# Question #67604

Jan 25, 2017

Trial and error process

Given equation

${C}_{3} {H}_{5} {\left(N {O}_{3}\right)}_{3} \to {N}_{2} + {O}_{2} + C {O}_{2} + {H}_{2} O$

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 ${H}_{2} O$ 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 ${H}_{2} O$ 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.

$4 {C}_{3} {H}_{5} {\left(N {O}_{3}\right)}_{3} \to 6 {N}_{2} + {O}_{2} + 12 C {O}_{2} + 10 {H}_{2} O$