\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\\ 4P+5O_2\underrightarrow{to}2P_2O_5\\ Vì:\dfrac{0,2}{4}=\dfrac{0,1}{2}\\ \Rightarrow Phản.ứng.hết\\ n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ V_{kk\left(đktc\right)}=\dfrac{100}{20}.V_{O_2\left(đktc\right)}=\dfrac{100}{20}.\left(0,25.22,4\right)=28\left(l\right)\)