Hướng dẫn:

${{m}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{\text{dd}}}.C%}{100}=\frac{200.4,6}{100}=9,2\left( g \right)$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{{{m}_{{{C}_{2}}{{H}_{5}}OH}}}{{{M}_{{{C}_{2}}{{H}_{5}}OH}}}=\frac{9,2}{46}=0,2\left( mol \right)$

${{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{6,72}{22,4}=0,3\left( mol \right)$

                 ${{C}_{2}}{{H}_{5}}OH+3{{O}_{2}}\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O$

Ban đầu:      0,2                             0,3                  (mol)

Phản ứng:     0,15                    $\leftarrow $ 0,3                     (mol)

Theo phương trình ta có: ${{n}_{{{C}_{2}}{{H}_{5}}OHpu}}=\frac{{{n}_{C{{O}_{2}}}}}{2}=\frac{0,3}{2}=0,15\left( mol \right)$

$%H=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OHpu}}}{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}.100=\frac{0,15}{0,2}.100=75\left( % \right)$

$\Rightarrow $Chọn đáp án A