a, \(PTHH:4P+5O_2\rightarrow2P_2O_5\)
Ta có:
\(n_P=\frac{6.10^{23}}{6.10^{23}}=1\)
\(n_{O2}=\frac{1.5}{4}=1,25\)
\(A_{O2}=1,25.6.10^{23}=7,5.10^{23}\)
b, \(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
\(\Rightarrow n_{P\left(spu\right)}=1-0,4=0,6\left(mol\right)\)
\(\Rightarrow m_P=0,6.31=18,6\left(g\right)\)