${{n}_{Ca{{C}_{2}}}}=\frac{{{m}_{Ca{{C}_{2}}}}}{{{M}_{Ca{{C}_{2}}}}}=\frac{16}{64}=0,25\left( mol \right)$

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

 0,25              $\to $ 0,25                           (mol)

 Theo phương trình ta có:

${{n}_{{{C}_{2}}{{H}_{2}}}}={{n}_{Ca{{C}_{2}}}}=0,25\left( mol \right)$

$\Rightarrow {{V}_{{{C}_{2}}{{H}_{2}}}}=22,4.{{n}_{{{C}_{2}}{{H}_{2}}}}=22,4.0,25=5,6\left( l \right)$

$\Rightarrow $Chọn đáp án B