# How do you balance HCl + Ba(OH)_2 -> BaCl_2 + H_2O?

May 29, 2016

Add a two in front of the $H C l$ and add a two in front of the ${H}_{2} O$ to get

$2 H C l + B a {\left(O H\right)}_{2} \to B a C {l}_{2} + 2 {H}_{2} O$

#### Explanation:

When Balancing acid-base reactions, typically you balance the elements which are the salt's cation and anion's first, here the barium and the chlorine, the oxygen next, and the hydrogen should now be balanced.

It is important to know all the valences of the species to make sure you have the salt right, look at the number of hydrogen and the acid and the number of hydroxides on the base for these numbers.

We have,

$H C l + B a {\left(O H\right)}_{2} \to B a C {l}_{2} + {H}_{2} O$

First we consider the barium, there is one on the LHS and one on the RHS, so we're good.

Second we consider the chlorine, there is one on the LHS and two on the RHS, so we need to have two moles/atoms of $H C l$ in the formula.

$\textcolor{red}{2} H C l + B a {\left(O H\right)}_{2} \to B a C {l}_{\textcolor{red}{2}} + {H}_{2} O$

Third we consider the oxygen, since there are two $O H$'s in the barium hydroxide, there are two oxygen on the LHS, we need two on the RHS so there must be two moles/atoms of ${H}_{2} O$ in the formula.

$2 H C l + B a {\left(O H\right)}_{\textcolor{red}{2}} \to B a C {l}_{2} + \textcolor{red}{2} {H}_{2} O$

Now we count up the hydrogen's (it should be balanced already).
There are two $H C l$ atoms/moles for two hydrogen and another two in the pair of $O H$'s in the barium hydroxide, for a total of four on the LHS. On the right all the hydrogen is in the ${H}_{2} O$, since two have two atoms/moles of ${H}_{2} O$, each of which contains two hydrogens, we have four hydrogens on the RHS, so they are balances.

This is the final form of the equation,
$2 H C l + B a {\left(O H\right)}_{2} \to B a C {l}_{2} + 2 {H}_{2} O$