Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of [tex]Cl_2[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Cl_2=3.9mol[/tex]
[tex]2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)[/tex]
As [tex]Na[/tex] is the excess reagent, [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product.
According to stoichiometry :
1 mole of [tex]Cl_2[/tex] gives = 2 moles of [tex]NaCl[/tex]
Thus 3.9 moles of [tex]Cl_2[/tex] will give=[tex]\frac{2}1}\times 3.9=7.8moles[/tex] of [tex]NaCl[/tex]
7.8 moles of NaCl result from the complete reaction of 3.9 mol of [tex]Cl_2[/tex]
