Đáp án đúng: B
24 gam.
${{n}_{C{{O}_{2}}}}\,<\,\frac{{11}{,2}}{{22}{,4}}\,{=}\,{0}{,5}\,{=}\,{{{n}}_{{NaOH}}}$⟹tạo muối Na2CO3 và dư NaOH
$\displaystyle \left\{ \begin{array}{l}{2}{{{n}}_{{N}{{{a}}_{{2}}}{C}{{{O}}_{{3}}}}}\,{+}\,{{{n}}_{{NaOH}\,{sau}}}{=}\,{{{n}}_{{NaOH}\,{trc}}}\,=\,0,5\\106{{n}_{N{{a}_{2}}C{{O}_{3}}}}\,+\,40{{n}_{NaOH}}\,=\,25,2\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{{{n}}_{{N}{{{a}}_{{2}}}{C}{{{O}}_{{3}}}}}\,{=}\,{0}{,2}\\{{{n}}_{{NaOH}}}\,{=}\,{0}{,1}\end{array} \right.$
$\begin{array}{l}\Rightarrow \,{{{n}}_{{MgC}{{{O}}_{{3}}}}}\,{=}\,{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0}{,2}\,{mol;}\,{{{n}}_{{Mg}}}\,{=}\,{{{n}}_{{{{H}}_{{2}}}}}\,{=}\,{0}{,3}\,{mol}\\\Rightarrow \,{m}\,{=}\,{0}{,2}{.84}\,{+}\,{0}{,3}{.24}\,{=}\,{24}\,{gam}\end{array}$