Hướng dẫn:

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

Ban đầu:         0,01     0,02                                       (mol)

Phản ứng:       0,01$\to $0,01        0,01                     (mol)

Sau phản ứng: 0          0,01                0,01                (mol)

${{n}_{S{{O}_{2}}}}=\frac{{{V}_{S{{O}_{2}}}}}{22,4}=\frac{0,224}{22,4}=0,01\left( mol \right)$

${{n}_{Ca{{\left( OH \right)}_{2}}}}={{V}_{Ca{{\left( OH \right)}_{2}}}}.{{C}_{{{M}_{Ca{{\left( OH \right)}_{2}}}}}}=2.0,01=0,02\left( mol \right)$

Theo phương trình phản ứng:

${{n}_{Ca{{\left( OH \right)}_{2}}_{du}}}=0,02-0,01=0,01\left( mol \right)$

${{n}_{CaS{{O}_{3}}}}={{n}_{S{{O}_{2}}}}=0,01\left( mol \right)$

${{m}_{CaS{{O}_{3}}}}={{n}_{CaS{{O}_{3}}}}.{{M}_{CaS{{O}_{3}}}}=0,01.120=1,2\left( g \right)$

$\Rightarrow $ Chọn đáp án C