How to balance HNO3 +Cu ------->Cu(NO3)2+NO+H2O ? please show me steps
This is a redox equation and is best balanced by following redox rules.
3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O
http://www.chemteam.info/Redox/Redox.html
Thank you DrBob, but i am only a form 4 student in hk,i havent learnt about anything of redox.is there any SUPER SHORT CUT for balance simple and complex equation?in fact i have asked several people,some of them said that take LCM of the atom counts,but i don't know take which side LCM in which condition.
If the equation is a redox equation, and it is relatively complicated, as this one is, the best way is to learn how to do it by following the redox rules. You will get that in due time. If it is not complicated then trial and error is about the only way to do it. With experience you learn where to start. Here are a couple of examples. One is very simple and the other uses the LCM method that some other tutor spoke about.
Na + H2O ==> NaOH + H2
I look at this and Na and O are ok. H is not balanced and whatever I do to make it balance messes up the Na and O but that's the name of the game. Hence the name of trial and error. So I put a 2 in front (note: NEVER change the subscripts to make it balance.) of NaOH to look like this.
Na + H2O ==>2NaOH + H2.
It isn't balanced yet but I have an even number of H atoms on the right (4 of them) and I can make that balance with a 2 for H2O to become
Na + 2H2O ==> 2NaOH + H2
Note that balances the O also with two atoms on the left and two on the right. That leaves only Na to balance. It may not appear that way to you but I deliberately tried to balance everything EXCEPT the Na. Why? Because that stands alone and I can use any number for Na I wish. So I try to balance all but the "single" atom. Now I count on the right to see how many Na atoms I need. That is two on the right so I place a 2 for Na on the left and th final equation is
2Na + 2H2O ==> 2NaOH + H2
Now this one, which just happens to be a redox equation but it can be balanced by inspection (trial and error).
Al + O2 ==> Al2O3.
When we see an equation like this, we know we will NEVER balance it by placing whole numbers for O2. That's because we have 2 O on the left and 3 on the right. So we "cheat" a little and think 6. By placing a 2 in front of Al2O3 and 3 in front of O2, it looks this way.
Al + 3O2 ==> 2Al2O3
See that easily balances the oxygen atoms with 6 O atoms on the left and 6 on the right. That leaves just Al (another case where I deliberately balanced all EXCEPT the single material). I have 4 Al atoms on the right so we need 4 on the left and the final equation is
4Al + 3O2 ==> 2Al2O3.
I hope this helps get you started. The only way to learn how to do this is practice, practice, practice.
thank you very much,your teaching really makes me a good start. then how about balance S +H2SO4 ---> SO2 +H2O ??
S + H2SO4 ----> SO2 +H2O
the left has 4O and the right has 3O
let's go for 12
S + 3H2SO4 ---> 4SO2 + 4H2O
nope, how about
S + 2H2SO4 ---> 3SO2 + 2H2O
Looks good.
C2H6 + O2 --> CO2 = H2O
2na+2h2o=2naoh+h2
To balance the chemical equation HNO3 + Cu → Cu(NO3)2 + NO + H2O, you need to adjust the number of atoms on both sides of the equation. Here are the steps to balance this equation:
Step 1: Count the number of atoms on both sides of the equation.
HNO3: 1 Nitrogen (N), 3 Oxygen (O), and 1 Hydrogen (H)
Cu: 1 Copper (Cu)
Cu(NO3)2: 1 Copper (Cu), 2 Nitrogen (N), and 6 Oxygen (O)
NO: 1 Nitrogen (N), 1 Oxygen (O)
H2O: 2 Hydrogen (H) and 1 Oxygen (O)
Step 2: Start by balancing the most complex molecule or polyatomic ion first. In this case, it is Cu(NO3)2.
Place a coefficient of 2 in front of Cu(NO3)2 to make the number of Copper atoms balanced.
The equation becomes: HNO3 + Cu → 2Cu(NO3)2 + NO + H2O
Step 3: Now, count the number of atoms on both sides again and focus on other elements.
HNO3: 1 Nitrogen (N), 3 Oxygen (O), and 1 Hydrogen (H)
Cu: 1 Copper (Cu)
Cu(NO3)2: 2 Copper (Cu), 4 Nitrogen (N), and 12 Oxygen (O)
NO: 1 Nitrogen (N), 1 Oxygen (O)
H2O: 2 Hydrogen (H) and 1 Oxygen (O)
Step 4: Balance Nitrogen (N) atoms by adding coefficients in front of HNO3 and NO.
Add coefficient of 2 in front of HNO3: 2HNO3 + Cu → 2Cu(NO3)2 + NO + H2O
Add coefficient of 2 in front of NO: 2HNO3 + Cu → 2Cu(NO3)2 + 2NO + H2O
Step 5: Count the number of atoms on both sides again and focus on Oxygen (O) and Hydrogen (H).
HNO3: 2 Nitrogen (N), 6 Oxygen (O), and 2 Hydrogen (H)
Cu: 1 Copper (Cu)
Cu(NO3)2: 2 Copper (Cu), 4 Nitrogen (N), and 12 Oxygen (O)
NO: 2 Nitrogen (N) and 2 Oxygen (O)
H2O: 2 Hydrogen (H) and 2 Oxygen (O)
Step 6: Balance Oxygen (O) atoms by adding coefficients in front of H2O and Cu(NO3)2.
Add coefficient of 6 in front of H2O: 2HNO3 + Cu → 2Cu(NO3)2 + 2NO + 6H2O
Add coefficient of 6 in front of Cu(NO3)2: 2HNO3 + Cu → 6Cu(NO3)2 + 2NO + 6H2O
Step 7: Finally, count the number of atoms on both sides and make sure they are balanced.
HNO3: 2 Nitrogen (N), 6 Oxygen (O), and 2 Hydrogen (H)
Cu: 1 Copper (Cu)
Cu(NO3)2: 12 Copper (Cu), 24 Nitrogen (N), and 72 Oxygen (O)
NO: 2 Nitrogen (N) and 2 Oxygen (O)
H2O: 12 Hydrogen (H) and 12 Oxygen (O)
Now the equation is balanced: 2HNO3 + Cu → 6Cu(NO3)2 + 2NO + 6H2O.