a, PTHH : 4P + 5\(O_2\) -> 2\(P_2O_5\) (1)
b, \(n_P=\frac{24,8}{31}=0,8\) (mol)
Theo (1) : \(n_{O_2}=\frac{5}{4}.n_P=\frac{5}{4}.0,8=1\) (mol)
\(V_{O_2}=1.22,4=22,4\) (lít)
c,Theo (1): \(n_{P_2O_5}=\frac{1}{2}.n_P=\frac{1}{2}.0,8=0,4\) (mol)
\(m_{P_2O_5}=0,4.142=56,8\) (g)