\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ Vì:\dfrac{0,1}{4}< \dfrac{1}{5}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=1-\dfrac{5}{4}.0,1=0,875\left(mol\right)\\ m_{O_2}=0,875.32=28\left(g\right)\\ n_{P_2O_5}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right);n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,1 0,125 0,05
Ta có: \(\dfrac{0,1}{4}< \dfrac{1}{5}\) ⇒ P hết, O2 dư
\(m_{O_2dư}=\left(1-0,125\right).32=28\left(g\right)\)
\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
