\(m_{hh}=\dfrac{110g}{mol}\)
\(=>M_{O2}=\dfrac{110.43,46\%}{100}=48g\)
=>\(M_P=62\)g
\(m_{O2}=\dfrac{48}{16}=3mol\)
\(m_P=\dfrac{62}{31}=2mol\)
\(=>CTHH:P_2O_3\)
c2 :
\(m_O=\dfrac{110.43,46}{100}=48\left(g\right)\\
m_P=110-48=62\left(g\right)\\
n_O=\dfrac{48}{16}=3\left(mol\right)\\
n_P=\dfrac{62}{31}=2\left(mol\right)\)
-> CTHH là P2O3