Question

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

Answer 1

`2Na + 2H_2O -> 2NaOH +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 `Na+H_2O->NaOH+H_2`, there are
`1-Na-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.

Answer 2

`2"Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g))`

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

`"Na"_ ((s)) + "H"_ 2"O"_ ((l)) -> "NaOH"_ ((aq)) + "H"_ (2(g))`

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

`"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))`

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

`2"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))`

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.