# Question #67604

##### 1 Answer

**Trial and error process**

Given equation

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

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.