số mol P,O là :
\(n_P=\frac{1,86}{31}=0,06\left(mol\right)\)
\(n_{O_2}=\frac{1,792}{22,4}=0,08\left(mol\right)\)
PTHH :\(4P+5O_2\rightarrow2P_2O_5\)
\(\frac{n_{O_2}}{n_{O_2}}=\frac{0,08}{5}>\frac{n_P}{n_P}=\frac{0,06}{4}\)
tính theo P và O2 dư
\(n_{dư}=n_{đề}-n_{pt}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ n_{dư}=0,08-0,016\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ n_{dư}=0,064\)
\(m_O=n.M=0,064.\left(16.2\right)=2,048\left(g\right)\)
\(m_{P_2O_5}=n.M=0,06.2.\left(31.2+16.5\right)=17,04\left(g\right)\)