Hướng dẫn:

${{n}_{CO}}=\frac{{{V}_{CO}}}{22,4}=\frac{6,72}{22,4}=0,3\left( mol \right)$ 

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

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

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

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

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

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

$\Rightarrow {{m}_{CaC{{O}_{3}}}}={{n}_{CaC{{O}_{3}}}}.{{M}_{CaC{{O}_{3}}}}=0,3.100=30\left( g \right)$ 

$\Rightarrow $Chọn đáp án A