balancing equations: Can anyone help me?

Question

#1) Cu + HNO3 ----------------> Cu(NO3)2 + NO + H20
AND
#2) S + HNO3 --------------> H2SO4 + NO2 + H2O

Answer 1

Well, I can give you the answers. The logic is a bit complicated but I could explain it with some time.

3Cu +8HNO3 = 3Cu(NO3)2 + 2 NO + 4 H20
1S + 6HNO3 = H2SO4 + 6 NO2 + 2 H20


Here is the logic for the first one:

Forget the Cu since you can take as many as you need and nothing else comes along with it.

1. Hydrogen on the right side is always even so HNO3 must be even.

2. If HNO3 is even, then N is even on the right side so NO must be even, too.

3. Since HNO3 is even, O is even on the right side so H20 must be even, too.

4. If H20 is even, then H comes in multiples of 4 so HNO3 must be a multiple of 4, i.e. 4, 8 12, etc.

Four does not work. Eight works by juggling the nitrogen balance between Cu(NO3)2 and NO using the oxygen balance to tell you when you are there.

The logic on the second one is similar. Try it.




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