How do you balance these equations
2nan3(s)-> 2na(s)+3n2(g)

The equation is already balanced in the question, but to reach this we must first assume that all chemicals are in a 1:1:1 molar ratio:
NaN3(s) --> Na(s) + N2(g)
Then we count the number of each atom on each side of the reaction:
Atom                     Left hand side                    Right hand side
  Na                                 1                                           1
   N                                  3                                           2
Clearly this is not balanced in terms of N, so to balance this we need to make it so the number of N atoms is the lowest common multiple of 2 and 3, which is 6. Therefore, we use 2 moles of NaN3 and 3 moles of N2 to get 6 N atoms on each side:
2NaN3(s) --> Na(s) + 3N2(g)
Counting atoms:
Atom                     Left hand side                    Right hand side
  Na                                 2                                           2
   N                                  6                                           6
Therefore this balances for all atoms and is the correct balanced equation.
Hope this helps!

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