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

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Apgt512 Share
Nov 16, 2015

Answer:

#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.

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Nov 12, 2016

Answer:

#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.

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