Hướng dẫn:

${{n}_{C}}=\frac{{{m}_{C}}}{{{M}_{C}}}=\frac{0,9}{12}=0,075\left( mol \right)$

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

  0,075      $\to $     0,075          (mol)

Theo phương trình ta có: ${{n}_{C{{O}_{2}}}}={{n}_{C}}=0,075\left( mol \right)$

$\Rightarrow {{V}_{C{{O}_{2}}}}=22,4.{{n}_{C{{O}_{2}}}}=22,4.0,075=1,68\left( l \right)$

$\Rightarrow $ Chọn đáp án B