\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a)\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{17}{32}=0,53\left(mol\right)\)
Ta có: \(\dfrac{n_P}{4}:\dfrac{n_{O_2}}{5}=\dfrac{0,4}{4}< \dfrac{0,53}{5}\Rightarrow O_2dư\)
Theo PTHH ta có:\(n_{O_2}\left(pư\right)=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\Rightarrow n_{O_2}\left(dư\right)=n_{O_2}-n_{O_2}\left(pư\right)=0,53-0,5=0,03\left(mol\right)\)
b)Chất được tạo thành là\(P_2O_5\)
Theo PTHH ta có:\(n_{P_2O_5}=\dfrac{2}{4}n_P=\dfrac{2}{4}.0,4=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.\left(31.2+16.5\right)=28,4\left(g\right)\)