${{P}_{2}}{{O}_{5}}+6NaOH\to 2N{{a}_{3}}P{{O}_{4}}+3{{H}_{2}}O$

0,05$\leftarrow $ 0,3                          (mol)

${{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,3.1=0,3\left( mol \right)$

Theo phương trình phản ứng :

$\begin{align}  & {{n}_{{{P}_{2}}{{O}_{5}}}}=\frac{{{n}_{NaOH}}}{6}=\frac{0,3}{6}=0,05\left( mol \right) \\ & \Rightarrow {{m}_{{{P}_{2}}{{O}_{5}}}}={{n}_{{{P}_{2}}{{O}_{5}}}}.{{M}_{{{P}_{2}}{{O}_{5}}}}=0,05.142=7,1\left( g \right) \\\end{align}$

$\Rightarrow $Chọn đáp án C