How to balance equations?

Answer 1

That depends on the kind of equation you have in mind...

Many can be balanced by INSPECTION or by trial and error.
Others involving a transfer of electrons require a more sophisticated process called the half-reaction method.

Must know more from you first.

Answer 2

this is in terms of just unbalanced equations as a general ok? i won't take into account redox reaction...

for most equations, it can simply be balanced just by inspection, meaning all you have to do is look at the equation and do some mental addition and multiplication to balance it out... once you're used to balancing, doing this would come naturally but in the meantime, the best way to go is always to list the individual elements under the equation so you can monitor the number of electrons

ok... let's start with a simple equation:

KClO3 -> KCL + O2

use the listing technique i told you about earlier so you can easily account for and monitor the number of electrons... you can also do trial and error at this list

so it'd be:
product (KCLO3)
K=1
Cl=1
O=3

reactant(KCl+O2)
K=1
Cl=1
O=2

notice that the main imbalance is with the O2, so what you do is add a coefficient or a number before the whole element/compound on bothe sides to try to even it out... the skill for common multiples is very handy here

since the O in the product side has 3 electrons, and the O in the reactant side has two, it'd make sense to add a two before KCLO3 and a 3 before the O2 so that when you multiply the coefficient with the subscripts of O, you'd get a total number of 6 for O on both sides..

do you understand so far?

moving on, observe how while the O is balanced, the K and CL in the 2KCLO3 is now imbalanced... again, to balance it out, just add a coefficient to the other side of the equation, in this case before the KCL...

since the K and Cl in the products are now greater by 2, just add a 2 on the KCl in the reactants side to even the values out...

in the end, you should get 2KCLO3 -> 2KCl + 3O2

just some last reminders:
given that you have a compound with a parenthesis, for example: Fe(OH)2

remember that the subscript after the parenthesis is for all the elements in the parenthesis... so if you're going to place a coefficient before the compound, remember that you have to multiply the coefficient with the subscript to get the correct value of the elements within the parenthesis...

ex. you're gonna add a coefficient of 2 before Fe(OH)2, while the value of Fe will become 2, the values of O and H would be 4 (coeffient 2 x subscript 2 =4)

another issue is if an element appears twice in one side:

ex.CO2 + H2O -> H2CO3

(its already a balanced eq but i couldn't think of a simpler eq at the moment that has an element appearing twice in one side)

O appears twice in the reactant side... the listing method is handy for these type of equations because given this situation, to get the value of the repeated element, you ahve to add their final values...

ex. you add a 2 in CO2, making the O=4, then you add a 5 to H2O making the O=5, for you to get the correct value of the O in the reactant side, add the value of O in CO2 and in H2O...

in this case, the total value of O then for the reactants would be 9... since in H2CO3, O has a coeffient of 3, you just have to multiply it with a coefficient that would make the value of O=9

i don't really know if you got that coz it's really hard to teach this without a first hand demo... just in case there're some confusions and/or questions, feel free to email/contact me...

i hope this helps... good luck!

(sorry that it's so long... :D)