${{n}_{{{C}_{2}}{{H}_{2}}}}=\frac{{{V}_{{{C}_{2}}{{H}_{2}}}}}{22,4}=\frac{4,48}{22,4}=0,2\left( mol \right)$

$Ca{{C}_{2}}+2{{H}_{2}}O\to {{C}_{2}}H{{ & }_{2}}+Ca{{\left( OH \right)}_{2}}$

 0,2                $\leftarrow $  0,2                           (mol)

 Theo phương trình ta có:

${{n}_{Ca{{C}_{2}}}}={{n}_{{{C}_{2}}{{H}_{2}}}}=0,2\left( mol \right)$

$\Rightarrow {{m}_{Ca{{C}_{2}}}}={{n}_{Ca{{C}_{2}}}}.{{M}_{Ca{{C}_{2}}}}=0,2.64=12,8\left( g \right)$

$\Rightarrow $Chọn đáp án C