a,
$n_{P}$ = $\frac{m}{M}$ = $\frac{6,2}{31}$ = $0,2$ mol
PTHH :$4P$ + $5O_{2}$ $\xrightarrow{t^o}$ $2P_{2}O_{5}$
$4$ : $5$ : $2$ (mol)
$0,2$ : $0,25$ : $0,1$ (mol)
b, $m_{p_{2}O_{5}}$ = $n . M$ = $0,1 . 142 = 14,2g$
Mk ko bik lm câu C sorii bn rất nhìu ọ :<<