\(\begin{array}{l}
2)\\
a)\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
b)\\
nP = \frac{{18,6}}{{31}} = 0,6\,mol\\
= > n{P_2}{O_5} = 0,3\,mol\\
m{P_2}{O_5} = 0,3 \times 142 = 42,6g\\
c)\\
n{O_2} = \frac{{0,6 \times 5}}{4} = 0,75\,mol\\
V{O_2} = 0,75 \times 22,4 = 16,8l\\
3)\\
a)\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
b)\\
nP = \frac{{6,2}}{{31}} = 0,2\,mol\\
n{O_2} = \frac{{6,72}}{{22,4}} = 0,3\,mol\\
\frac{{0,2}}{4} < \frac{{0,3}}{5}
\end{array}\)
=>$O_2$ dư
$nO_2$ dư=0,3-0,25=0,05mol
c)
nP=0,1mol
$nP_2O_5$=0,1x142=14,2 g
