"write net ionic equations for the following reactions:
a. the ionization of HClO3 in water
HClO3 + H2O yields H3O + ClO3
net ionic = H + H2O yields H3O"
Well, you're on the right track; you have chloric acid dissociating fully in water, as you should, because it is a strong acid; good job! Now, the thing to be careful of is to make sure your charges are on your cations and anions. I don't know if you forgot to include them or omitted them because you were unfamiliar with the format of the forums, but you can add both superscripts and subscripts by using the buttons under UBBC tags.
Now, for the full equation, with the charges written in, it's almost as you have it above:
HClO3 + H2O yields H3O+ + ClO3-
An interesting point of note, and something that I'm sure is well beyond what your teacher cares about, chloric acid is fairly unstable and only stable in a certain concentration range; if it's too concentrated, it destabilizes and decomposes.
To writ the net ionic equation, you have to cancel out any ions that are spectators; personally, I think the choice of the above equation is a poor one, as the chlorate anion is not dissociated until in water, but that's another matter, I guess. If your teacher assumes that you are starting with anhydrous material (which is technically impossible because of the instability that I mentioned), the net ionic would actually be just the full equation, because the chloric acid isn't dissociated yet. However, I do not know what your teacher is telling you, whether to assume that the chloric acid is dissociated already; if that's the case, then you cancel out the chlorate anions on both sides of the equation and are left with something similar to what you have:
H+ + H2O --> H3O+
> b. NH3 functioning as an Arrhenius base
You have again correctly used the definition of an Arrhenius base (something with an OH- group), and so your equation, aside from not having charges, is correct.
The net ionic with charges is:
NH3 + H2O <--> NH4+ + OH-
> c. CaCO3 + 2HCl yields CaCl2 + H2O + CO2
This is calcium carbonate (chalk, eggshells, etc.) being added to hydrochloric acid. In this case, because of solubility rules (and general knowledge), you know that calcium carbonate is insoluble in water, so there's no chance of it being ionized. Calcium chloride, one of the products, is soluble in water, though, so you wind up with Ca+2 and 2 Cl2-. The HCl, presumably an aqueous solution, is written as H+ + Cl-, so your full equation looks like:
CaCO3 + 2H+ + 2Cl- --> Ca+2 + 2Cl- + H2O + CO2
To get your net ionic equation, you just cross out your spectator ions. In this case, the spectator ions (the ones that are on both sides of the equation, not participating in the reaction), are comprised only of the chloride ion. So, you cross it out, and you're left with the net ionic equation:
CaCO3 + 2H+ --> Ca+2 + H2O + CO2
Now, if you wanted to get fancy, you can add the intermediate, carbonic acid. See, when you add acid to a carbonate, it doesn't go directly to carbon dioxide and water; you first produce carbonic acid, H2CO3. Carbonic acid is an unstable acid, though, so it undergoes this reaction spontaneously:
H2CO3 --> H2O + CO2
In your answer, you have calcium and oxygen appearing alone; the calcium ion can appear alone, but O-2 is not a species that is common to see. Matter of fact, I don't know of any reaction that uses it free, much less within the scope of general chemistry. Make sure if you've got oxygen that it's bound to a polyatomic ion or another atom.
> d. how many liters of CO2 form at STP if 5.0g of CaCO3 are treated with excess hydrochloric acid?
Well, now that you have the net ionic equation:
CaCO3 + 2H[sup+[/sup] --> Ca+2 + H2O + CO2
you can calculate the number of liters of carbon dioxide that are formed. You can accomplish this because you know that one mol of CO/sub] is formed for every mol of CaCO3 that is consumed. So if you can find out the number of mols of calcium carbonate originally used. In fact, you have the number of grams of calcium carbonate, so you can find the number of mols of it that you have. 5.0g of calcium carbonate divided by the molecular weight will give you the number of mols. That number of mols of calcium carbonate that is used is the same number of mols of CO2 that you have. Now you use the ideal gas law:
P is pressure, in atmospheres
V is volume, in liters
n is the number of mols of gas
R is the ideal gas constant, 0.08206 L*atm/mol*K
T is the temperature, in Kelvin
STP indicates that P = 1 L and T = 298 K, so those are filled in. R is a constant, and you just found n, the number of mols of carbon dioxide that are formed during this reaction. Now the only unknown that you have is V, the volume, which is what you're looking for in the first place.
Hope this helps. Good luck!
BTW, are you in general chemistry or AP? Just curious.