# How do you balance ClO_2 + H_2O = HClO_3 + HCl?

Dec 27, 2015

$6 C l {O}_{2}$ + $3 {H}_{2} O$ $\to$ $5 H C l {O}_{3}$ + $H C l$

#### Explanation:

This is quite difficult for a beginner especially if you're doing the trial and error method. But this kind of equation can also be solved using the algebraic method to find the coefficients of each compound in the equation.

By Trial and Error:

$C l {O}_{2}$ + ${H}_{2} O$ $\to$ $H C l {O}_{3}$ + $H C l$

$6 C l {O}_{2}$ + $3 {H}_{2} O$ $\to$ $5 H C l {O}_{3}$ + $H C l$

Make initial inventory of the elements composing the compounds in both reactants and products of the equation.

Reactants
Cl=$\cancel{1}$6
O=$\cancel{3}$$\cancel{13}$15
H=$\cancel{2}$6

Products
Cl=$\cancel{2}$6
O=$\cancel{3}$15
H=$\cancel{2}$6

1. Put 6 in front of the 1st reactant and recount the number of moles of each element since coefficient has been changed;
2. To balance the chlorine on the product side, put 5 as coefficient for the 1st compound of the products;
3. Then, recount the number of moles for each element of the products,in which case chlorine is already 6; balance with the reactants side;
4. Back to the reactants side,balance the hydrogen by putting 3 as coefficient for the 2nd reactant; then recount the number of moles and observed whether the inventory of both the reactants and products are balanced.