Hướng dẫn:

${{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$ 

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

  0,15            $\leftarrow $ 0,15        (mol)

Theo phương trình ta có:

${{n}_{{{C}_{thep}}}}={{n}_{C{{O}_{2}}}}=0,15\left( mol \right)$ 

${{m}_{{{C}_{thep}}}}={{n}_{{{C}_{thep}}}}.{{M}_{{{C}_{thep}}}}=0,15.12=1,8\left( g \right)$ 

$\Rightarrow %C=\frac{{{m}_{{{C}_{thep}}}}}{{{m}_{thep}}}.100=\frac{1,8}{95,8}.100=1,88%$ 

 Chọn đáp án B