# chemical equations

I need help figuring out how to write these chemical reactions into chemical equations
Copper(II)oxide + hydrochloric acid -> copper(II)chloride + water

CuO2+ HCl -> CuCl + H2O?

Aluminum + copper(II) chloride-> copper + aluminum chloride
Al+CuCl -> Cu + AlCl

Aluminum + hydrochloric acid -> hydrogen + aluminum chloride
Al+HCl -> H + AlCl

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1. I've helped you with one; now I see a bunch more of the same kind. What's the problem? What do you not understand about all of this. Do you the valences? Do you know how to guesstimate what the valences are? Do you know the polyatomic ions and their valaences?

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posted by DrBob222
2. The roman numeral confuses me and I don't know what some of the formulas are-once I have the equaiton I can balance it. So I guess my problem is getting to the equation

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3. The Roman numeral TELLS you the valence of the ion. Cu(I) means Cu^+1. Cu(II) means Cu^2+.
Fe(III) tells you Fe^3+. Couldn't be easier than that.

If it has no Roman numeral, look for it on the periodic table. If it is group I (or 1) (the left most column) it is +1. If in group II (or 2) it is +2. If in group 3 (or 13) it is +3. If in group 15 usually -3, if 16 it is -2, if in group 17 is is -1. Those are the simple ones. Practice makes all of this easy and once you've used them several times you begin to see the pattern and to get them memorized. For polyatomic ions here is a simple table (the table also contains many of the ions you get from the discussion above about the groups).
http://www.sciencegeek.net/APchemistry/APpdfs/CommonIons.pdf
Take this chart (I suggest printing it out) and see how it fits with the next post I do which will be to balance one of the equations you posted..

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posted by DrBob222
4. Aluminum + copper(II) chloride-> copper + aluminum chloride

Al of course is aluminum. For coper(II) chloride it tells you the valence of Cu is +2. Chloride from the table I gave you is -1. All compounds are zero so you must make the + charges and the - charges add up to zero. So with Cu as +2 and Cl as -1, you must have CuCl2 [+2 + (2*-1) = 0].
So I write Al + CuCl2 --> AlCl3 + Cu
How did I know it was AlCl3. You look on the periodic table, Al is in 13 so it is +3 and Cl is -1 so you must have 3 of those -1 charges to equal -3 so it will add to the _3 to make zero.
Then balance. The secret here is to practice until these numbers get imbedded in yo
Hope this helps. Post any other follow ups here.

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posted by DrBob222

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