\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
tỉ lệ 4 : 5 : 2
n(mol) 0,2--->0,25-------->0,1
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ V_{kk}=5,6:\dfrac{1}{5}=28\left(l\right)\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=28\left(l\right)\)