July 09, 2020, 04:49:47 PM
Forum Rules: Read This Before Posting

### Topic: balancing equations  (Read 4050 times)

0 Members and 1 Guest are viewing this topic.

#### A7X

• Guest ##### balancing equations
« on: September 11, 2004, 08:43:15 AM »
can someone please tell me how to balance the following equations:

C6H12 + 02 ----> CO2 + H20
H2O2 -----> H2O + O2
SO2 + NaOH ---> NaSO3 + H20
FeSO4 ----> Fe2O3 + SO3 + SO2
NH3 + CuO ----> N2 + Cu + H2O

#### jdurg

• Banninator
• Retired Staff
• Sr. Member
• • Posts: 1366
• Mole Snacks: +106/-23
• Gender: • I am NOT a freak. ##### Re:balancing equations
« Reply #1 on: September 11, 2004, 02:55:44 PM »
In order to balance equations you have the count the number of each type of atom on the left side and make sure it equals the same thing on the right side.  I'll show you how to solve one of them, but the rest you'll have to attempt/do yourself.

NH3 + CuO -> N2 + Cu + H2O.

Here, I would start with balancing the hydrogen and nitrogen atoms since they are both contained in the initial reactant; ammonia.  So we'll start with hydrogen.  In the products we have two hydrogen atoms in the water and in the reactants we have three hydrogen atoms in the ammonia.  So we need to balance these things out.  If we multiply the ammonia by two, we'll have six hydrogen atoms, and if we multiply the water by three we'll have six hydrogen atoms.  So this gives us;

2NH3 + CuO -> N2 + Cu + 3H2O.

If we do an atom count, we'll get the following;
Reactants:  2N, 6H, 1Cu, 1O
Products:  2N, 6H, 1Cu, 3O.

So you can see that we are almost balanced.  The hydrogen, copper and nitrogen are balanced, which leaves us with the oxygen which is not balanced right.  In the products, we have three oxygen atoms but we only have one in the reactants.  So we'll need to multiply the reactant that contains oxygen by 3 in order to balance the oxygen.  This will also multiply the copper by 3, so we'll need to balance that in the reactants as well.  This will give us;

2NH3 + 3CuO -> N2 + 3Cu + 3H2O.

If we do an atom count, we'll get the following;
Reactants:  2N, 6H, 3Cu, 3O
Products:  2N, 6H, 3Cu, 3O.

Now all the products and reactants are in perfect balance; the equation is in its lowest whole number coefficients; the equation is balanced.  Now you can use the same process to balance the others. "A real fart is beefy, has a density greater than or equal to the air surrounding it, consists