Hướng dẫn:

${{n}_{P}}=\frac{{{m}_{P}}}{{{M}_{P}}}=\frac{3,1}{31}=0,1\left( mol \right)$

$4P+5{{O}_{2}}\xrightarrow{{{t}^{0}}}2{{P}_{2}}{{O}_{5}}$

0,1            $\to $     0,05        (mol)

Theo phương trình ta có: ${{n}_{{{P}_{2}}{{O}_{5}}}}=\frac{2.{{n}_{P}}}{4}=\frac{2.0,1}{4}=0,05\left( mol \right)$

$\Rightarrow {{m}_{{{P}_{2}}{{O}_{5}}}}={{n}_{{{P}_{2}}{{O}_{5}}}}.{{M}_{{{P}_{2}}{{O}_{5}}}}=0,05.142=7,1\left( g \right)$

$\Rightarrow $ Chọn đáp án A