a) Ta có:
\(\left\{{}\begin{matrix}n_P=\frac{12,4}{31}=0,4\left(mol\right)\\n_{O2}=\frac{17}{32}=0,53125\left(mol\right)\end{matrix}\right.\)
\(4P+5O_2\rightarrow2P_2O_5\)
Vì \(n_{O2}=\frac{5}{4}n_P\Rightarrow\) O2 dư
\(\Rightarrow n_{O2\left(dư\right)}=0,53125-\frac{5}{4}n_P=0,03125\left(mol\right)\)
Chất tạo thành là P2O5
\(n_{P2O5}=\frac{1}{2}n_P=0,2\left(mol\right)\)
\(\Rightarrow m_{P2O5}=0,2.\left(31.2+16.5\right)=28,4\left(g\right)\)