- Chọn C.

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

- Ta có: $5{{n}_{{{O}_{2}}}}<4{{n}_{P}}.$Suy ra, Photpho còn dư sau phản ứng.

Theo PTHH: ${{n}_{{{P}_{2}}{{O}_{5}}}}=\frac{2}{5}\times {{n}_{{{O}_{2}}}}=\frac{2}{5}\times 0,05=0,02\left( mol \right).$

$\to {{m}_{{{P}_{2}}{{O}_{5}}}}={{n}_{{{P}_{2}}{{O}_{5}}}}.M=0,02\times 142=2,84\left( gam \right).$