Understanding reaction products

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Homework Statement

KHSO4 + KOH→
SO4 + H2O→
P2O5 + H2O→
BaO + H2O→
CrO + SO2→
C6H12O6 + O2 (excess)→
CH4 + O2 (limited)→
NH3 + CO2 + H2O →
NH3 + HCl→

Homework Equations

Chemical equations

The Attempt at a Solution

KHSO4 + KOH→ KOH + KHSO4 Does this stay the same? since the first ion for both molecules are K+?
SO4 + H2O→ HSO3 + O2 This seems correct but just making sure.
P2O5 + H2O→ P(OH)5 + H2O This seems entirely incorrect and I'm not sure how to do this. I think the H2O splits to H+ and OH- but then the O from P2O5 would seem strange to combine with OH-.
BaO + H2O→ Ba(OH)2 What does BaO and H2O produce when reacted?
CrO + SO2→ CrO + SO2 Both seem to have O for the second ion.
C6H12O6 + 9O2 (excess)→ 6CO2 + 6H2O If this is correct, then good. Aside from that, what does excess mean?
CH4 + 3O2 (limited)→ CO2 + 2H2O
NH3 + CO2 + H2O → NH3 + HCO3 I really doubt this is correct but I don't understand how to do three or more reactants.
NH3 + HCl→ NCl + 2H2 This seems strange. Also is does the hydrogen in NH3 have a charge of 1-?

To sum it up, I'm not clear on how to determine the products from a given reaction, or understand the kind of reaction it is (synthesis, decomposition, displacements...). What is the way to understand which does what? Unless I am over thinking it.

Another part is what products are formed if the first ions or second ions of compounds, for double displacement/replacement, are same?

Edit: The purpose of these equations were for balancing the net ionic equations. Do I have to do acid-base reactions for it too?

 

[Borek]

KHSO4 + KOH→ KOH + KHSO4 Does this stay the same? since the first ion for both molecules are K+?

acd/base

SO4 + H2O→ HSO3 + O2 This seems correct but just making sure.

Are you sure it was SO₄ on the left? Not SO₃ or SO₂ or SO₄^2-?

P2O5 + H2O→ P(OH)5 + H2O This seems entirely incorrect and I'm not sure how to do this. I think the H2O splits to H+ and OH- but then the O from P2O5 would seem strange to combine with OH-.
BaO + H2O→ Ba(OH)2 What does BaO and H2O produce when reacted?

To some extent these questions are similar. This is about metal oxides and nonmetal oxides reacting with water. There are some simplified rules that help predict products.

CrO + SO2→ CrO + SO2 Both seem to have O for the second ion.

Again, simple rule that states what happens when metal oxide reacts with nonmetal oxide.

C6H12O6 + 9O2 (excess)→ 6CO2 + 6H2O If this is correct, then good. Aside from that, what does excess mean?

Have you checked if it is balanced?

Excess means in this case that there is enough oxygen to fully oxidize carbon.

CH4 + 3O2 (limited)→ CO2 + 2H2O

And there is not enough oxygen here, so one of the products must be different.

Other two ignored, enough hints for one post.

Edit: The purpose of these equations were for balancing the net ionic equations.

So why there are no ions in your approach?

 

[312213]

KHSO4 + KOH → K2SO4 + H2O
2K+ + HSO4- + OH- → 2K+ + SO4^2- + H2O
HSO4- + OH- → SO4^2- + H2O
When acid and base react, salt and water is formed. When putting in net ionic equation, the first three molecules are aqueous, so they split into ions, and the H2O is liquid, so it stays together. The 2K+ are canceled from both sides leaving HSO4- + OH- → SO4^2- + H2O. Is this done correctly for net ionic equation?

For balancing net ionic equations, do I split up the soluble molecules into ions if they aren't double replacement/displacement?

SO4 + H2O →
Yes, I am sure that it is SO4. I wasn't sure about this either so I asked to make sure. What I was thinking was that SO4 cannot exist as a molecule and instead is supposed to be SO4^2-. Assumed that the SO4 is instead SO3, would the reaction be:
SO3 + H2O → H2SO4
SO3 + H2O → 2H+ + SO4^2-
Would this be the correct net ionic equation, since the first two are liquid and since H2SO4 is a strong acid, it splits up into ions? Nonmetal oxide with water produces an acid.

P2O5 + H2O → H3PO4
P2O5 + H2O → 3H+ + PO4^3-
For the most part, same as above reasoning.

BaO + H2O → Ba(OH)2
Ba2+ + O2- + H2O → Ba2+ + 2OH-
O2- + H2O → 2OH-
Metal oxide with water produces a base. BaO and Ba(OH)2 are aqueous so they split up and the Ba2+ cancel leaving O2- + H2O → 2OH-. Is this correct?

CrO + SO2 → CrSO3
Metal oxide with nonmetal oxide produces an oxysalt.

C6H12O6 + 6O2 (excess)→ 6CO2 + 6H2O
I forgot the oxygen from the C6H12O6.

CH4 + O2 (limited)→ 1/3CO2 + 2/3H2O
I am still not sure how to write limited in the equation. Would the limited mean I have only one mole of oxygen? Which is a third of the required oxygen? Which reduces the products by a third? And does the limited make CH4 have a 1/3 coefficient because only a third of a mole of CH4 is required with one mole of oxygen?

 

[Borek]

HSO4- + OH- → SO4^2- + H2O. Is this done correctly for net ionic equation?

Yes.

For balancing net ionic equations, do I split up the soluble molecules into ions if they aren't double replacement/displacement?

Kidn of reaction doesn't matter, if it is net ionic, you split it into ions.

What I was thinking was that SO4 cannot exist as a molecule

That's why it doesn't make sense.

SO3 + H2O → H2SO4
SO3 + H2O → 2H+ + SO4^2-

First one is correct, as for the second one - H₂SO₄ is a strong acid, HSO₄^- is much weaker, so - especially in more concentrated solution - sulfuric acid is not fully dissociated. That's why it is hard to say if your reaction is correct or not.

Nonmetal oxide with water produces an acid.

P2O5 + H2O → H3PO4
P2O5 + H2O → 3H+ + PO4^3-

First OK (balancing won't hurt). Second - phosphoric acid is generally much weaker than othe rmineral acids.

[BaO + H2O → Ba(OH)2

OK

Ba2+ + O2- + H2O → Ba2+ + 2OH-
O2- + H2O → 2OH-
Metal oxide with water produces a base. BaO and Ba(OH)2 are aqueous so they split up and the Ba2+ cancel leaving O2- + H2O → 2OH-. Is this correct?

No such thing as aqueous BaO, it immediately reacts with water - see the reaction above.

CH4 + O2 (limited)→ 1/3CO2 + 2/3H2O
I am still not sure how to write limited in the equation. Would the limited mean I have only one mole of oxygen?

Please reread what I wrote earlier.