Question

balance the below equations by adding appropriate coefficients
a. Mg(NO3)+K3PO4~Mg3(PO4)2+KNO3

Answer 1

To balance the equation \( \text{Mg(NO}_3)_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KNO}_3 \), we need to ensure that there are equal numbers of each type of atom on both sides of the equation.

The unbalanced equation is:

\[
\text{Mg(NO}_3)_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KNO}_3
\]

### Step-by-step balancing:

1. **Identify all the elements** in the equation: Mg, N, O, K, P.

2. **Count atoms on both sides:**
- Reactants:
- Mg: 1
- N: 2
- O: 6 (2 from \( \text{Mg(NO}_3)_2 \))
- K: 3
- P: 1

- Products:
- Mg: 3
- N: 1
- O: 8 (from \( \text{Mg}_3(\text{PO}_4)_2 \) and \( \text{KNO}_3 \))
- K: 1
- P: 2

### Balancing the equation:

1. **Balance magnesium (Mg):** There are 3 magnesium atoms on the product side and only 1 in the reactants. So, place a coefficient of 3 in front of \( \text{Mg(NO}_3)_2 \):

\[
3 \text{Mg(NO}_3)_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KNO}_3
\]

2. **Update counts:**
- Reactants:
- Mg: 3
- N: 6 (3 from each \( \text{Mg(NO}_3)_2 \))
- O: 18 (3 from each \( \text{Mg(NO}_3)_2 \))
- K: 3
- P: 1

- Products (still the same):
- Mg: 3
- N: 1
- O: 8
- K: 1
- P: 2

3. **Balance phosphorus (P):** There are 2 phosphorus atoms in the products and only 1 in the reactants. Therefore, put a coefficient of 2 in front of \( \text{K}_3\text{PO}_4 \):

\[
3 \text{Mg(NO}_3)_2 + 2 \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + \text{KNO}_3
\]

4. **Update counts again:**
- Reactants:
- Mg: 3
- N: 6
- O: 18 (from \( 3 \text{Mg(NO}_3)_2 \))
- K: 6 (2 from \( \text{K}_3\text{PO}_4 \))
- P: 2 (from \( 2 \text{K}_3\text{PO}_4 \))

- Products:
- Mg: 3
- N: 1
- O: 8 (4 from \( \text{Mg}_3(\text{PO}_4)_2 \) and 3 from \( \text{KNO}_3 \))
- K: 1
- P: 2

5. **Balance potassium (K):** We also need to adjust the \( \text{KNO}_3 \) to balance the 6 potassium atoms from the reactants by placing a coefficient of 6 in front:

\[
3 \text{Mg(NO}_3)_2 + 2 \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + 6 \text{KNO}_3
\]

6. **Final counts:**
- Reactants:
- Mg: 3
- N: 6
- O: 18
- K: 6
- P: 2

- Products:
- Mg: 3
- N: 6
- O: 18 (2*4 from \( \text{Mg}_3(\text{PO}_4)_2 \) and 6 from \( 6 \text{KNO}_3 \))
- K: 6
- P: 2

Now the equation is balanced.

### Final balanced equation:

\[
3 \text{Mg(NO}_3)_2 + 2 \text{K}_3\text{PO}_4 \rightarrow \text{Mg}_3(\text{PO}_4)_2 + 6 \text{KNO}_3
\]