a. PTHH: \(4P+5O_2\rightarrow2P_2O_5\\ 0,2mol:0,25mol\rightarrow0,1mol\)
b. Ta có:
\(m_P=6,2\left(g\right)\)
\(\Rightarrow n_P=\dfrac{m_P}{M_P}=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(V_{O_2}=n_{O_2}.22,4=0,25.22,4=5,6\left(l\right)\)
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,1.142=14,2\left(g\right)\)
a) 4P + 5O2 \(\underrightarrow{to}\) 2P2O5
b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}\times0,2=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25\times22,4=5,6\left(l\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1\times142=14,2\left(g\right)\)
