$n_{Na}=\dfrac{4,6}{23}=0,2\ (mol)$
$n_{Cl_2}=\dfrac{6,72}{22,4}=0,3\ (mol)$
$2Na+Cl_2\xrightarrow{t^o} 2NaCl$
Ta có:
$\dfrac{n_{Na}}{2}=\dfrac{0,2}{2}=0,1<\dfrac{n_{Cl_2}}{1}=\dfrac{0,3}{1}=0,3$
$\Rightarrow Na$ hết
Ta có: $n_{NaCl}=n_{Na}=0,2\ (mol)$
$\Rightarrow m_{muối}=m_{NaCl}=0,2.58,5=11,7\ (g)$