\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)
Theo PT: \(n_{Na_2O\left(LT\right)}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow m_{Na_2O\left(LT\right)}=0,1.62=6,2\left(g\right)\)
\(\Rightarrow H=\dfrac{4,96}{6,2}.100\%=80\%\)