Respuesta :
Answer: [tex]Na[/tex] is in excess.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Cl_2=\frac{20.5g}{71g/mol}=0.29mol[/tex]
[tex]\text{Moles of} Na=\frac{20.5g}{23g/mol}=0.89moles[/tex]
[tex]2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)[/tex]
According to stoichiometry :
1 mole of [tex]Cl_2[/tex] require 2 moles of [tex]Na[/tex]
Thus 0.29 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{1}\times 0.29=0.58moles[/tex] of [tex]Na[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Na[/tex] is the excess reagent.
