Posted by bob on Sunday, February 8, 2009 at 8:47pm.
2 Ag2 O --> 4 Ag + O2
first try all but O
C6 H6 + blankO2 --> 6 C O2 + 3 H2 O
but you need an even number of O so multiply everything so far by 2
2 C6 H6 + blank O2 --> 12 C O2 + 6 H2 O
so we need 15 O2
2 C6 H6 + 15 O2 --> 12 C O2 + 6 H2 O
The easiest way is by trial and error.
I think Bob Pursley helped you before and we want you to know how to do these yourself. Here is how you do the first one by trial and error.
Ag2O ==> Ag + O2
I look at the Ag2O on the left and see immediately that there are 2 Ag atoms on the left and only 1 on the right. So we fix that by adding a coefficient of 2 for Ag on the right (NOTE:You may change ONLY the coefficients when you balance equations. Changing the subscripts is a No, No.). So now we have this.
Ag2O ==> 2Ag + O2
Next is O and I see only 1 atom of O on the left but 2 on the right. I can fix that by multiplying Ag2O by 2 (that is, placing a 2 as a coefficient for Ag2O) and it looks like this.
2Ag2O ==> 2Ag + O2
WELL, we balanced the oxygen ok BUT that messed up the Ag which we had balanced before we started messing with oxygen. But that's the part of trial and error. We can fix the Ag by changing the 2Ag on the right to 4Ag. So the equation now looks like this.
2Ag2O ==> 4Ag + O2 and it should be balanced BUT you ALWAYS check it.
I see 4 Ag on the left and 4 on the right.
I see 2 O on the left and 2 on the right. So it balances. Done. I'll leave the second one for you but someone can check it for you if you care to post your numbers.