\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
a. PTHH: \(4P+5O_2-t^o->2P_2O_5\)
Theo PT ta có: \(n_{O_2}=\dfrac{0,1.5}{4}=0,125\left(mol\right)\)
=> \(V_{O_2}=0,125.22,4=2,8\left(l\right)\)
b. \(V_{kk}=\dfrac{2,8}{20\%}=14\left(l\right)\)
nP=3,131=0,1(mol)
a. PTHH: 4P+5O2−to−>2P2O5
Theo PT ta có: nO2=0,1.54=0,125(mol)
=> VO2=0,125.22,4=2,8(l)
b.