How to balance chemical equations with multiple elements?

[transgalactic]

[itex]

Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2
[/itex]i got the same problem with this one
[itex]
MgO + H_3PO_4 -> Mg_3(PO_4)_2 + H_2O
[/itex]
each time i balance one element i get the other which i balance before
to be dis balanced

what is the general way of solving such things
??

 

[CompuChip]

I suggest starting with the elements that only occur in one part, usually you end with C, H and O because they occur most.
For example, for the first one,
a Ca₃(PO₄)₂) + b H₃PO₄ -> c Ca (H₂ PO₄)₂

First start with setting a = 1. You only have Ca in one element on the left, so you need c = 3 to get the amount of Ca on the right correctly. Then you have also fixed an amount of PO₄ on both sides, and you will have to use b to compensate that. Then you have used all your freedom, so the H should automatically be balanced (if not, the reaction is not possible). Finally, check if you have fractions somewhere, and multiply through by some number to get integers everywhere.

If you are more mathematically inclined, you can write down a set of equations:
For Ca: 3a = c
For PO4: 2a + b = 2
For H: 3b = 4c
and solve that. Afterwards, again multiply a, b and c by some number to get rid of possible fractions (e.g. if you find a = 1, b = 1/2 and c = 1/5 multiply by 10 and use a = 10, b = 5, c = 2).

 

[Borek]

Please read this balancing by inspection tutorial.

 

[transgalactic]

I suggest starting with the elements that only occur in one part, usually you end with C, H and O because they occur most.
For example, for the first one,
a Ca₃(PO₄)₂) + b H₃PO₄ -> c Ca (H₂ PO₄)₂

First start with setting a = 1. You only have Ca in one element on the left, so you need c = 3 to get the amount of Ca on the right correctly. Then you have also fixed an amount of PO₄ on both sides, and you will have to use b to compensate that. Then you have used all your freedom, so the H should automatically be balanced (if not, the reaction is not possible). Finally, check if you have fractions somewhere, and multiply through by some number to get integers everywhere.

If you are more mathematically inclined, you can write down a set of equations:
For Ca: 3a = c
For PO4: 2a + b = 2
For H: 3b = 4c
and solve that. Afterwards, again multiply a, b and c by some number to get rid of possible fractions (e.g. if you find a = 1, b = 1/2 and c = 1/5 multiply by 10 and use a = 10, b = 5, c = 2).

i can't see how you get this equations..
why you put po4 as one instead doing for "p" ad oxiden
why in the second one you have only 2

 

[symbolipoint]

transgalactic, CompuChip is trying to generalize your first example and used a , b, and c as counting variables (or coefficients?) for each compound.

Myself, starting in your Ca example, I would start with the Ca atoms first. The compound on the leftside has 3 Ca atoms, so you probably want 3 Ca atoms on the rightside. How many calcium dihydrogenphosphates would you need on the rightside? You would best pick 3 of these. Next, account for the phosphorus atoms. Do you have the same number of atoms of phosphorus on both sides? If not, then you have another adjustment to make. WORK WITH IT!

 

[Borek]

why you put po4 as one instead doing for "p" ad oxiden

Becasue they are always combined in the same way. You will have two linearly dependent equations - no new information, more places to make a mistake.

why in the second one you have only 2

Must be a typo, shoud be 2c.