Unit 2 Chemistry

[Sanguinne]

At school, we just completed the chapter on acids and bases but Im still confused by what I am suppose to do in an ionic equation. What am I suppose to cancel out and why do I cancel it out?


Also could someone give me the ionic equation for this:
Fe₂O₃(s) + 6HCl(aq) -----> 2FeCl₃(aq) + 3H₂O(l)
and
ZnO(s) + H₂SO₄(aq) ----> ZnSO₄(aq) + H₂O(l)

help is greatly appreciated

[Kuroyuki]

At school, we just completed the chapter on acids and bases but Im still confused by what I am suppose to do in an ionic equation. What am I suppose to cancel out and why do I cancel it out?


Also could someone give me the ionic equation for this:
Fe₂O₃(s) + 6HCl(aq) -----> 2FeCl₃(aq) + 3H₂O(l)
and
ZnO(s) + H₂SO₄(aq) ----> ZnSO₄(aq) + H₂O(l)

help is greatly appreciated

How I got taught to do ionic equations was the expand all substances that are aq into ions. Then cancel anything that appears on both sides of the equation. You should be left with the ionic equation.

[lzxnl]

Why do you cancel the ions out? Let's have a look.
If you react AgNO3 with BaCl2 in solution, the silver nitrate and barium chloride exist as solvated ions, not as ionic compounds. The only reaction that is really taking place is the precipitation of AgCl, so the only real reaction is Ag+(aq) + Cl-(aq) => AgCl(s)

Similarly in acid base reactions, if you react HCl with Na2CO3, the sodium ion doesn't do anything. Also HCl(aq) doesn't exist like that; it's actually H+(aq) and Cl-(aq), so the actual reaction taking place is H+ and CO3 2-, forming H2CO3, which decomposes to form CO2 and H2O.
However, you can only simplify things down if you have dissolved ionic substances or something is a strong acid/base. In the first example, iron(III) oxide is insoluble in water, but the HCl becomes H+ and Cl-, so the ionic equation would be Fe2O3 (s) + 6H+(aq) => 2Fe 3+(aq) + 3H2O(l)
Something similar happens with the second reaction. You can omit the sulfate in sulfuric acid.

[vox nihili]

You need to look for the change that has occurred. If an aqueous substance results, we're not interested in that substance because technically no change has occurred to form it. If there is a precipitate, write out the equation for the precipitate.

The ionic equation for both is: 2H⁺ + O^2- —> H₂O

That's my understanding of it. Otherwise, there's really nothing changing. The movement from Fe₂O₃ and 6HCl to FeCl₃  isn't really doing anything other than taking away the oxide (yes?) ions and the protons and whacking them together. You start off with Fe³⁺ ions and Cl^- ions, and end up with them at the end! Nothing's changed in regards to those ions.

[vox nihili]

You need to look for the change that has occurred. If an aqueous substance results, we're not interested in that substance because technically no change has occurred to form it. If there is a precipitate, write out the equation for the precipitate.

The ionic equation for both is: 2H⁺ + O^2- —> H₂O

That's my understanding of it. Otherwise, there's really nothing changing. The movement from Fe₂O₃ and 6HCl to FeCl₃  isn't really doing anything other than taking away the oxide (yes?) ions and the protons and whacking them together. You start off with Fe³⁺ ions and Cl^- ions, and end up with them at the end! Nothing's changed in regards to those ions.

Yeah, my bad. I thought they started off aqueous! Oops. The premise is right though. When you're doing ionic equations, look for the ions that change state, and include them. Piss off the rest

[Yacoubb]

Fe₂O₃(s) + 6HCl(aq) -----> 2FeCl₃(aq) + 3H₂O(l)
2Fe³⁺+3O^2- + 6H⁺ + 6Cl^- ----> 2Fe³⁺ 6Cl^- + 3H₂O
Ionic Equation: 3O^2-_(aq) + 6H⁺_aq) -----> 3H₂O_(l)

ZnO(s) + H₂SO₄(aq) ----> ZnSO₄(aq) + H₂O(l)
Zn²⁺ + O^2- + 2H⁺ + SO₄^2- -----> Zn²⁺ + So₄^2- + H₂O
Ionic Equation: 2H⁺_(aq) + O^2-_(aq) ---> H₂O_(l)

[vox nihili]

Don't think that's it Yacoubb. The dissociation is what you want to show.

