a) 4P + 5O2 \(\underrightarrow{to}\) 2P2O5
b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}\times0,2=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25\times22,4=5,6\left(l\right)\)
\(\Rightarrow V_{KK}=\dfrac{5,6}{20\%}=28\left(l\right)\)
c) Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1\times142=14,2\left(g\right)\)
d) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_P=\dfrac{4}{5}n_{O_2}\)
Theo bài: \(n_P=\dfrac{2}{3}n_{O_2}\)
Vì \(\dfrac{2}{3}< \dfrac{4}{5}\) ⇒ O2 dư.
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1\times142=14,2\left(g\right)\)
a) 4P + 5O2 to→ 2P2O5
b) nP=6,231=0,2(mol)
Theo PT: nO2=54nP=54×0,2=0,25(mol)
⇒VO2=0,25×22,4=5,6(l)
⇒VKK=5,620%=28(l)
c) Theo PT: nP2O5=12nP=12×0,2=0,1(mol)
⇒mP2O5=0,1×142=14,2(g)
d) nO2=6,7222,4=0,3(mol)
Theo PT: nP=45nO2
Theo bài: nP=23nO2
Vì 23<45 ⇒ O2 dư.
Theo PT: nP2O5=12nP=12×0,2=0,1(mol)