To balance the chemical equation [tex]\( \text{Li}_2\text{O} + \text{HPO}_4 \rightarrow \text{Li}_3\text{PO}_4 + \text{H}_2\text{O} \)[/tex], we need to ensure that the number of atoms of each element is the same on both sides of the equation.
Let's go through the steps to balance it:
1. Write down the initial count of atoms for each element on both sides of the equation:
- Left side (Reactants):
- Lithium (Li): 2 (from 1 [tex]\(\text{Li}_2\text{O}\)[/tex])
- Phosphorus (P): 1 (from 1 [tex]\(\text{HPO}_4\)[/tex])
- Oxygen (O): 1 (from [tex]\(\text{Li}_2\text{O}\)[/tex] ) + 4 (from [tex]\(\text{HPO}_4\)[/tex]) = 5
- Hydrogen (H): 1 (from 1 [tex]\(\text{HPO}_4\)[/tex])
- Right side (Products):
- Lithium (Li): 3 (from 1 [tex]\(\text{Li}_3\text{PO}_4\)[/tex])
- Phosphorus (P): 1 (from 1 [tex]\(\text{Li}_3\text{PO}_4\)[/tex])
- Oxygen (O): 4 (from [tex]\(\text{Li}_3\text{PO}_4\)[/tex]) + 1 (from [tex]\(\text{H}_2\text{O}\)[/tex]) = 5
- Hydrogen (H): 2 (from 1 [tex]\(\text{H}_2\text{O}\)[/tex])
2. Start with an element that appears in only one reactant and one product. In this case, we can start with Lithium (Li):
Currently, we have:
- Left: 2 Li
- Right: 3 Li
To balance the Li atoms, we need to adjust the coefficients. We can achieve this by multiplying [tex]\(\text{Li}_2\text{O}\)[/tex] by 3 and [tex]\(\text{Li}_3\text{PO}_4\)[/tex] by 2:
[tex]\[
\text{3Li}_2\text{O} + \text{HPO}_4 \rightarrow \text{2Li}_3\text{PO}_4 + \text{H}_2\text{O}
\][/tex]
Now we have:
- Left: 6 Li (from 3 [tex]\(\text{Li}_2\text{O}\)[/tex])
- Right: 6 Li (from 2 [tex]\(\text{Li}_3\text{PO}_4\)[/tex])
3. Next, balance Phosphorus (P):
Currently, we have:
- Left: 1 P
- Right: 2 P
To balance the P atoms, we need to adjust the coefficient of [tex]\( \text{HPO}_4 \)[/tex] by multiplying it by 2:
[tex]\[
\text{3Li}_2\text{O} + 2\text{HPO}_4 \rightarrow \text{2Li}_3\text{PO}_4 + \text{H}_2\text{O}
\][/tex]
Now we have:
- Left: 2 P (from 2 [tex]\(\text{HPO}_4\)[/tex])
- Right: 2 P (from 2 [tex]\(\text{Li}_3\text{PO}_4\)[/tex])
4. Finally, balance Oxygen (O) and Hydrogen (H):
- Oxygen (O):
- Left: 3 (from 3 ([tex]\(\text{Li}_2\text{O}\)[/tex]) + 8 (from 2 ([tex]\(\text{HPO}_4\)[/tex]) = 8)
- Right: 8 (from 2 ([tex]\(\text{Li}_3\text{PO}_4\)[/tex]) + 1 (from [tex]\(\text{H}_2\text{O}\)[/tex])
To balance the Oxygen atoms, we will need 3 molecules of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{3Li}_2\text{O} + 2\text{HPO}_4 \rightarrow \text{2Li}_3\text{PO}_4 + 3\text{H}_2\text{O}
\][/tex]
Now we have:
- Left: 8 O
- Right: 8 O
- Hydrogen (H):
- Left: 2 H (from 2 ([tex]\(\text{HPO}_4\)[/tex]))
- Right: 6 H (from 3 ([tex]\(\text{H}_2\text{O}\)[/tex]))
Therefore, with the given coefficients, all elements have the same number of atoms on both sides of the equation, achieving a balanced state.
Thus, the final balanced chemical equation is:
[tex]\[
\boxed{\text{3Li}_2\text{O} + 2\text{HPO}_4 \rightarrow \text{2Li}_3\text{PO}_4 + 3\text{H}_2\text{O}}
\][/tex]
