Balance these equation please?

Question

C8H18 + O2 yields CO2 + H20

Cu + HNO3 yields Cu (NO3) 2 + NO + H20

Answer 1

2C18H18 + 45 O2 ---> 36 CO2 + 18 H2O
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O

Answer 2

Are you familiar with Least Common Multiples? This is vital in balancing equations. If you can catch on to that, the rest of this will be easy.

You can start with any element you'd like, but I'll start with hydrogen because you've got the biggest number of them. You are going to have some multiple of 18 total hydrogen atoms on each side of the equation because the LCM of 18 (the H's on the left) and 2 (same thing on the right) is 18. Start with 18 and see if that works. That's already done for you on the left, so adjust the right side accordingly:
C8H18 + O2------->CO2 + 9H2O

Now, that won't work, because now you're going to end up with an odd number of O's on the right and an even number on the left, so try the next multiple of 18:
2C8H18 + O2----->CO2 + 18H2O

That looks better. You now have 16 C's on the left (2 molecules of 8 atoms each), so make 16 on the right:
2C8H18 + O2----->16CO2 + 18H2O

Now the only thing out of balance is oxygen. You have a total of 50 O's on the right ((16 X 2) + 18), so put 50 on the left:
2C8H18 + 25O2----->16CO2 + 18H2O

Check your (er, my) work: 16 C's, 36 H's, 50 O's on the left, 16 C's, 36 H's, 50 O's on the right.

Following this process, you should be able to do the next one on your own.

Answer 3

C8H18 + O2 yields CO2 + H20
10 C8H18 + 80 O2 yields 80 CO2 + 9 H20

I dont get the NO3 in parenthesis in the 2nd equation could you specify that?