\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)
4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
\(\dfrac{0,1}{4}< \dfrac{0,2}{5}\) => O2 dư, Photpho đủ
\(n_{O_2}=0,2-0,04=0,16\left(mol\right)\)
\(m_{P_2O_5}=\) 0,05 . 142 = 7,1 ( g )