Oxidation is rust, iron with oxygen.

xD that was stupid of me. That makes sense, thank you so much! So I’m assuming that in this reaction, Fe (II) and OH – are aqueous?

hmm, I do have one more question that I’m simply not sure how to solve, it doesn’t have anything to do with precipitates, but if anyone can help it’d still be really appreciated. I’ll type it up here just in case:

When 70.0 milliliters of 3.0-molar Na2CO3 is added to 30.0 milliliters of 1.0-molar NaClO4, the resulting concentration of Na+ is:

A. 2.0 M
B. 2.4 M
C. 4.0 M
D. 4.5 M
E. 7.0 M

I can definitely do the math myself, I’m just not sure how to set up the equations or what steps to go through in order to find the answer…

I’m thinking hydroxides would be aqueous, but I’m not sure why I’m thinking that.

@Adirondackwannabe I’m looking at http://www.csudh.edu/oliver/chemdata/solrules.htm and the only information I’m seeing about hydroxides is that when paired with a salt, they are only slightly soluble; in the notes I took during class hydroxides weren’t even mentioned! But, for some reason, I feel like OH- should be aqueous as well.

It’s been a few years since I took AP Chem but I’ll take a shot at it:

Iron(II) is Fe2+ and hydroxide is OH-, so I think the reaction that would occur is

Fe2+ + 2OH- -> Fe(OH)2, which is, according to my google search, a precipitate called Iron(II) hydroxide. (http://en.wikipedia.org/wiki/Iron(II)_hydroxide) That’s a damn hard question.

@Mariah ohhhh, okay. I had written out the equation as Fe2+ + OH- > FeO + H. That’s probably where I went wrong then, I shouldn’t have separated the OH!

whoops, the products in that reaction shouldn’t be crossed out xD sorry!

For your second question, I think the general steps you’ll need to do are:

Use the definition of molarity to determine how much of each compound is in each solution
Determine the reaction taking place. Apparently there’ll be leftover sodium ions.
Figure out from your equation what percentage of the sodium will go into another compound and what percentage will be free ions.
Then determine the concentration of those ions in your final solution.

Let me know if you need more help. I’m a little iron-oxide at chemistry now (rusty – get it get it?) but I’ll try. Here on Fluther we’re not allowed to give you the answers to your homework but we can lead you there.

Oh yeah…and happy mole day.

bleh. So complicated! I’ve gotten the first step done I believe. We’ll see how it goes…Thanks so much, both of you, for your help :)

No prob. Welcome to Fluther! :)

Haha, it’s great to be here! Fluther is fantastic, I love the community <3

I see your questions have already been answered, but I took AP Chem just last year and loved it, so I figured I’d just offer up a couple of tips about determining reactions since you had trouble with it in your first question. I hope that it helps!

It is a general rule that compound cations and anions will not be separated in a reaction, except in very complicated reactions and specific exceptions; if you are unsure or are unaware of any exceptions regarding to that reaction, treat them as you would a single/simple cation or anion.
In cases like the reaction you gave above in which you have only unpaired anions and cations in the reaction, it is a safe bet to never separate compound cations (i.e. NH4+) or compound anions (i.e. in OH-,NO3—, CrO4–2), even though very few cations are compounds, and not all anions are compounds (i.e. H-). This is also generally true for single- and double-replacement reactions, and even in cases of oxidation-reduction, separation of the cation or anion is not common except in very complex reactions. However, as it is with most general rules, there are a few exceptions, but you most likely shouldn’t be learning about many of those until late in your class.

Also, it is very important to remember to balance the charges. It is the key to telling if a reaction is oxidation-reduction or not. In the way that you wrote the reaction, one way that you can tell that neither oxidation nor reduction could have occurred is that the H+ part of the original OH1— charge was not included in the completed reaction, which made the charges unbalanced.

Also, a purely oxidation reaction will always look something like this:
Zn ——> Zn2+ + 2e -
And a purely reduction reaction will always look something like this:
F2 + 2e−——> 2F−
These are half-reactions of oxidation-reduction reactions. If it looks like a regular reaction but oxidation or reduction occur within the reaction, it is an oxidation-reduction reaction.

It may seem confusing now, but once you get the hang of reactions, it will come naturally to you. I’m not sure if any of this is helpful to you or not, but I figured I’d give it a shot, and I hope that it will help you to be better able to determine reaction types and outcomes! Feel free to PM me if you have questions now or in the future. :)

@Fly That’s incredibly useful, thank you so much! There were quite a few tips in there that I had either forgotten or simply didn’t know; what you wrote was very, very helpful. Thank you again!

It’s precipitation because positive ions and negative ions will always precipitate into a salt, with various solubilities varying on the Ksp.

Huh, okay. I didn’t even know that, @Rarebear xD thank you!