Hướng dẫn:

${{n}_{C}}=\frac{{{m}_{C}}}{{{M}_{C}}}=\frac{4,8}{12}=0,4\left( mol \right)$

$C+{{O}_{2}}\xrightarrow{{{t}^{0}}}C{{O}_{2}}$ (1)

0,4       $\to $     0,4          (mol)

$C{{O}_{2}}+2NaOH\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O$         (2)

0,4                  $\to $0,4                           (mol)

Theo phương trình (1) có: ${{n}_{C{{O}_{2}}}}={{n}_{C}}=0,4\left( mol \right)$

Theo phương trình (2) có: ${{n}_{N{{a}_{2}}C{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,4\left( mol \right)$

$\Rightarrow {{m}_{N{{a}_{2}}C{{O}_{3}}}}={{n}_{N{{a}_{2}}C{{O}_{3}}}}.{{M}_{N{{a}_{2}}C{{O}_{3}}}}=0,4.106=42,4\left( g \right)$

$\Rightarrow $Chọn đáp án D