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

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

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Answer 1 · 2015-11-16T19:37:11

$2 N a + 2 H_{2} O \to 2 N a O H + H_{2}$ $2Na + 2H_2O -> 2NaOH +H_2$

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 $N a + H_{2} O \to N a O H + H_{2}$ $Na+H_2O->NaOH+H_2$ , there are
$1 - N a - 1$ $1-Na-1$
$2 - H - 3$ $2-H-3$
$1 - O - 1$ $1-O-1$
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 · 2016-11-12T01:57:29

$2 {Na}_{(s)} + 2 H_{2} O_{(l)} \to 2 {NaOH}_{(a q)} + H_{2 (g)}$ $2"Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g))$

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

${Na}_{(s)} + H_{2} O_{(l)} \to {NaOH}_{(a q)} + H_{2 (g)}$ $"Na"_ ((s)) + "H"_ 2"O"_ ((l)) -> "NaOH"_ ((aq)) + "H"_ (2(g))$

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

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

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

So you now have

${Na}_{(s)} + 2 H_{2} O_{(l)} \to 2 {NaOH}_{(a q)} + H_{2 (g)}$ $"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))$

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

$2 {Na}_{(s)} + 2 H_{2} O_{(l)} \to 2 {NaOH}_{(a q)} + H_{2 (g)}$ $2"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))$

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

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

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