balancing equations...dont understand

[Loz_metalhead]

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My teacher is really bad...I just dont no how you do it, please help. Im in year 11.


does it have anything to do with the valencies...I think it would. Here is a question and the answer...can you tell me how it is worked out. Thanks

Al+0(2)=Al(2)O(3)

then once you balance it its...4Al(2) + 3O(2)---->2Al(2)o(3)

[Juddy2250]

Hey there, Im in year 12 and i do Chemistry. Yes it does have to do with valencies.

Al+0(2)=Al(2)O(3)

then once you balance it its...4Al(2) + 3O(2)---->2Al(2)o(3

Oxygen has a valency of 2-, Aluminium has a valency of 3+, so to balance the valencies ie have no plus or minus' you need 2Al & 3O (2x3+ + 3x2- =0) This means on the right hand side you have
Al(2)O(3).
The next part is you trying to get the whole equation balanced. By juggling the numbers you will get the answer provided. It is something you just need to practice. What your trying to do is get the same number of each type of element on both sides of the arrow.

Good luc and contact me if you need anymore help

[theChemCoach]

Originally Posted by Loz_metalhead

My teacher is really bad...I just dont no how you do it, please help. Im in year 11.


does it have anything to do with the valencies...I think it would. Here is a question and the answer...can you tell me how it is worked out. Thanks

Al+0(2)=Al(2)O(3)

then once you balance it its...4Al(2) + 3O(2)---->2Al(2)o(3)

Don't worry, this has nothing to do with valencies.

Firstly, your reaction is
Al(s) + O₂ ---> Al₂O₃(s)

• Now, we have in science this law, called the Conservation of Mass, that states that matter (elements) cannot be created or destroyed

• applied to a chemical reaction... all of the matter that appears in the reactants, MUST appear in the products.

• put even simply, all the atoms that appear on the left hand side (LHS) of the reaction (reactants) must equal all the atoms that appear on the right hand side (RHS) of the reaction

• to do this we BALANCE THE REACTION, by placing numbers (coeffecients) infront of the elements and compounds

• so in our example:
  Al(s) + O₂ ---> Al₂O₃(s)
  on the LHS we have one atom of aluminium and 2 atoms of oxygen,
  and on the RHS we have two atoms of aluminium and 3 atoms of oxygen.
  because they have different amounts on both sides, we have to 'balance' both side, so they have equal atoms of both

STEP ONE:

• place a 2 infront of aluminium on LHS (so now both sides have 2 atoms of aluminium)

2Al(s) + O₂ ---> Al₂O₃(s)
STEP TWO

• place a 1.5 infront of oxygen gas on right hand side, so mathematically we have (1.5x2) 3 atoms of oxygen gas on LHS (and 3 on RHS)
  2Al(s) + 3/2O₂ ---> Al₂O₃(s)
  
  .....BUT...if we double all the coefficients (i.e. the numbers infront of the atoms and compounds, we can get rid of that fraction, and have a whole number instead
  4Al(s) + 3O₂ ---> 2Al₂O₃(s)
  and there you have it, a balanced chemical reaction.

• notice that on the LHS we have 4 Al atoms and 6 oxygen atoms, and on RHS we have 4 Al atoms and 6 oxygen atoms....hence all the matter that appears in the reactants...appears in the products.

Also note, how i balanced the metal (Al) first, then a balanced the non-metal (O) second, you should always follow this rule.


Best of Luck,
George

[petraaa]

i dont no if this is the right spot to post this....its my first!

ok, im in year twelve and STILL haven't been able to get a handle upon doing ionic equations.

is there a simple set of rules or guidelines that i can follow in order to finally understand them?

[Omium]

Way to revive a 3 year old thread .

Ionic Equations just involve breaking up the reaction into its ions.

Give me some examples and ill explain em to ya.