Question #67604

1 Answer
Jan 25, 2017

Trial and error process

Given equation

#C_3H_5 (NO_3)_3 -> N_2 + O_2+ CO_2 + H_2O#

If we inspect both sides of the given equation, we see that in LHS no.of H-atom is odd whereas it is even in RHS.So to make no.of H-atoms in both sides even we are to multiply LHS by 2.But to equate no. of H-atoms of both sides we are to multiply #H_2O# of RHS by 5. Now if we compare the no.of O-atom on both sides. we see there is a mismatch.In LHS no. of O-atoms is even but it is odd in RHS.So multiplication factor 2 in LHS is replaced by 4 and that of #H_2O# at RHS is replaced by 10.

Now the odd-even mismatch on O and H atoms on both sides will not exist and we can proceed to adjust other multiplication factors.And finally we get the following balanced equation.

#4C_3H_5 (NO_3)_3 ->6N_2 + O_2+ 12CO_2 + 10H_2O#