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How do I balance formulas?

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Ca(OH)2+H3PO Ca3(PO4)2+HOH
 
Writing the water as HOH is useful when balancing double displacement reactions that involve an acid and a base, like this one.  It's useful because it helps you see that this isn't a reaction involving individual atoms; it's one that involves four different polyatomic ions switching partners.
 
Notice on the left side, you've got Ca2+ and OH in the first compound, and H+ and PO43– in the second.
 
On the right side, you've got Ca2+ and  PO43– in the first compound, and H+ and OH in the second.
 
Now it's just a matter of balancing the ions.  Do them one at a time, strictly one at a time.  That means, when you're balancing one of them, you don't look at the others, you just make sure the one you're balancing is balanced.  Start with any ion.  I usually start on the left side.
 
There is one Ca2+ ion on the left and three on the right.  Add a 3 in front of the calcium-containing compound on the left.
 
3 Ca(OH)2+H3PO Ca3(PO4)2+HOH
 
Now look at the hydroxides: you now have six of them on the left and only one on the right, so add a 6 in front of the HOH. 
 
3 Ca(OH)2+H3PO Ca3(PO4)2+6 HOH
 
Now check the H+ ions.  You need to be careful not to include the hydrogens that are in the OH ions when balancing the H+ ions.  There are three H+ ions on the left and six of them on the right, so put a 2 in front of the phosphoric acid.
 
3 Ca(OH)2+2 H3PO Ca3(PO4)2+6 HOH
 
Now it's time to check the phosphate ions.  There are two on the left and two on the right.  As usual for a double-displacement reaction: if you balance the first three correctly, you're usually done, but it's always worth checking.
 
Hint: don't bother with tables of elements under the reaction.  It's a waste of precious time, especially if you're doing a timed exam.  Just balance the elements (or the ions, when possible) one at a time, as above.
 
 
 
 
Shari's explanation is good.  The equation is balanced.  I've replicated her work in the form of a spreadsheet showing the individual element count.  I use this general approach on difficult equations and find it helps with the solution while minimizing errors.  This response is long, but it works for even complex equations and minimizes errors.
 
Look at the table below.  I start by write the equation on top, normally without the coefficients if they are unknown (use pencil).  I then write each element in a column on the left (Ca, O, H, P) and then count the number of times that element appears in a single molecule of each compound.  In this equation, this is what my initial chart would look like.  I arbitrarily assigned "1" for each coefficient, knowing that I need at least 1 of every molecule.  That is the minimum.
 
Look over the table of numbers and note that the number of times that element appears in one molecule of each of the reactants and products.  For example, Oxygen (O) appears twice in Ca(OH)2 and four times in H3PO4.  I write each number directly below its compound, and then sum each element on both the reactant and product sides.  I can see that the equation is not balanced for oxygen, since it appears 6 times on the left (reactants) and 7 times on the right (products).  In fact, none of the elements are balanced.
 
           1 Ca(OH)2 + 1 H3PO4 ==> 1 Ca3(PO4)2 + 1 H2O

     Ca        1     +    0  =   1           3       +       0     = 3
      O         2     +    4  =   6           8       +       1     = 9
      H         2     +    3  =   7           0       +       2     = 2
      P         0      +    1 =   1           2        +      0     = 2
 
 
The next step is to change one of the coefficients to help balance the numbers. I usually start with the most complex formula and assume it has a coefficient of "1" and work from there. In this case, Ill choose Ca3(PO4)2. There has to be at least one molecule, so I pencil in the "1." I then go through each element and make changes to the reactants to try and balance each element.  I immediately know I need to start with at least 3 Ca(OH)2 molecules, since it is the only source of Ca, and it only has 1 Ca per molecule.  Pencil in a "3" for Ca(OH)2.  The Calciums are now balanced.  I then change the element count for this molecule, which results in 6 H's and 6 O's.
 
