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`

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}*{\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}* 3.9=7.8moles` of
`NaCl`

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