Đáp án đúng: A
17,5 gam.
$\begin{array}{l}{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0}{,2}\,{mol;}\,{{{n}}_{{O}{{{H}}^{{-}}}}}\,{=}\,{0}{,3}\,{mol}\\\left\{ \begin{array}{l}{{{n}}_{{CO}_{{3}}^{{2-}}}}\,{+}\,{{{n}}_{{HCO}_{{3}}^{{-}}}}\,{=}\,{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0}{,2}\,{mol}\\{2}{{{n}}_{{CO}_{{3}}^{{2-}}}}\,{+}\,{{{n}}_{{HCO}_{{3}}^{{-}}}}\,{=}\,{{{n}}_{{O}{{{H}}^{{-}}}}}\,{=}\,{0}{,3}\,{mol}\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}{{{n}}_{{CO}_{{3}}^{{2-}}}}\,{=}\,{0}{,1}\,{mol}\\{{{n}}_{{HCO}_{{3}}^{{-}}}}\,{=}\,{0}{,1}\,{mol}\end{array} \right.\end{array}$
Cô cạn:${2HCO}_{{3}}^{{-}}\,\xrightarrow{{{{t}}^{{o}}}}\,{CO}_{{3}}^{{2-}}\,{+}\,{C}{{{O}}_{{2}}}\,{+}\,{{{H}}_{{2}}}{O}$
${{{n}}_{{r}{.khan}}}\,{=}\,{{{m}}_{{N}{{{a}}^{{+}}}}}\,{+}\,{{{m}}_{{{{K}}^{{+}}}}}\,{+}\,{{{m}}_{{CO}_{{3}}^{{2-}}}}\,{=}\,{17}{,5}\,{gam}$
