$a/$
$PTHH :$
$4P+5O2→2P2O5$
$b/$
$n_{P}=12,4/31=0,4mol$
$n_{O_{2}}=14,56/22,4=0,65mol$
$4P+ 5O2→2P2O5 $
$theo$ $pt :$ $ 4$ $5 $
$Theo$ $đbài :$ $0,4 $ $0,65$
ta có tỷ lệ :
$\frac{0,4}{4}<\frac{0,65}{5}$
⇒Sau phản ứng O2 dư
$theo$ $pt :$
$nP2O5=1/2.nP=1/2.0,4=0,2mol$
$⇒m_{P_{2}O_{5}}=0,2.142=28,4g$