# How would you balance: Na + H2O --> NaOH + H2?

Nov 16, 2015

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

#### Explanation:

You have to count the number of atoms in both sides of the reaction to be equal, to make it a little easier you can make a list. So in the equation $N a + {H}_{2} O \to N a O H + {H}_{2}$, there are
$1 - N a - 1$
$2 - H - 3$
$1 - O - 1$
So you multiply NaOH and H2O by 2, so you have 4 hydrogens in both sides, and then multiply Na by 2 to be equal to the NaOH you previously multiplied.

Nov 12, 2016

$2 {\text{Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H}}_{2 \left(g\right)}$

#### Explanation:

I'm assuming that you're supposed to balance this chemical equation by inspection. You could also balance it by using oxidation numbers, but I don't think that you're supposed to go that route here.

So, your unbalanced chemical equation looks like this

${\text{Na"_ ((s)) + "H"_ 2"O"_ ((l)) -> "NaOH"_ ((aq)) + "H}}_{2 \left(g\right)}$

Notice that you have $3$ atoms of hydrogen on the products' side, but only $2$ on the reactants' side.

Here is where a little experience can come in handy. You can multiply the sodium hydroxide by $2$ to get a total of $4$ atoms of hydrogen on the products' side.

This will allow you to simply double the number of water molecules to get $4$ atoms of hydrogen on the reactants' side.

So you now have

${\text{Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H}}_{2 \left(g\right)}$

Balance the atoms of sodium by multiplying sodium metal by $2$

$2 {\text{Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H}}_{2 \left(g\right)}$

And now check to see if the atoms of oxygen are balanced. You have $2$ atoms of hydrogen on the reactants' side and $2$ on the products' side, which means that the equation is balanced.