This is a little tricky to balance since your left hand side is a rather large molecule and placing a coefficient before it would affect all those atoms.
First step is to tally all the atoms involved.
#C_3H_5(NO_3)_3# #rarr# #CO_2# + #H_2O# + #N_2# + #O_2# (unbalanced)
based on the subscipts, we have
left side:
#C# = 3
#H# = 5
#N# = 3
#O# = 9
Notice that since the #NO_3# was enclosed by a parenthesis followed by the subscript 3, I have to multiply the number of atoms inside the parenthesis by 3. (e.g. #N_"1 x 3"# ; #O_"3 x 3"#)
right side:
#C# = 1
#H# = 2
#N# = 2
#O# = 2 + 1 + 2 (do not add this up yet)
Since the molecules on the right side are far more simpler than the molecule on the left side, I'm going to pick an element from the right side that I think would be the key to balance this whole equation.
#C_3H_5(NO_3)_3# #rarr# #CO_2# + #color (blue) 5H_2O# + #N_2# + #O_2#
left side:
#C# = 3
#H# = 5
#N# = 3
#O# = 9
right side:
#C# = 1
#H# = 2 x #color (blue) 5# = 10
#N# = 2
#O# = 2 + (1 x #color (blue) 5#) + 2
Notice that since #H_2O# is a substance, I would also need to multiply the #O# by 5. Now that there are 10 atoms of #H# on the right side, I would need to also have 10 atoms of #H# on the left side.
#color (red) 2C_3H_5(NO_3)_3# #rarr# #CO_2# + #5H_2O# + #N_2# + #O_2#
left side:
#C# = 3 x #color (red) 2# = 6
#H# = 5 x #color (red) 2# = 10
#N# = 3 x #color (red) 2# = 6
#O# = 9 x #color (red) 2# = 18
right side:
#C# = 1
#H# = 2 x 5 = 10
#N# = 2
#O# = 2 + (1 x 5) + 2
Again notice that all atoms on the left side are multiplied by 2 because they are bonded to one another. Now let's balance the rest of the equation.
#2C_3H_5(NO_3)_3# #rarr# #color (green) 6CO_2# + #5H_2O# + #color (orange) 3N_2# + #O_2#
left side:
#C# = 3 x 2 = 6
#H# = 5 x 2 = 10
#N# = 3 x 2 = 6
#O# = 9 x 2 = 18
right side:
#C# = 1 x #color (green) 6# = 6
#H# = 2 x 5 = 10
#N# = 2 x #color (orange) 3# = 6
#O# = (2 x #color (green) 6#) + (1 x 5) + 2
Now the only atom left to balance is the #O#. Notice that if you add all the O atoms on the right side, you will come up with a sum of 19. So what to do? Use your knowledge of fractions.
#2C_3H_5(NO_3)_3# #rarr# #6CO_2# + # 5H_2O# + #3N_2# + #color (magenta) (1/2)O_2#
left side:
#C# = 3 x 2 = 6
#H# = 5 x 2 = 10
#N# = 3 x 2 = 6
#O# = 9 x 2 = 18
right side:
#C# = 1 x 6 = 6
#H# = 2 x 5= 10
#N# = 2 x 3 = 6
#O# = (2 x 6) + (1 x 5) + (2 x #color (magenta) (1/2)#) = 18
The equation is now balanced.
But if you want whole number coefficients instead of fractions, you can always multiply the WHOLE equation by 2.
#cancel 2# [#2C_3H_5(NO_3)_3# #rarr# #6CO_2# + #5H_2O# + #3N_2# + #1/ cancel 2O_2#]
=
#4C_3H_5(NO_3)_3# #rarr# #12CO_2# + #10H_2O# + #6N_2# + #O_2#
