Hướng dẫn:

${{n}_{C{{l}_{2}}}}=\frac{{{V}_{C{{l}_{2}}}}}{22,4}=\frac{1008}{22,4}=45\left( mol \right)$

$2NaCl+2{{H}_{2}}O\to C{{l}_{2}}+{{H}_{2}}+2NaOH$

90                   45                          (mol)

Theo phương trình ta có: ${{n}_{NaCl}}=2.{{n}_{C{{l}_{2}}}}=45.2=90\left( mol \right)$

$\Rightarrow {{m}_{NaCl}}={{n}_{NaCl}}.{{M}_{NaCl}}=90.58,5=5265\left( g \right)=5,265\left( kg \right)$

Chọn đáp án D