Question

According to the following reaction, how many moles of iron(III) chloride will be formed upon the complete reaction of 0.371 moles iron with excess chlorine gas?

iron (s) + chlorine (g) iron(III) chloride (s)

Now, I have 3Fe(s) + 3Cl -> FeCl3, I cannot seem to get the right answer with my formulated equation

Answer 1

Your equation isn't balanced. Your 2 Fe's are missing from the right side and your Cl should be diatomic. (It's safe to say that all halogens are diatomic, they simply aren't stable otherwise.

It should be:

\[2Fe(s) + 3Cl^2(g) = 2FeCl _{3}\]

You have 0.371 moles of iron.

the ratio of FE to FeCl IS 2:2 (equal)

therefore, you will have 0.371 moles of FeCl

If you want the weight, just multiply that by the molar weight of FeCl.

Hope this helps! c:

Answer 2

Thank you. What I still do not understand is why is it a 2Fe and 3Cl^2. is it going by the valence electrons? the charges of the groups? I'm guessing it is a 2Fe because of the Cl^2, but why is it Cl^2 ?

Answer 3

Sorry for the late response!

For halogens, (atoms with 7 valence electrons) it is customary for them to be diatomic. All materials "lean" towards a noble gas octet. (8 electrons in their last shell)

As halogens are "desperate" to be stable, (have an octet) they often combine into a diatomic form, fulfilling that need for an octet and hence becoming stable. Even if you started with single atoms of halogens, they would soon combine if given the chance. This is way you almost always see Halogens in a \[x _{2}\] form. (where x is a halogen.)

You have 2 Fe so that you have enough Fe for the diatomic chlorine. In this reaction, there is only one product, so once iron(III) chloride is formed, you should not have anything else on the right hand side.

I hope this clarifies things. c: