$n_{P}=7,44/31=0,24mol$
$n_{O2}=6,6/22,4=33/112mol$
$a/$
Ta có:
$ 4P + 5O2\overset{t^o}{\rightarrow}2P2O5$
Theo pt: $4 mol$ $5 mol
Theo đbài: $0,24 mo$l $\dfrac{33}{112}mol$
⇒Sau phản ứng P dư
Theo pt:
$n_{P pư}=5/4.n_{O2}=4/5.\dfrac{33}{113}=\dfrac{132}{565}mol$
$⇒n_{P dư}=0,24-\dfrac{132}{565}=0,00636mol$
$⇒m_{P dư}=0,00636.31=0,19716g$
$b/$
$m_{P2O5}=142.33/280=16,74g$