3/
$n_{O2}=3,36/22,4=0,15mol$
$2KClO3→2KCl+3O2$
$theo$ $pt :$
$n_{KClO3}=2/3.n_{O2}=2/3.0,15=0,1mol$
$⇒m_{KClO3}=0,1.122,5=12,25g$
$4/$
$n_{P}=12,4/31=0,4mol$
$n_{O2}=17/32=0,53125mol$
$a/pthh :$
$4P + 5O2→2P2O5$
$theo$ $pt :$ $ 4 mol$ $5 mol
$theo$ $đbài: $ $0,4 mol $ $ 0,53125mol$
ta có tỷ lệ :
$\frac{0,4}{4}<\frac{0,53125}{5}$
⇒Sau phản ứng O2 dư
theo pt :
$n_{O2 pư}=5/4.n_{P}=5/4.0,4=0,5mol$
$⇒n_{O2 dư}=0,53125-0,5=0,03125mol$
$⇒m_{O2 dư}=0,03125.32=1 g$
$b/$
theo pt :
$n_{P2O5}=n_{P}=0,4mol$
$⇒m_{P2O5}=0,4.142=56,8g$