Hướng dẫn:

${{n}_{CaC{{O}_{3}}}}=\frac{{{m}_{CaC{{O}_{3}}}}}{{{M}_{CaC{{O}_{3}}}}}=\frac{30}{100}=0,3\left( mol \right)$

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

     0,3        $\leftarrow $    0,3                                            (mol)

$C{{O}_{2}}+Ca{{\left( OH \right)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$                   (2)

 0,3                     $\leftarrow $   0,3                                    (mol)

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

Theo phương trình (1) có : ${{n}_{C}}={{n}_{C{{O}_{2}}}}=0,3\left( mol \right)$

$\Rightarrow {{m}_{C}}={{n}_{C}}.{{M}_{C}}=0,3.12=3,6\left( g \right)$

$\Rightarrow $ Chọn đáp án D