text{O}{{\text{H}}^{\text{-}}}}}\,\text{=}\,\text{0}\text{,5}\,\text{mol}\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}{{\text{n}}_{\text{CO}_{\text{3}}^{\text{2-}}}}\,\text{=}\,\text{0}\text{,2}\,\text{mol}\\{{\text{n}}_{\text{HCO}_{\text{3}}^{\text{-}}}}\,\text{=}\,\text{0}\text{,1}\,\text{mol}\end{array} \right.\end{array}$
Cô cạn:$\text{2HCO}_{\text{3}}^{\text{-}}\,\xrightarrow{{{\text{t}}^{\text{o}}}}\,\text{CO}_{\text{3}}^{\text{2-}}\,\text{+}\,\text{C}{{\text{O}}_{\text{2}}}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}$
${{\text{n}}_{\text{r}\text{.khan}}}\,\text{=}\,{{\text{m}}_{\text{N}{{\text{a}}^{\text{+}}}}}\,\text{+}\,{{\text{m}}_{{{\text{K}}^{\text{+}}}}}\,\text{+}\,{{\text{m}}_{\text{CO}_{\text{3}}^{\text{2-}}}}\,\text{=}\,\text{29}\text{,7}\,\text{gam}$
