\(\begin{array}{l}
n_{Na}=\frac{1}{23}(mol)\\
n_{Cl_2}=\frac{1}{71}(mol)\\
2Na+Cl_2\xrightarrow{t^o}2NaCl\\
\frac{n_{Na}}{2}>n_{Cl_2}\to Na\,du\\
Theo\,PT:\,n_{NaCl}=n_{Na(pu)}=2n_{Cl_2}=\frac{2}{71}(mol)\\
\to\begin{cases} m_{NaCl}=\frac{2}{71}.58,5≈1,65(g)\\ m_{Na(du)}=\left(\frac{1}{23}-\frac{2}{71}\right).23≈0,35(g) \end{cases}
\end{array}\)