\(4P+5O_2\rightarrow2P_2O_5\)
\(V_{O_2}=20\%.V_{kk}=20\%.5,6=1,12\left(l\right)\)
\(\Rightarrow n_{O_{2\left(đb\right)}}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{P\left(đb\right)}=\frac{10}{31}\left(mol\right)\)
\(\frac{n_{P\left(đb\right)}}{n_{P\left(PTHH\right)}}\) \(\frac{n_{O_{2\left(đb\right)}}}{n_{O_{2\left(PTHH\right)}}}\)
\(\frac{10}{\frac{31}{4}}>\frac{0,05}{5}\)
⇒P dư, O2 hết
4P + 5O2 -> 2P2O5
nkk=5,6.22,4=0,25(mol)
nO2=0,25/5=0,05(mol)
nP=10/31(mol)
VÌ 0,05.4/5<10/31 nên P dư
