- Chọn B.

- Ta có: ${{n}_{{{O}_{2}}}}=\frac{4,48}{22,4}=0,2\left( mol \right)$

- PTHH:              C         +     ${{O}_{2}}$      $\xrightarrow{{{t}^{o}}}$    $C{{O}_{2}}.$

                           0,3(mol)       0,2(mol)

                Pư:     0,2(mol)      0,2(mol)     $\to $  0,2(mol).

Khối lượng khí cacbonic sinh ra: ${{m}_{C{{O}_{2}}}}=0,2\times 44=8,8\left( gam \right).$