hi!i am a f4 student in hk.i have know the basic concept of balance equation.but some too complex equation i cant balance.is there any general short cut for balancing any simple or complex chemical equation?

Answer 1

just look for least common multiples of the atom counts

For example, if you want to make a hydrogen balloon, a good reaction is lye and aluminum (try it!)

Al + NaOH = Al(NaO)₃ + H₂

The problem is that the right side needs
H₂
(NaO)₃

Since you're mixing 2's and 3's, you will want to make multiples so that things get used up in 6's

2Al + 3NaOH = 2Al(NaO)₃ + 3H₂

With more complex reactions, things do get more involved, but usually the numbers are relatively small, even with things like burning gasoline (octane):

2C₈H₁₈(g)+25O₂(g) = 16CO₂(g)+18H₂O(g)

Answer 2

so what side should i look for the LCM?

Answer 3

how about HNO3 +Cu ------->Cu(NO3)2+NO+H2O,PLEASE SHOW ME STEPS.

Answer 4

Note that Cu(NO3)2 requires 2 NO3 groups. So, the first step would be to make

2HNO3 +Cu ------->Cu(NO3)2+NO+H2O

Now you have the (NO3)2 taken care of.

On the right we still have NO+H2O = H2NO2, so we need another H2NO3 on the left:

3HNO3 +Cu ------->Cu(NO3)2+NO+H2O

Now we are left with an extra O on the left. So, for every 3 NO3 on the left we get 8 O on the right. Sounds like 24 is a good LCM

8 NO3 + Cu ----> Cu(NO3)2 + NO + 4H2O
We've used up all the H, but now need to fill in for the NO3 stuff.

3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O

That was pretty trial-and-error. Maybe a web search on balancing reaction equations or stoichiometry will yields a more methodical method.