$n_P=\frac{3,1}{31}=0,1(mol)$
$n_{O_2}=\frac{4,48}{22,4}=0,2(mol)$
a)PT: $4P+5O_2→2P_{2}O_5$
b)Xét $\frac{n_P(bđầu)}{n_P(pt)}=\frac{0,1}{4}=0,025<\frac{n_{O_{2}}(bđầu)}{n_{O_{2}}(pt)}=\frac{0,2}{5}=0,04$
=>$P$ hết $O_{2}$ dư
c)$n_{P_{2}O_5}=2n_P=2.0,1=0,2(mol)$
$m_{P_{2}O_5}=n_{P_{2}O_5}.M_{P_{2}O_5}=0,2.(31.2+16.5)=28,4(g)$