\(n_P=\frac{3,1}{31}=0,1\left(mol\right);n_{O_2}=\frac{3}{22,4}=\frac{15}{112}\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
(mol)____0,1________0,05
Tỉ lệ: \(\frac{0,1}{4}< \frac{\frac{15}{112}}{5}\rightarrow\) Oxi dư
\(m_{P_2O_5}=142.0,057,1\left(g\right)\)
\(n_P=\frac{m_P}{M_P}=\frac{3,1}{31}=0,1\left(mol\right)\)
\(V_{O_2}=\frac{n_{O_2}}{22,4}=\frac{3}{22,4}\simeq0,13\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Lập tỉ lệ: \(\frac{0,1}{4}< \frac{0,13}{5}\)
=> P PỨ hết, O2 dư
=> Chọn số mol của P
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(0,1\) \(0,125\) \(0,05\) \(\left(mol\right)\)
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1\left(g\right)\)
