How ro Balance the Equation : -
Copper + Nitric Acid -------> Copper Nitrite + Nitric Oxide + Water
In the reaction of Copper and Nitric acid, Copper nitrate, nitric oxide and water are produced.
Step I:Write the unbalanced chemical equation for the given reaction.
Cu + HNO3 → Cu(NO3)2 + NO + H2O
Step II: List the number of atoms of the various elements present in the unbalanced equation in the form of a table.
Element | Number of atoms on the reactant side (L.H.S) | Number of atoms on the product side (R.H.S) |
Cu | 1 | 1 |
H | 1 | 2 |
N | 1 | 3 |
O | 3 | 8 |
Here, we can see that only Copper has an equal number of atoms on both sides of the equation.
Step III: In the next step, select a compound, which contains the maximum number of atoms. In this case, the compound will be copper nitrate (it has 1 atoms of Cu, 2 atoms of N, and 6 atoms of O). From this compound, select the element, which has the maximum number of atoms i.e., oxygen in this case. To balance the number of oxygen atoms, we can multiply copper nitrate present on the right hand side by 3 (as shown below).
Oxygen atoms | Number of atoms on L.H.S | Number of atoms on R.H.S |
Before balancing | 3 | 8 |
To balance | 3x8=24 | 8x3=24 |
Now, the equation becomes:
Cu + 8HNO3 → 3Cu(NO3)2 + NO + H2O
Again, compare the number of atoms of the various elements present in the chemical equation (as shown in the table below).
Element | Number of atoms on the reactant side (L.H.S) | Number of atoms on the product side (R.H.S) |
Cu | 1 | 1 |
H | 8 | 2 |
N | 8 | 6 + 1 = 7 |
O | 24 | 24 |
Step IV: As the atoms of oxygen are balanced, we will now balance copper.
Copper atoms | Number of atoms on L.H.S | Number of atoms on R.H.S |
Before balancing | 1 | 3 |
To balance | 1x3 = 3 | 3 |
Now, the equation becomes:
3Cu + 8HNO3 → 3Cu(NO3)2 + NO + H2O
Make the table again to compare the number of atoms of the elements on both sides of the equation.
Element | Number of atoms on the reactant side (L.H.S) | Number of atoms on the product side (R.H.S) |
Cu | 3 | 3 |
H | 8 | 2 |
N | 8 | 6 + 1 = 7 |
O | 24 | 24 |
We can see that the atoms of copper and oxygen are balanced.
Step V: Now, only the atoms of hydrogen and nitrogen are unbalanced. We will first balance the atoms of hydrogen.
Hydrogen atoms | Number of atoms on L.H.S | Number of atoms on R.H.S |
Before balancing | 8 | 2 |
To balance | 8 | 2x4 = 8 |
Now, the equation becomes:
3Cu + 8HNO3 → 3Cu(NO3)2 + NO + 4H2O
Let us again prepare a table to compare the number of atoms of the elements on both sides of the equation.
Element | Number of atoms on the reactant side (L.H.S) | Number of atoms on the product side (R.H.S) |
Cu | 3 | 3 |
H | 8 | 8 |
N | 8 | 6 + (1x2) = 8 |
O | 24 | 24 |
It can be observed that the chemical equation is balanced now.
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O