how will i balance a chemical equation?

Question

can u please tell me in the easiest way on how to balance a chemical equation.
our professor in chemistry already teach that, but still i feel difficult
can u please give me the easiest way and give at least 5 examples to balance a chemical equation?
please/

Answer 1

You don't need to keep worrying about every element while balancing. Just follow this simple tecnique:
Step 1 : Balance the metals first, i.e. make sure that the no. of metal atoms are equal.
Step 2 : Balance the non-metals except for oxygen and hydrogen.
Step 3 : Balance hydrogen and oxygen..

Examples:
1) Rusting of Fe:
Fe + O2 + x.H2O = Fe2O3.xH2O
Balancing Fe:
2Fe + O2 + x.H2O = Fe2O3.xH2O
Balancing O2:
2Fe + 3O2 + x.H2O = 2Fe2O3.xH2O ( incase when number of atoms are not all even or odd, we take their LCM, like LCM of 2 & 3 is 6, thus we make both side 6 )
Now, again balancing Fe:
4Fe + 3O2 + x.H2O = 2Fe2O3.xH2O
This is the final equation.

2) Photographic plate:
AgBr = Ag + Br2
Since metal is already balanced, we take Br:
2AgBr = 2Ag + Br2
This is the final equation.

3) Formation of NH3:
N2 + H2 = NH3
Balancing non-metal, N2:
N2 + H2 = 2NH3
Balancing H2: (here there are 6 atoms of H2 on rhs and two on lhs. thus we multiply 2 with 3 to get the result)
N2 + 3H2 = 2NH3
This is the final equation.

Sorry, don't have enough time to give other examples. Try this tecnique on any other recation and you'll find it useful

Answer 2

co2+h20 ---> c6h12o6 + o2
1st pick the biggest (plentiest) things. c6h12o6
make it to be (1)
when c6h12o6 is 1 u can make co2 to be 6 right?
then because the hydrogen (h) in rihgtside ih 12 u only have 2 in left u need to multiple it to 6 right? so we got 6h2o
so
we have
6co2+ 6 h2o ---> c6h12o6 + n o2
now find n
count o in left, there are 18 o and we got 6 o in rightside we still need 12 o. so it divide by 2 =6
so n is 6
6co2+ 6 h2o ---> c6h12o6 + 6 o2
there also another example that not this easy. but i think its enougt for now

Answer 3

need examples of problems to help.