a, PTHH: $\text{4P + $5O_2$ $\xrightarrow{t^o}$ $2P_2$$O_5$}$

Ta có: $\text{$n_{O_2}$ = $\dfrac{13,44}{22,4}$ = 0,6 mol}$

b, Theo PTHH $\rightarrow$ $\text{$n_{P_{(p.ư)}}$ = $\dfrac{4}{5}$. $n_{O_2}$ = 0,48 mol}$

$\rightarrow$ $\text{$m_{P_{(p.ư)}}$ = 0,49. 31 = 14,88 gam}$

c, Theo PTHH $\rightarrow$ $\text{$n_{P_2O_5}$ = $\dfrac{2}{5}$. $n_{O_2}$ = 0,24 mol}$

$\rightarrow$ $\text{$m_{P_2O_5}$ = 0,24. 142 = 34,08 gam}$

$\text{#Kudo}$

a, 4P+5O2 -> 2P2O5 ( ĐK nhiệt độ)

b, nO2= 13,44/22,4= 0,6 mol

          4P+      5O2 ->      2P2O5

         0,48        0,6               0,24               mol

mP= 0,48.31= 14,88 g

c, mP2O5= 0,24. ( 31.2+16.5)= 34,08 g