$n_P=\dfrac{12,4}{31}=0,4mol \\n_{O_2}=\dfrac{24}{32}=0,75mol \\a.PTHH :$
$4P + 5O_2\overset{t^o}\to 2P_2O_5$
Theo pt : 4 mol 5 mol
Theo đbài : 0,4 mol 0,75 mol
$Tỉ\ lệ : \dfrac{0,4}{4}<\dfrac{0,75}{5}$
⇒Sau phản ứng O2 dư
$Theo\ pt : \\n_{O_2\ pư}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,4=0,5mol \\⇒n_{O_2\ dư}=0,75-0,5=0,25mol \\⇒m_{O_2\ dư}=0,25.16=4(g) \\b.Theo\ pt : \\n_{P_2O_5}=\dfrac{1}{2}.n_{O_2}=\dfrac{1}{2}.0,4=0,2mol \\⇒m_{P_2O_5}=0,2.142=28,4g$