Show us how far you've gotten and we can help you get through the parts you're stuck on.  If we just did it for you, you'd never learn how to do it on your own.  So just post how far you've gotten and what part of the balancing is getting you stuck.  

Well, remember, you can always start out by balancing it with fractions as coefficients.  Then you can just use some simple mathematics to eliminate the fractions and get a whole number.  It's also important to try and see which is the 'key' compound in the reaction.  Every reaction has one compound which really controls the progress of the balancing.  In this case, it would be KClO3 since it contains K, Cl, and O in its formula.  It's only missing the H which is easy to balance since H is only present in the water on the right side.  So you'd start out by eliminating pretty much everything but the KClO3.  So I would start like this:

KClO3 -> ClO2 + KCl

With this step the potassium is balanced, but the chlorine is one too many on the right and the oxygen is one two few.  So to correct this we will bring in the HCl portion and the water.  This gives:

KClO3 + HCl -> KCl + H2O + ClO2

Now the potassium is balanced, the oxygen is balanced, and the chlorine is balanced.  But the hydrogen is short one on the left side.  So we'll add another molecule of HCl to give us:

KClO3 + 2HCl -> KCl + H2O + ClO2

Now the potassium is balanced, the hydrogen is balanced, and the oxygen is balanced.  Sadly, the chlorine is off with 3 on the left side and only two on the right.  So now we'll bring in our final part of this equation which is the chlorine gas.  This can be balanced by using a 1/2 coefficient which gives us:

KClO3 + 2HCl -> KCl + H2O + ClO2 + (1/2)Cl2

Now everything is balanced.  So we just need to multiply every coefficient by two to get a balanced, whole number coefficient for each substance.

2KClO3 + 4HCl --> 2KCl + 2H2O + 2ClO2 + Cl2

It is now balanced.  

Jdurg,

Thanks so much for the help. I never thought of using a 1/2 coef. and then just multiplying up. That will help alot with these harder ones, well, hard in my eyes.

Thank you again!