\(\begin{array}{l}
a)\\
4Na+O_2\xrightarrow{t^o}2Na_2O\\
n_{Na}=\frac{6,02.10^{23}}{6,02.10^{23}}=1(mol)\\
n_{Na_2O}=\frac{1}{2}n_{Na}=0,5(mol)\\
m_{Na_2O}=0,5.62=31(g)\\
b)\\
2ca+O_2\xrightarrow{t^o}2CaO\\
n_{CaO}=\frac{3,01.10^{23}}{6,02.10^{23}}=0,5(mol)\\
n_{Ca}=n_{CaO}=0,5(mol)\\
m_{Ca}=0,5.40=20(g)
\end{array}\)