PTHH :
4P + $5O_{2}$ ------$\frac{ t0 }{}$-------->$n_{P_{2}O_{5}}$
a/ $n_{P}$ =$\frac{12,4}{31}$ = 0,4(mol)
theo pt :
$n_{O_{2}}$= $\frac{5}{4}$ $n_{P}$ = 0,5 (mol)
$V_{O_{2}}$=0,5 x 22,4=11,2 (l)
b/theo pt:
$n_{P_{2}O_5}$ =$\frac{2}{4}$ $n_{P}$ =0,25(mol)
⇒$m_{P_{2}O_{5}}$=0,25 x 142=35,5