Gọi số mol etilen và axetilen lần lượt là a, b (mol)

${{n}_{B{{r}_{2}}}}={{V}_{B{{r}_{2}}}}.{{C}_{{{M}_{B{{r}_{2}}}}}}=0,25.1=0,25\left( mol \right)$

${{C}_{2}}{{H}_{4}}+B{{r}_{2}}\to {{C}_{2}}{{H}_{4}}B{{r}_{2}}$                             (1)

  a   $\to $  a                       (mol)

${{C}_{2}}{{H}_{2}}+2B{{r}_{2}}\to {{C}_{2}}{{H}_{2}}B{{r}_{4}}$                           (2)

   b  $\to $2.b                      (mol)

Theo phương trình (1) và (2) có :

${{n}_{B{{r}_{2}}}}={{n}_{{{C}_{2}}{{H}_{4}}}}+2.{{n}_{{{C}_{2}}{{H}_{2}}}}=a+2.b=0,25\left( mol \right)$          (*)

${{m}_{hh}}={{m}_{{{C}_{2}}{{H}_{4}}}}+{{m}_{{{C}_{2}}{{H}_{2}}}}={{n}_{{{C}_{2}}{{H}_{4}}}}.{{M}_{{{C}_{2}}{{H}_{4}}}}+{{n}_{{{C}_{2}}{{H}_{2}}}}.{{M}_{{{C}_{2}}{{H}_{2}}}}$

$\Rightarrow 28.a+26.b=5,5\left( g \right)$                                         (**)

Từ (*) và (**) có: a= 0,15 (mol)

                             b=0,05 (mol)

$\Rightarrow {{m}_{{{C}_{2}}{{H}_{4}}}}={{n}_{{{C}_{2}}{{H}_{4}}}}.{{M}_{{{C}_{2}}{{H}_{4}}}}=0,15.28=4,2\left( g \right)$

$%{{m}_{{{C}_{2}}{{H}_{4}}}}=\frac{{{m}_{{{C}_{2}}{{H}_{4}}}}}{{{m}_{hh}}}.100=\frac{4,2}{5,5}.100=76,36\left( % \right)$

$\Rightarrow %{{m}_{{{C}_{2}}{{H}_{2}}}}=100-%{{m}_{{{C}_{2}}{{H}_{2}}}}=100-76,36=23,64\left( % \right)$

$\Rightarrow $Chọn đáp án D