4P+5O2=to=>2P2O5
\(n_P=\frac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH, ta có: \(\frac{0,2}{4}< \frac{0,3}{5}\)=>O2 dư
\(n_{O_2\left(dư\right)}=0,3-\left(\frac{0,2.5}{4}\right)=0,05\left(mol\right)\)
\(m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
\(n_{P_2O_5}=\frac{2}{4}.n_P=\frac{2}{4}.0,2=0,1\left(mol\right)\)
\(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)