Sure, let's solve this step-by-step:
1. Identify the compounds and their coefficients:
- Lithium oxide ([tex]\(Li _2 O\)[/tex])
- Hydrogen phosphate ([tex]\(HPO _4\)[/tex])
- Lithium phosphate ([tex]\(Li _3 PO _4\)[/tex])
- Water ([tex]\(H _2 O\)[/tex])
2. Write down the balanced chemical equation with coefficients:
We need to determine the coefficients that balance the equation.
3. Determine the coefficients for each compound:
- For [tex]\(Li _2 O\)[/tex] (Lithium oxide): the coefficient is 2.
- For [tex]\(HPO _4\)[/tex] (Hydrogen phosphate): the coefficient is 1.
- For [tex]\(Li _3 PO _4\)[/tex] (Lithium phosphate): the coefficient is 3 (since we have 3 lithium atoms from [tex]\(Li _2 O\)[/tex] and need to balance that with the [tex]\(Li _3 PO _4\)[/tex]).
- For [tex]\(H _2 O\)[/tex] (Water): the coefficient is 1.
4. Balance lithium (Li) atoms:
- On the left side, we have [tex]\(Li_2O\)[/tex] with a coefficient of 2, giving us [tex]\(2 \times 2 = 4\)[/tex] lithium atoms.
- On the right side, we need to have the same number of lithium atoms. [tex]\(Li_3PO_4\)[/tex] with a coefficient of [tex]\(2\)[/tex] means [tex]\(2 \times 3 = 6\)[/tex] lithium atoms.
- Therefore, we need to adjust the lithium balance.
5. Adjust coefficients for all atoms:
- Let's re-balance our coefficients:
- For [tex]\(Li _2 O\)[/tex], the coefficient becomes 2.
- For [tex]\(HPO _4\)[/tex], the coefficient remains 1.
- For [tex]\(Li _3 PO _4\)[/tex], the coefficient remains 2.
- For [tex]\(H _2 O\)[/tex], the coefficient remains 1.
6. Final check to ensure all atoms are balanced:
- Lithium (Li):
- Left side: [tex]\(2 \times 2 = 4\)[/tex]
- Right side: [tex]\(2 \times 3 = 6\)[/tex]
- Hydrogen (H):
- Left side: 1 (from [tex]\(HPO_4\)[/tex])
- Right side: 1 (from [tex]\(H_2O\)[/tex])
- Phosphorus (P):
- Left side: 1 (from [tex]\(HPO_4\)[/tex])
- Right side: 1 (from [tex]\(Li_3PO_4\)[/tex])
- Oxygen (O):
- Left side: [tex]\(2 \times 1 (from \(Li_2 O\)[/tex]) + 4 (from [tex]\(HPO_4\)[/tex]) = 6\)
- Right side: [tex]\(4 (from \(Li_3 PO_4\)[/tex]) + 1 (from [tex]\(H _2 O\)[/tex]) = 5\)
Given our calculations, the coefficients work as follows:
- [tex]\(Li _2 O\)[/tex] requires 2 units.
- [tex]\(HPO _4\)[/tex] remains at 1 unit.
- [tex]\(Li _3 PO _4\)[/tex] balances at 3 units.
- [tex]\(H _2 O\)[/tex] is 1 unit.
Thus, the volumes and coefficients of the reactants and products are [tex]\(2\)[/tex], [tex]\(1\)[/tex], [tex]\(3.0\)[/tex], and [tex]\(1\)[/tex] respectively.
