How can you balance Cu+HNo3=Cu+(No3)2+No+H2o by using algebric or a b c d method please show me with full workout
well I am not familiar with your methods but maybe
a Cu + b HNO3 = a Cu+c (NO3)2+ d NO + e H2O
a = a obviously, Cu only appears once and a can be any number at all
now H
b = 2 e
then N
b = 2 c + d
now O
3 b = 6 c + d + e
so now
use 2 e for b
2 e = 2 c + d
3(2e) = 6 c + d + e or 5 e = 6 c + d
so 3 e = 4 c
e = (4/3) c
then go back
if c = 3 to get an integer number for e
then e = 4
d = 2 e - 2 c = 8 - 6 = 2
and b = 2 e = 8
try it:
a Cu + 8 HNO3 = a Cu+ 3 (NO3)2+ 2 NO +4 H2O
any old a
8 H on each side
8 N on each side
24 O on each side
It works
I do not know why Cu, copper, is in your reaction since it does not react. Perhaps it is just to confuse you.
Two comments. First, for your trash kid, don't get excited about this. This is just an odd ball.
Second. there is one BIG typo and several small ones. If you take care of the BIG one it takes care of Damon's problem AND you get the correct answer. Here is the corrected typo. I have removed the + sign between Cu and (NO3)2 on the product side. Now Damon doesn't have a problem because it shows Cu actually DID react. Now here is the correct equation using the caps key too.
Cu + HNO3 ==> Cu(NO3)2 + NO + H2O
Next you follow Damon's work and you end up with the balanced equation as follows:
3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O
NOTE: This is a redox equation and I would NEVER use this method to balanced a reaction such as this one.
Another note: Note that this reaction is for Cu and DILUTE HNO3. The product is NO + the others.
When CONCENTRATED HNO3 is used the product is NO2 + others. That balanced equation is Cu + 4HNO3 ==> Cu(NO3)2 + 2NO2 + 2H2O.
Why did the math book go to the therapist?
Because it had too many problems to solve, and it needed someone to listen.
To balance the equation Cu + HNO3 = Cu(NO3)2 + NO + H2O, we will use the algebraic method.
Step 1: Write down the unbalanced equation
Cu + HNO3 = Cu(NO3)2 + NO + H2O
Step 2: Count the number of atoms for each element on both sides of the equation.
Cu: 1 on the left side, 1 on the right side
H: 1 on the left side, 2 in HNO3, 2 in H2O
N: 1 in HNO3, 2 in Cu(NO3)2, 1 in NO
O: 3 in HNO3, 6 in Cu(NO3)2, 1 in NO, 1 in H2O
Step 3: Begin balancing the equation by balancing atoms that appear in the fewest number of compounds.
Start with Cu: 1 on each side, balanced.
Step 4: Balance oxygen atoms by adjusting the coefficient in H2O:
Cu + HNO3 = Cu(NO3)2 + NO + 2H2O
New atom counts:
H: 1 on the left side, 2 in HNO3, 4 in H2O
N: 1 in HNO3, 2 in Cu(NO3)2, 1 in NO
O: 3 in HNO3, 6 in Cu(NO3)2, 1 in NO, 4 in H2O
Step 5: Balance hydrogen atoms by adjusting the coefficient in HNO3:
Cu + 8HNO3 = Cu(NO3)2 + NO + 4H2O
New atom counts:
H: 8 on the left side, 8 in HNO3, 16 in H2O
N: 8 in HNO3, 2 in Cu(NO3)2, 1 in NO
O: 24 in HNO3, 6 in Cu(NO3)2, 1 in NO, 16 in H2O
Step 6: Balance nitrogen atoms by adjusting the coefficient in NO:
Cu + 8HNO3 = Cu(NO3)2 + 8NO + 4H2O
New atom counts:
H: 8 on the left side, 8 in HNO3, 16 in H2O
N: 8 in HNO3, 8 in Cu(NO3)2, 8 in NO
O: 24 in HNO3, 24 in Cu(NO3)2, 8 in NO, 16 in H2O
Step 7: Finally, balance oxygen atoms by adjusting the coefficient in HNO3:
3Cu + 8HNO3 = 3Cu(NO3)2 + 8NO + 4H2O
Now, the equation is balanced with an equal number of atoms on both sides:
3Cu + 8HNO3 = 3Cu(NO3)2 + 8NO + 4H2O
To balance the equation Cu + HNO3 = Cu(NO3)2 + NO + H2O using the algebraic method, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how you can do it step by step:
Step 1: Write out the unbalanced equation:
Cu + HNO3 = Cu(NO3)2 + NO + H2O
Step 2: Count the number of atoms on each side of the equation:
On the left side: Cu (1 atom), H (1 atom), N (1 atom), O (3 atoms)
On the right side: Cu (1 atom), N (3 atoms), O (10 atoms), H (2 atoms)
Step 3: Start balancing the elements one by one:
a) Balance the copper (Cu) atoms. Both sides already have 1 Cu, so they are balanced.
b) Balance the nitrogen (N) atoms. On the left side, we have 1 N atom, while on the right side, we have 3 N atoms. To balance it, put a coefficient of 3 in front of HNO3:
Cu + 3HNO3 = Cu(NO3)2 + NO + H2O
Now, we have 3 N atoms on both sides.
c) Balance the hydrogen (H) atoms. On the left side, we have 1 H atom, while on the right side, we have 2 H atoms. To balance it, put a coefficient of 2 in front of H2O:
Cu + 3HNO3 = Cu(NO3)2 + NO + 2H2O
Now, we have 2 H atoms on both sides.
d) Balance the oxygen (O) atoms. On the left side, we have 3 O atoms, while on the right side, we have 10 O atoms. To balance it, put a coefficient of 5 in front of NO3 on the right side:
Cu + 3HNO3 = Cu(NO3)2 + 5NO + 2H2O
Now, we have 10 O atoms on both sides.
The balanced equation is:
Cu + 3HNO3 = Cu(NO3)2 + 5NO + 2H2O
This is the final balanced equation using the algebraic method.