Hướng dẫn :

${{C}_{12}}{{H}_{22}}{{O}_{11}}+12{{O}_{2}}\to 12C{{O}_{2}}+11{{H}_{2}}O$

${{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=\frac{51,3}{342}=0,15mol$

Theo phương trình : ${{n}_{C{{O}_{2}}}}=12{{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=0,15.12=1,8mol$

                                    ${{n}_{{{H}_{2}}O}}=11{{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=0,15.11=1,65mol$

$\Rightarrow {{m}_{sp}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}=1,8.44+1,65.18=108,9g$

                      $\Rightarrow $ Chọn đáp án A.