\(n_P=\frac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\frac{6,4}{32}=0,2\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
(mol)______4____5_______2_
(mol)_____0,16__0,2_____0,08_
Tỉ lệ: \(\frac{0,2}{4}>\frac{0,2}{5}\rightarrow\) P dư \(0,2-0,16=0,04\left(mol\right)\Leftrightarrow m_P=31.0,04=1,24\left(g\right)\)
\(m_{P_2O_5}=0,08.142=11,36\left(g\right)\)