PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,5\left(mol\right)\) \(\Rightarrow V_{O_2}0,5\cdot22,4=11,2\left(l\right)\)
a) \(4P+5O_2\underrightarrow{t\text{°}}P_2O_5\)
b)\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Từ PTHH: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\)
\(\Rightarrow\)\(V_{O_2}=0,5.22,4=11,2\left(l\right)\)