Chemistry: balancing equations?

___ Na + ___ H2O → ___ NaOH + ___ H2

As it stands now the Na and O atoms are already balanced, but the H atoms are not, so you start by looking at the H atoms. Notice that there will always be an even number of H atoms on the left of the arrow because of the "2" in "H2O". On the right-hand side, you've got one H atom in NaOH and two H atoms in H2. If the left side is always even, the right side must also be even, and the only way that can happen is for there to be an even number of H atoms from NaOH, and so an even number of NaOH molecules. So start by using the smallest even number for the NaOH coefficient:

___ Na + ___ H2O → 2 NaOH + ___ H2

Now you have two sodium atoms on the right, so you must have two sodium atoms on the left:

2 Na + ___ H2O → 2 NaOH + ___ H2

You also have two O atoms on the right, so you must have two O atoms on the left:

2 Na + 2 H2O → 2 NaOH + ___ H2

Now you have a total of four H atoms on the left, so you must have a total of four H atoms on the right. You've already got two H atoms in NaOH, so you need two more H atoms, so you need only one molecule of H2:

2 Na + 2 H2O → 2 NaOH + H2

Now everything is balanced.

Or, starting over, it might have been easier to look at it this way:

___ Na + ___ H2O → ___ NaOH + ___ H2

Na and O are already balanced. Looking at H, you have two H atoms on the left, so you need two H atoms on the right. You must take one H atom with the NaOH, so you need just one more H atom on the right. The only other source of H atoms is H2, but you need only one atom, so use a coefficient of 1/2:

Na + H2O → NaOH + 1/2 H2

Now the equation is completely balanced, but the rule forbidding fractional coefficients has been violated, so multiply everything by 2 to get rid of the fraction:

2 Na + 2 H2O → 2 NaOH + H2