Hey Jamesey!
Not sure so much about resources specifically on equation writing, but I'll give you a couple of pointers that hopefully may help you:
Firstly - try not to rote-learn rules such as "acid + metal carbonate --> salt + carbon dioxide + water"; rote-learning these will make it more difficult to learn and also you may misapply the rule in writing your equations. I see this in my students all the time who rote-learn formulae and when trying to solve a problem, "pick formulae off a shelf" and try apply them and see where they go, which often leaves them stuck.
Many chemical reactions are one of the following:
- acid-base (where there is a proton transfer from one species to another species)
- redox (where there is an electron transfer between species); combustion reactions in oxygen are a particular type of redox reaction
- precipitation reactions (where two aqueous species 'combine' to form a solid product)
Acid-BaseThe trick here is to actually see the proton transfer.
For example - say you want to write an equation of a reaction between MgO and HCl.
The first thing you need to pick out is that MgO is an ionic solid that contains the Mg
2+ and O
2- ion, which is a pretty damn strong base; you can 'stick' two protons (H
+) onto this ion to form H
2O.
The second thing is that HCl is an acid, and can donate its proton to MgO (the base).
Now, what happens when HCl donates a proton? It becomes Cl-.
What happens when O
2- accepts protons? It becomes H
2O.
In the meantime, the Mg
2+ is liberated into solution with the Cl- ions; MgCl
2 is soluble in water.
So we are left with MgO + HCl --> MgCl2 + H2O. No formulae needed here!
To balance the equation - remember that you need 2 H
+ ions to protonate the O
2- all the way to water.
So the equation would be:
MgO + 2 HCl --> MgCl2 + H2O
States - most oxides are solid (including MgO), HCl is presumably in solution, as is MgCl2, and water is obviously a liquid.
MgO (s) + 2 HCl (aq) --> MgCl2 (aq) + H2O (l).
This seems a little long, but it's so intuitive that you'll be able to go through this thought process quickly.
Another example: HCl and CaCO
3.
HCl is again the proton donor. CaCO3 is an ionic solid with Ca
2+ and CO
32- ions, the latter being a base. You can stick two protons onto it to form H
2CO
3.
Now, if you remember Coca Cola (which contains H
2CO
3), once you open the top it fizzes; that is because H
2CO
3 is pretty unstable and will just dissociate into H
2O and CO
2.
When HCl gives off a proton, it becomes Cl-.
When CO
32- takes up two protons, it becomes H
2CO
3 which just decomposes to H
2O and CO
2.
Meanwhile, the Ca
2+ ion just wanders off and does its own thing, joining Cl
- in swimming around solution. You basically have a CaCl
2 salt solution.
Your equation would just be then HCl + CaCO
3 --> CaCl
2 + CO
2 + H
2O.
Redoxhttp://www.chemguide.co.uk/inorganic/redox/equations.htmlChemguide.co.uk is the BIBLE for Year 12 Chemistry in general. I used it religiously when I was doing Chemistry at school!
Did you want me to take you through combustion reactions of hydrocarbons specifically?
PrecipitationThis is basically two ions combining to form a solid. The best thing here is to know your solubility table here.
For example if you mixed BaCl
2 and Na
2SO
4, you look at the ions and you should recognise that Ba
2+ and SO
42- will not coexist in solution; they will precipitate out. So instantly you have the ionic equation -
Ba
2+ (aq) + SO
42- (aq) --> BaSO
4 (s)
If you want the overall equation, then slot in the spectator ions where appropriate, and note that they 'associate' together in aqueous solution after the reaction - as NaCl (aq) on this occasion:
BaCl
2 (aq) + Na
2SO
4 (aq) --> BaSO
4 (s) + 2 NaCl (aq)
Hope that helped somewhat!