Balance the equation..
__Cu+__HNO3 -> __Cu(HNO3)2+__NO+__H2O
You must know two things:
a. what are the products? The ones you have are not correct.
b. is HNO3 dilute or concentrated. The products are different for each.
However, I think you meant the following:
3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + 4H2O.
To balance the given chemical equation: Cu + HNO3 -> Cu(HNO3)2 + NO + H2O,
1. Start by counting the number of atoms on each side of the equation.
On the left side (reactants):
- One copper (Cu) atom
- One nitrogen (N) atom
- Three oxygen (O) atoms
- One hydrogen (H) atom
On the right side (products):
- One copper (Cu) atom
- Two nitrogen (N) atoms
- Eight oxygen (O) atoms
- One hydrogen (H) atom
2. Begin the balancing process by adjusting the coefficients (whole numbers) in front of each compound to balance the number of atoms of each element on both sides of the equation.
Let's start with copper (Cu). Since both sides already have one Cu atom, the coefficient for Cu does not change.
Cu + HNO3 -> Cu(HNO3)2 + NO + H2O
Next, balance the nitrogen (N) atoms:
On the left side, there is one N atom in HNO3, and on the right side, there are two N atoms in Cu(HNO3)2 and one N atom in NO. To obtain two N atoms on the left side, we need to put a coefficient of 2 in front of HNO3.
Cu + 2HNO3 -> Cu(HNO3)2 + NO + H2O
Now, look at the oxygen (O) atoms: On the left side, there are three O atoms in HNO3, and on the right side, there are two O atoms in NO and eight O atoms in Cu(HNO3)2 and one O atom in H2O. To achieve the same number of O atoms on both sides, we need to put a coefficient of 4 in front of HNO3.
Cu + 2HNO3 -> Cu(HNO3)2 + NO + H2O
Finally, balance the hydrogen (H) atoms:
On the left side, we have two H atoms in 2HNO3, and on the right side, we have two H atoms in H2O. Both sides are already balanced in terms of hydrogen atoms.
The final balanced equation is:
Cu + 2HNO3 -> Cu(HNO3)2 + NO + H2O