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balancing equations
Posted: Wed Aug 20, 2008 10:57 am
by JoshPoshMosh
I need help with the following chemical equations. If someone good at chem could make sure the formulae is right and balance it then thx

copper + sulfuric acid --------> copper(II) sulfate + water + sulfur dioxide
Cu + HSO4 --------> Cu2SO3 + H2O + SO2
Sodium carbonate + hydrochloric acid --------> sodium chloride + water + carbon dioxide
NaCO3 + HCl --------> NaCl + H2O + CO2
nitric acid solution + tin (IV) oxide --------> tin nitrate + water
HNO3 + Sn2O4 --------> SnNO3 + H2O
carbon dioxide gas + lithium oxide --------> lithium carbonate
CO3 + Li2O --------> Li2CO3
ammonium carbonate + phosphoric acid ----> ammonium phosphate + carbon dioxide + water
(NH4)2CO3 + H3PO4 ----> (NH4)PO4 + CO2 + H20
Re: balancing equations
Posted: Wed Aug 20, 2008 11:03 am
by iGod
First one is copper with concentrated sulfuric acid
Cu(s) + 2H2SO4 -> CuSO4 + SO2(g) + 2H20
I'll post the rest in a few
actually nevermind i hate chemistry.
Re: balancing equations
Posted: Wed Aug 20, 2008 11:04 am
by takolin
JoshPoshMosh wrote:I need help with the following chemical equations. If someone good at chem could make sure the formulae is right and balance it then thx

copper + sulfuric acid --------> copper(II) sulfate + water + sulfur dioxide
Cu + HSO4 --------> Cu2SO3 + H2O + SO2
Sodium carbonate + hydrochloric acid --------> sodium chloride + water + carbon dioxide
NaCO3 + HCl --------> NaCl + H2O + CO2
nitric acid solution + tin (IV) oxide --------> tin nitrate + water
HNO3 + Sn2O4 --------> SnNO3 + H2O
carbon dioxide gas + lithium oxide --------> lithium carbonate
CO3 + Li2O --------> Li2CO3
ammonium carbonate + phosphoric acid ----> ammonium phosphate + carbon dioxide + water
(NH4)2CO3 + H3PO4 ----> (NH4)PO4 + CO2 + H20
Sulphuric acid: H2SO4
Copper sulphate: CuSO4
Sodium carbonate: Na2CO3
Tin(IV)oxide: SnO2
Tin nitrate: Sn(NO3)4
Carbon dioxide gas: CO2
Ammonium phosphate: (NH4)3PO4
Staring with the correct molecule tends to help.
Last part isn't hard any more.
Re: balancing equations
Posted: Wed Aug 20, 2008 3:54 pm
by crazyskwrls
JoshPoshMosh wrote:
Sodium carbonate + hydrochloric acid --------> sodium chloride + water + carbon dioxide
Na2CO3 + 2HCl --------> 2 NaCl + H2O + CO2
nitric acid solution + tin (IV) oxide --------> tin nitrate + water
4HNO3 + SnO2 --------> Sn(NO3)4 + 2H2O
carbon dioxide gas + lithium oxide --------> lithium carbonate
CO2 + Li2O --------> Li2CO3
ammonium carbonate + phosphoric acid ----> ammonium phosphate + carbon dioxide + water
3(NH4)2CO3 + 2H3PO4 ---->
2(NH4)3PO4 + 3CO2 + 3H20
NAO pay me baitch
Re: balancing equations
Posted: Wed Aug 20, 2008 4:26 pm
by Asteroid
how do you guys do this. i got chem hnrs and maybe ap next year and didnt get that at all
Re: balancing equations
Posted: Wed Aug 20, 2008 4:41 pm
by iGod
I found the first one in my 3rd year textbook.
Re: balancing equations
Posted: Wed Aug 20, 2008 5:47 pm
by crazyskwrls
thats wat u get for spending a year in ap chem with a depressed teacher

Re: balancing equations
Posted: Wed Aug 20, 2008 5:50 pm
by Asteroid
crazyskwrls wrote:thats wat u get for spending a year in ap chem with a depressed teacher

my sister had a guy named mr abramz, everyone said he was a retard.
EVERYONE
Re: balancing equations
Posted: Wed Aug 20, 2008 6:54 pm
by F-22
get the elements table find the + and - for each compound then add them up. chemistry is a fun class.
Re: balancing equations
Posted: Wed Aug 20, 2008 7:05 pm
by takolin
Asteroid wrote:how do you guys do this. i got chem hnrs and maybe ap next year and didnt get that at all
Step1: Making sure your molecules have the correct composition.
EG If you place a single charged kation (+charge) with double negative anion (-charge) you should realise that you need 2kations to balance the molecule.
EG:
Sodium: Na. It'll become Na+
Oxygen: O > O2-.
Thus sodiumoxide will be Na2O.
1st line: +1
2nd line: +2
3rd line: +3
5th line: -3
6th line: -2
7th line: -1
the others need to be learned by hearth because it's slight more complicated.
EG Iron: Fe can be Fe2+ or Fe 3+.
Step 2: writing the equation down.
Sodium carbonate + hydrochloric acid --------> sodium chloride + water + carbon dioxide
That gives:
Na2CO3 + HCl => NaCl + H2O + CO2
Step 3: Puzzling with the help of a pencil.
You see 2 sodium left and 1 right. Thus you write a 2 in front of NaCl.
Thus you'll see 2 chlorides right, leading to a adding a 2 in front of HCl.
This gives 2 hydrogens left and you see 2 on the right.
And because there are 3oxygens and 1 crabon left and right, you know that the equation fits.
This works for all substitution and acid/base reactions.
Reactions involving oxidations and reductions are solved somewhat different but I'm sure they'll explain that later.
Re: balancing equations
Posted: Wed Aug 20, 2008 7:38 pm
by Dystopia
Balancing is easy, you first might want to turn it into a molecular formula instead of a word, so its easier.
Then note down each ELEMENT in the compound.*If there is a polyatomic compound note it only IF the same polyatomic compound is on the other side, if not write the elements.
For each side, the reactants and the products count how many elements or on each side and write it side by side. Then say there is Cu on one side a Cu2Cl on the other side, that means the first Cu on the Reactants side (left side, before the arrow) also needs to be to so you put a 2 before the Cu so its 2Cu + blablahblah --> Cu2Cl + blablablah
Like that you just balance one element, you many need to change the values because when you put a number before the compound it effects the whole thing like 2Cu2Cl, here you have 4 Cu and 2 Cl, you see?
I'll give you one of my own questions because the others don't seem right.
2Fe (s) + 3H2SO3 (aq) --> 3H2 (g) + Fe2(SO3)3
So make a little table.
R P
Fe 1 2
H 2 2
SO3 1 3
So, you can see that SO3 on the Products side doesn't match the the reactants side, so put a three behind the whole compound [H2SO3] but this makes Hydrogen 6, so on the products side put a 3 behind the H2 (3*2), and now all is left is the Fe, its two on the products side so simple put a two behind it on the reactants side
Easy enough?