Đáp án đúng: A
29,89.
Để kết tủa X có lượng lớn nhất thì ${{{n}}_{{C}{{{O}}_{{2}}}}}\,\,{=}\,\,{{{n}}_{{Ca(OH}{{{)}}_{{2}}}}}\,{=}\,{0,2}\,{mol}$
$\begin{array}{l}\left\{ \begin{array}{l}{84}{{{n}}_{{MgC}{{{O}}_{{3}}}}}\,{+}\,{197}{{{n}}_{{BaC}{{{O}}_{{3}}}}}\,{=}\,{28,1}\,{gam}\\{{{n}}_{{MgC}{{{O}}_{{3}}}}}\,{+}\,{{{n}}_{{BaC}{{{O}}_{{3}}}}}\,{=}\,{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0,2}\,{mol}\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}{{{n}}_{{MgC}{{{O}}_{{3}}}}}\,{=}\,{0,1}\,{mol}\\{{{n}}_{{BaC}{{{O}}_{{3}}}}}\,{=}\,{0,1}\,{mol}\end{array} \right.\,\\\Rightarrow \,{a}\,{=}\,\frac{{0,1}{.84}}{{28,1}}{.100}{\scriptstyle{}^{{o}}\!\!\diagup\!\!{}_{{o}}\;}\,{=}\,{29,89}{\scriptstyle{}^{{o}}\!\!\diagup\!\!{}_{{o}}\;}\end{array}$
