Balancing Chemical Equations

Date: 11/21/1999 at 20:10:33
From: Katie
Subject: Balancing Chemical Equations

My question is, how do you balance the following equations:

        HCI + Ca(OH)2 = CaCI2 + H2O

             KI + CI2 = KCI + I2

        CaSO4 + AIBr3 = CaBr3 = CaBr2 + AI2(SO4)3

                 H2O2 = H2O + O2

             Na + H2O = NaOH + H2

            C2H6 + O2 = Co2 + H2O

     Mg(NO3)2 + K3PO4 = Mg3(PO4)2 + KNO3

Date: 11/22/1999 at 00:57:45
From: Doctor Ian
Subject: Re: Balancing Chemical Equations


Hi Katie, 

The answer to your question is: by trial and error.

Do you understand what it means to 'balance' an equation? It means 
that when you add up all the atoms of each type on one side of the 
equation, and then on the other, all the atoms are accounted for. 
Atoms don't disappear in chemical reactions, nor do they appear. 
Whatever you start with, you have to end up with.

Let's look at one of your examples:

     C2H6 + O2 = CO2 + H2O

Assume that there is one molecule of each type, and write down the 
number of atoms of each type that appear on each side of the equation:

         C2H6 + O2  =  CO2 + H20

     2*C, 6*H, 2*O     1*C, 3*O, 2*H

This can't be right, because it would mean that one carbon atom and 
four hydrogen atoms disappeared, and one oxygen atom appeared out of 
nowhere.

But suppose we had two CO2 molecules on the right side:

         C2H6 + O2  =  2 CO2 + H20

     2*C, 6*H, 2*O     2*C, 5*O, 2*H

Now we have carbon working out okay (two carbon atoms on each side), 
but the oxygen and hydrogen don't add up. Suppose we have two more H2O 
molecules on the right side?

         C2H6 + O2  =  2 CO2 + 3 H20

     2*C, 6*H, 2*O     2*C, 7*O, 6*H

Now everything seems to add up except oxygen. If we had three O2 
molecules on the left side, we'd need one more on the right side. But 
if we had four O2 molecules on the left side, we'd have one extra on 
the right side. And we can't have 3-1/2 molecules of something.

The problem is that we have an odd number of oxygen atoms on the right 
side, but we can only generate an even number on the left side. So, 
what if we double everything else?

        2 C2H6 + O2  =  4 CO2 + 6 H20

     4*C, 12*H, 2*O     4*C, 14*O, 12*H

Now we can have seven O2 molecules on the left side, and everything 
adds up:

       2 C2H6 + 7 O2  =  4 CO2 + 6 H20

     4*C, 12*H, 14*O     4*C, 14*O, 12*H

Can you follow the same kinds of steps to balance the remaining 
equations?

I hope this helps. Be sure to write back if you're still stuck, or if 
you have any other questions.

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/