I am having a hard time balancing equations in chemistry.Any techniques to do it ?

I know you don't just want the answers, but if you post a few you are having trouble with i can show you how i came about figuring out what to do first. I dont really know a standard procedure.

Added.
Well basically, the number in front gets distributed through the elements. So lets say you have 6H2O. You will get 12 H's and 6 O because you multiply 6*2 for the H's and 6*1 for the oxygens. You then compare the number of H's and O's on the left side of the equation and put constants like the 6 i have above in front of molecules/elements to balance out the numbers of elements on both sides of the equations. I can find some examples for you if you want.

So H2 + O2 yields H2O.

To balance this you would first note the number of H's and O's on both sides. Left side 2 H's and 2 O's. Right side 2 H's and 1 O. (you dont actually have to write this down, just the thinking process).

To balance it you first notice you could put a 2 infront of H2O to get 2 O's to equal the left side.

H2 + O2=2H2O


But then this gives you 4 H's. Since the H's and O's are separated on the left side you can fix that by placing a 2 in front of the H2 to get 4 H's.

2H2 + O2 yields 2H2O

Have a look at this link....
http://www.rockwood.k12.mo.us/eurekahs/academics/mcilwee/ch9notes.pdf

A huge tip before starting is use pencil! It's a trial and error kind of approach so don't be suprised if you have to attempt more than once.
First you want to start off with metals, then work your way through.
eg. Cr(ClO4)3 + K2CO3 > KClO4 + Cr2(CO3)3
First you want to balance Cr, then ClO4(treat it as a whole), then K, then CO3(also treat it as a whole)
Keep a tally on each side of the equation of each 4 species.
First move: since there are 2Cr on the right, put the coefficient of 2 infront of Cr(ClO4)3 on the left.
Second move: to balance the ClO4, there is now 6 on the left side. To get six on the right side, put the coefficient 6 infront of KClO4.
Third move: now there is 6 K on the right side, and 2K on the left side, to balance we put a 3 infront of K2CO3.
Fourth move: now there is 3 CO3 on the left side and already 3 on the right side. The equation is balanced
2Cr(ClO4)3 + 3K2CO3 > 6KClO4 + Cr2(CO3)3

The site is really good too, explains it in more detail than i just did and even has practise at the bottom