Here's how to balance the equation for the reaction between sodium phosphate (Na₃PO₄) and barium acetate (Ba(CH₃COO)₂):

1. Write the unbalanced equation:

Na₃PO₄ (aq) + Ba(CH₃COO)₂ (aq) → Ba₃(PO₄)₂ (s) + NaCH₃COO (aq)

2. Balance the barium (Ba) atoms:

2 Na₃PO₄ (aq) + 3 Ba(CH₃COO)₂ (aq) → Ba₃(PO₄)₂ (s) + NaCH₃COO (aq)

3. Balance the phosphate (PO₄) groups:

2 Na₃PO₄ (aq) + 3 Ba(CH₃COO)₂ (aq) → Ba₃(PO₄)₂ (s) + 6 NaCH₃COO (aq)

4. Balance the sodium (Na) atoms:

2 Na₃PO₄ (aq) + 3 Ba(CH₃COO)₂ (aq) → Ba₃(PO₄)₂ (s) + 6 NaCH₃COO (aq)

The balanced equation is:

2 Na₃PO₄ (aq) + 3 Ba(CH₃COO)₂ (aq) → Ba₃(PO₄)₂ (s) + 6 NaCH₃COO (aq)

Explanation:

* The reaction produces barium phosphate (Ba₃(PO₄)₂) which is an insoluble solid, indicated by "(s)".

* The other product, sodium acetate (NaCH₃COO), is soluble and remains in solution, indicated by "(aq)".

* The equation is balanced because there are the same number of atoms of each element on both sides of the equation.