How do you balance Na + H_2O -> NaOH + H_2?

Sep 11, 2017

$2 N a + 2 {H}_{2} O \to 2 N a O H + 2 {H}_{2}$

Explanation:

First, let's rewrite ${H}_{2} O$ as $H O H$. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the "new" equation is:

$N a + H O H \to N a O H + {H}_{2}$

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

Left:
Na - 1
H - 1
OH - 1

Right:
Na - 1
OH -1
H -2

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

$2 N a + 2 H O H \to N a O H + {H}_{2}$

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the ${H}_{2}$ is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

$2 N a + 2 H O H \to 2 N a O H + 2 {H}_{2}$

Changing back $H O H$ to ${H}_{2} O$, we get:

$2 N a + 2 {H}_{2} O \to 2 N a O H + 2 {H}_{2}$