Question

Sodium chloride, NaCl forms in this reaction between sodium and chlorine. 2Na(s) + Cl2(g) → 2NaCl(s) How many moles of NaCl result from the complete reaction of 3.9 mol of Cl2? Assume that there is more than enough Na.

Answer 1

Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of $Cl_2$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} Cl_2=3.9mol$

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

As $Na$ is the excess reagent, $Cl_2$ is the limiting reagent as it limits the formation of product.

According to stoichiometry :

1 mole of $Cl_2$ gives = 2 moles of $NaCl$

Thus 3.9 moles of $Cl_2$ will give=$\frac{2}1}\times 3.9=7.8moles$  of $NaCl$

7.8 moles of NaCl result from the complete reaction of 3.9 mol of $Cl_2$