\(n_P=\dfrac{5}{31}=0,16mol\)
\(V_{O_2}=\dfrac{V_{kk}}{5}=\dfrac{2,8}{5}=0,56l\)
\(n_{O_2}=\dfrac{0,56}{22,4}=0,025mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
\(\dfrac{0,16}{4}\)< \(\dfrac{0,25}{5}\) ( mol )
0,16 0,08 ( mol )
\(m_{P_2O_5}=0,08.142=11,36g\)