\(n_P=\frac{15,1}{31}=0,5\left(mol\right)\)
\(n_{O2}=\frac{16,8}{22,4}=0,75\left(mol\right)\)
\(PTHH4P+5O_2\rightarrow2P_2O_5\)
Tỉ lệ: \(\frac{0,5}{4}< \frac{0,75}{5}\Rightarrow\) O2 dư, P hết
\(n_{O2_{dư}}=0,75-\left(\frac{5}{4}.0,5\right)=0,125\left(mol\right)\)
\(\Rightarrow m_{P2O5}=0,5.\frac{1}{2}.142=35,5\left(g\right)\)