Đáp án đúng: C
46,5 gam.
$\begin{array}{l}{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0}{,25}\,{mol;}\,{{{n}}_{{NaOH}}}\,{=}\,\frac{{164}{.1}{,22}{.20}}{{100}{.40}}{=1}\,{mol}\\\Rightarrow \,{{{n}}_{{N}{{{a}}_{{2}}}{C}{{{O}}_{{3}}}}}\,{=}\,{0}{,25}\,{mol;}\,{{{n}}_{{NaOH}\,{sau}}}\,{=}\,{0}{,5}\,{mol}\\\Rightarrow \,{{{m}}_{{c}{.r}}}\,{=}\,{0}{,25}{.106}\,\,{+}\,{0}{,5}{.40}\,{=}\,{46}{,5}\,{gam}\end{array}$