\(n_{O_2}=\frac{0,1}{22,4}=\text{0,0045}\left(mol\right)\)
PTHH : \(4P+5O_2-^{t^o}\rightarrow2P_2O_5\)
Theo PT:4..........5........................
Theo ĐB:0,1....0,0045.....................
\(\Rightarrow\frac{0,1}{4}>\frac{0,0045}{5}\)
Vậy P dư, O2 phản ứng hết
Theo PT \(n_{P\left(p.ứ\right)}=\frac{0,0045.4}{5}=\text{0,0036}\left(mol\right)\)
\(\Rightarrow n_{P\left(dư\right)}=0,1-\text{0,0036}=\text{0,0964}\left(mol\right)\)
\(\Rightarrow m_{P\left(dư\right)}=\text{0,0964}.31=\text{2,9884}\left(g\right)\)