One molecule of Ca3(PO)4 also contains 8 Oxygen, zero Hydrogen, and 2 Phosphorus atoms, as indicated.  To get two Phosphorus atoms, I need to add two molecules of H3PO4, so I pencil in a "2" for that molecule.  That results in a new element count for that molecule, as you can see below.  Now we have 14 O's, 12 H's, and 2 P's.  This means that the Ca and P atoms are now balanced, but we need to correct the H's and O's.  We have a toatl of 12 H and 14 O atoms on the reactant side.  They have to exist since we assumed just one molecule of Ca3(PO4)2 and this requires the minimum of 3 Ca(OH)2 and 2 H3PO4 molecules.
 
On the product side, the only place hydrogen can go is into water, H2O.  Since we have a total of 12 hydrogens, and water has 2 H's per molecule, we need 6 molecules of water to balance the H's.  
 
This leave O.  We have 14 O atoms from the reactants.  The Ca3(PO)4 takes 8 of them, leaving 6.  But now that we have assigned a 6 to the H2O, we can now account for the O's (14 on both sides).
 
We're done!  As you can see, the element counts are the same on both sides of the equation.  Not all equations works are straightforward as this, as you will discover.  You can often get to near the end and find that something isn't working.  This often occurs when working with molecules that have even numbers of elements on one side, but odd numbers on the other.  Painful, but simple to fix.  Go back to the original molecule you set to "1," Ca3(PO4)2, and increase it to "2."  By iterating back and forth, you will soon see a way of balancing both sides of the equation.  It can be difficult sometimes, but just keeping increasing one of the molecule coefficient and rebalancing, you will eventually see a solution.  It is easy sometimes to increase by too much, and you will note that the coefficients have a common factor, such as 2.  Just divide by the common factor to get the lowest whole number coefficients.
 
For example, if I started this process by assigning a "2" to the Ca3(PO)4, my final equation would be 
6Ca(OH)2 + 4H3PO4 ==> 2Ca3(PO4)2 + 12H2O.  This is divisible by 2. so the correct equation is given by dividing each coefficient by "2."
 
 
         3 Ca(OH)2 + 2 H3PO4  ==>  1 Ca3(PO4)2  + 6 H2O

     Ca      3     +        0    =     3           3        +    0     =   3
     O        6      +       8    =   14           8       +     6     = 14
     H        6      +       6    =   12           0        +   12     = 12
     P        0      +       2     =    2           2        +    0      =   2
 
                                   reactant total                            product total
Balanced!   
 
Sorry this is so long.  Use an Excel spreadsheet for difficult problems and you'll have fun making the changes while everyone else complains.  :)
 
Bob
When you balance a chemical formula all elements present on the reactant side must equal the elements present on the product side.  It is easiest to start with those elements which only appear once on each side.  Additionally if compound appears on each side then balance it as a compound.  Lastly balance those elements which appear in multiple compounds.
 
For the equation above calcium (Ca) appears only once on both the reactant and product side.  There is one Ca on the reactant side but three on the product side so to balance you would need 3 Ca(OH)2.   Next look at phosphate (PO)4.  Since it appears as itself, unchanged, on both sides of the equation balance it as if it were a singular element.  On the reactant side you have only 1 PO4; however, on the product side you have 2.  Therefore you need 2 phosphoric acids (2 H3PO4).    The equation at this time would look like below:
 
3 Ca(OH)2  +  2 H3PO4  --------->   Ca3(PO4)2  +  HOH
 
 
The only thing left is to balance out the oxygen atoms and hydrogen atoms in the compound.  On the reactant side you have 14 oxygens and 12 hydrogens.   On the product side you have 8 oxygens in the calcium phosphate leaving 6 left to balance.  Therefore you need 6 water molecules to balance out this equation and it would look like the following:
 
3  Ca(OH)2  +  2  H3PO4    ----------->           Ca3(PO4)2  +  6 HOH