\(n_P=\frac{3,1}{31}=0,1\left(mol\right)\)
\(n_{O2}=\frac{5}{32}=0,15625\left(mol\right)\)
\(PTHH:4P+5O_2\rightarrow2P_2O_5\)
Trước _0,1 ___0,15625______
Phản ứng_0,1__0,125____
Sau_____ 0 ___0,03125___0,05
Vậy O2 dư.
\(n_{P\left(đb\right)}=\frac{3,1}{31}=0,1\left(mol\right)\)
\(n_{O_2\left(đb\right)}=\frac{5}{32}=0,15625\left(mol\right)\)
\(4P+5O_2\rightarrow2P_2O_5\)
Theo PTHH: 4mol 5mol
Ta có: \(\frac{n_{P\left(đb\right)}}{n_{P\left(PTHH\right)}}\) \(\frac{n_{O_2\left(đb\right)}}{n_{O_2\left(PTHH\right)}}\)
\(\Rightarrow\frac{0,1}{4}< \frac{0,15625}{5}\)
\(\Rightarrow\) P hết, O2 dư
