- Chọn B.

- Ta có: ${{n}_{{{O}_{2}}}}=\frac{V}{22,4}=\frac{13,44}{22,4}=0,6\left( mol \right);{{n}_{{{P}_{2}}{{O}_{5}}}}=\frac{28,4}{142}=0,2\left( mol \right).$

- PTHH: $4P+{{O}_{2}}\xrightarrow{{{t}^{o}}}2{{P}_{2}}{{O}_{5}}.$

- Ta có: ${{n}_{{{O}_{2}}}}>2{{n}_{{{P}_{2}}{{O}_{5}}}}.$ Suy ra, khí oxi còn dư sau phản ứng.

Theo PTHH: ${{n}_{P}}=2{{n}_{{{P}_{2}}{{O}_{5}}}}=2\times 0,2=0,4\left( mol \right).$

$\to {{m}_{P}}={{n}_{P}}.M=0,4\times 31=12,4\left( gam \right).$