[lzxnl]

Fe2O3(s) can NOT be written as 2Fe 3+ + 3O 2- as they are bound in an ionic lattice. The ions are not free. There are no O 2- (aq) ions; those can't actually exist as the oxide ion is a ridiculously powerful base. You dissolve sodium oxide and you won't find any oxide left anywhere.
For the first one, the ionic equation is the same as the original equation except with the removal of Cl- ions. For the second one, replace H2SO4 with 2H+ and you're done. Again, you can't write ZnO(s) as Zn 2+(aq) + O 2-(aq) because...you don't have separate zinc and oxide ions.

[Kuroyuki]

You should only expand the aq ones so in the second part.ZnO(s) + H2SO4(aq) ----> ZnSO4(aq) + H2O(l)
It will be ZnO +2H(+) + SO4(2-) -----> Zn(2+) + SO4(2-) + H2O
You can cancel the SO4
So the ionic eq will be
ZnO + 2H -----> Zn + H2O

[Yacoubb]

You should only expand the aq ones so in the second part.ZnO(s) + H2SO4(aq) ----> ZnSO4(aq) + H2O(l)
It will be ZnO +2H(+) + SO4(2-) -----> Zn(2+) + SO4(2-) + H2O
You can cancel the SO4
So the ionic eq will be
ZnO + 2H -----> Zn + H2O

My bad. I assumed all the materials were soluble. Zinc oxide is insoluble so it remains the same! Oops! Thanks for that though.

[lzxnl]

I'm going to be very nit-picky and say this.
Technically you can't actually split the sulfuric acid up into 2H+ and SO4 2- because although H2SO4 is a strong acid, HSO4- is a relatively weak acid with Ka of 0.01, i.e. incomplete dissociation. Therefore you'll have mostly HSO4-. What's really happening initially is that the H+ formed from H2SO4- reacts with the zinc oxide. As the levels of H+ get low, more of the HSO4- is able to dissociate into SO4 2- and H+ as the amount of H+ that can react with SO4 2- in a back-reaction decreases. Only now does the HSO4- actually form SO4 2-. So if the zinc oxide is in excess, you can split up the sulfuric acid like that, but if the sulfuric acid is in excess, you may end up with HSO4-. It's slightly dodgy how VCE assumes that H2SO4 is strong in both dissociation. Think of it this way. HSO4- has a negative charge. Does it REALLY want to give up another positive charge?
If you don't have enough zinc oxide, you may be left with hydrogen sulfate ions. It's like how if you burn carbon in insufficient oxygen, you may get some carbon monoxide.

[vox nihili]

My bad. I assumed all the materials were soluble. Zinc oxide is insoluble so it remains the same! Oops! Thanks for that though.

Did exactly the same thing above, so don't worry haha

[Sanguinne]

thanks for the help everyone

but is it a  requirement for me to simplify ionic equations?
example
Overall Reaction: H₂SO₄(aq) + 2KOH(aq)---> K₂SO₄(aq) + 2H₂O(l)

Ionic Equation: 2H⁺(aq) + 2OH^-(aq) ----> 2H₂O(l)
simplified into
H⁺(aq) + OH^-(aq) ----> H₂O(l)

[lzxnl]

If the question says so, yes. Your equation as written is fine (note my previous rant about sulfuric acid; if the acid is in excess your equation is in trouble), but it's like having a lot of extra information. We don't really care about the potassium and sulfate ions.