[chemhuff]

I tried this problem many times!  First and foremost, I concentrated on balancing the polyatomic ions, then the others.....but the Oxygen atoms keep coming out uneven!  On the reactant side I got 6 Oxygen and on the product side I got 7 Oxegen....  I tried using fractions but I just can't get it!  Here's the problem:

Pb(NO₃)₂  ----------->  PbO + NO₂ + O₂

Please help Mitch!!

Anne-Marie

P.S.  I'm supposed to add up the coefficients when it is balanced and submit it.

[jdurg]

Okay.  Start out by basically ignoring that oxygen molecule in the products and just attempt to balance the other products as best as you can.  So your equation would be:

Pb(NO3)2 --> PbO + NO2

This should be pretty easy to balance since you need another Nitrogen on the right side.  You would now right the equation as:

Pb(NO3)2 --> PbO + 2NO2

On the left hand side you have 1 Pb, 2 N, and 6 O.  On the right hand side you have 1 Pb, 2N, and 5 O.  So the lead and nitrogen are balanced, but the oxygen is one deficient on the right side.  Do you remember that O2 molecule we removed?  Well, one half of O2 is O, and that O2 molecule was originally listed as a product.  So if we use 1/2 O2, everything will be balanced.

Pb(NO3)2 --> PbO + 2NO2 + (1/2)O2

On the left there is 1Pb, 2N, and 6O.  On the right there is 1Pb, 2N, and 6O.  So it is balanced.  Now just multiply every compound by 2 and the fraction will go away.  Your final result will be:

2Pb(NO3)2 --> 2PbO + 4NO2 + O2

Do a final check on that and you'll see that it is all balanced.  

When you have a complex equation to balance, it is very important to start simple.  If one of the reactants or products is a pure element, then it becomes very easy to solve the equation.  Try and pick out the one set of compounds that will almost balance out perfectly.  In this case, the lead nitrate was our only reactant so that had to be used, and the lead oxide and nitrogen dioxide were the most similar to each other.  The oxygen could then be ignored for the time being until we balance out what remained.  If you take a look at a reaction in this manner, figuring it out becomes a lot simpler.  

[chemhuff]

THANKS A LOT!!  That was a lot easier.  Although in my text book it says to do the complex things first....but I've heard otherwise from different sources.

Anne-Marie

[AWK]

Salts of oxygen containing acids can always be written as oxide.acid anhydride (sometimes also water)
Pb(NO3)2=PbO.N2O5
N2O5= 2NO2 + 1/2O2
Final result
2Pb(NO3)2 = 2PbO + 4NO2 + O2

Ca(H2PO4)2 = CaO.P2O5.2H2O

[jdurg]

Salts of oxygen containing acids can always be written as oxide.acid anhydride (sometimes also water)
Pb(NO3)2=PbO.N2O5
N2O5= 2NO2 + 1/2O2
Final result
2Pb(NO3)2 = 2PbO + 4NO2 + O2

Ca(H2PO4)2 = CaO.P2O5.2H2O

I had never heard of doing it that way before, but that does make a helluva lot of sense.