Answer: True
Explanation:
[tex]\mathrm{The\ balanced \ chemical \ equation\ is\ :}\\$2 \mathrm{Na}{(s)}+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(s)[/tex]
For Na
[tex]\\$Given mass $=20.4 \mathrm{~g}$ \ \ Molecular mass $=23 \mathrm{~g}\ \ $Number of moles $=\frac{\text { Given mass }}{\text { Molecular mass }}$$\begin{aligned}&=\frac{20.4}{23} \\&=0.887 \text { moles }\end{aligned}$Stoichiometric ratio of $\mathrm{Na}$ is same as that of $\mathrm{NaCl}$ i.e., $1: 1$.$\therefore$ Moles of $\mathrm{NaCl}=0.887$ motes[/tex]
For NaCl
[tex]$\\\begin{aligned}\text { Molecular mass } &=23+35.5 \\&=58.5 \mathrm{~g}\end{aligned}$[/tex]
[tex]\begin{aligned}\\& $\therefore$ Mass of $\mathrm{NaCl}$ produced will be=\text { No.of moles } \times \text { Molecular mass } \\&=0.887 \times 58.5 \\&=51.9 \\&\simeq 52 \mathrm{~g} \quad\end{aligned}$[/tex]
Therefore, it is true that when 20.4g of sodium metal are mixed with chlorine gas, 52.0 g of sodium chloride is produced
