Chemistry: Need help BALANCING EQUATIONS

I understand what you're supposed to do: the atoms of each element should be equal on the reactant and product side. I do a "guess and check" thing which seems to be the method most people use, but I need help working my way through problems as they are taking me so long to do. For example:

K2CrO4 + Al(NO3)3 --> KNO3 + Al2(CrO4)3

K2CrO4 + Al(NO3)3 --> KNO3 + Al2(CrO4)3

The problem some students have is that they try and balance EVERY ATOM individually instead of balancing groups of atoms.

In the above equation you have two main groups of atoms. You have a CrO4 group and an NO3 group. Balance those first.

It is a bit of guess and check as to which group to balance first, but if you pick the wrong one just do the other one.

I'm going to pick the CrO4 first. Can you see that you have 3 of those groups on the right hand side in the formula Al2(CrO)4)3? So put a 3 IN FRONT OF the K2CrO4

3K2CrO4 + Al(NO3)3 --> KNO3 + Al2(CrO4)3 Now the CrO4's are balanced.

Now look what you did. You now have 6 K's in the 3K2CrO4 right? Well add a 6 in front of the KNO3

3K2CrO4 + Al(NO3)3 --> 6KNO3 + Al2(CrO4)3 Now all you have to do is balance the Al's. You have 2 Al's on the right side and only one on the left so put a 2 in fron tof the Al(NO3)3

3K2CrO4 + 2Al(NO3)3 --> 6KNO3 + Al2(CrO4)3 The NO3 group is now automatically balanced.

OK? Once you get used to balancing the polyatomic ions first (the groups) you will have more success.

Sometimes you don't have any to balance thought. e.g.

Fe + O2 --------------> Fe2O3 Here the problem is the Oxygen. you have 2 O's on the left and 3 on the right. I always tackle that first. The way to balance that is to take the 3 from the right and put it in front of the O2 on the left, and the 2 from the left and put it in front of the Fe2O3 on the right You end up with: