\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(n_{O_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)
\(4Na+O_2\underrightarrow{^{^{t^0}}}2Na_2O\)
\(4..........1\)
\(0.2.....0.02\)
\(LTL:\dfrac{0.2}{4}>\dfrac{0.02}{1}\Rightarrow Nadư\)
\(m_{Na\left(dư\right)}=\left(0.2-0.08\right)\cdot23=2.76\left(g\right)\)
\(m_{Na_2O}=0.04\cdot62=2.48\left(g\right)\)