Question

balance the reaction, do not leave any fractions, then answer the questions about the reaction. well start with a hard one.
ba(c₂h₃o₂)₂ + na₃po₄ → nac₂h₃o₂ + ba₃(po₄)₂(s)
according to the guidelines for balancing in chapter 12, we should start with a polyatomic ion. it doesnt matter which one. if we start with acetate, c₂h₃o₂⁻¹, there are 2 on the left and one on the right so we put a 2 in front of the nac₂h₃o₂.
ba(c₂h₃o₂)₂ + na₃po₄ → 2 nac₂h₃o₂ + ba₃(po₄)₂(s)
then, we play off of the molecule that we just worked on, and try to balance the na. but, we run into a problem - too many na on the left. so, we start with a different polyatomic ion the phosphate, po₄⁻³. there are 2 on the right and 1 on the left so we put a 1 in front of the na₃po₄.
ba(c₂h₃o₂)₂ + 2 na₃po₄ → nac₂h₃o₂ + ba₃(po₄)₂(s)
then, we play off of the molecule that we just worked on, and try to balance the na. there are 6 on the left and 1 on the right so we put a 3 in front of the nac₂h₃o₂. now po₄⁻³ and na⁺¹ are balanced.
ba(c₂h₃o₂)₂ + 4 na₃po₄ → 5 nac₂h₃o₂ + ba₃(po₄)₂(s)
then, we play off of the molecule that we just worked on, and try to balance the c₂h₃o₂⁻¹. there are 6 on the right and 2 on the left so we put a 6 in front of

Explanation:

Step1: Balance phosphate ions

Since there are 2 phosphate ($\text{PO}_4^{3 - }$) ions on the right - hand side (in $\text{Ba}_3(\text{PO}_4)_2$) and 1 on the left - hand side (in $\text{Na}_3\text{PO}_4$), we put a 2 in front of $\text{Na}_3\text{PO}_4$.
The equation becomes: $\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2+2\text{Na}_3\text{PO}_4
ightarrow\text{NaC}_2\text{H}_3\text{O}_2+\text{Ba}_3(\text{PO}_4)_2(s)$

Step2: Balance sodium ions

After putting 2 in front of $\text{Na}_3\text{PO}_4$, the number of $\text{Na}$ atoms on the left - hand side is $2\times3 = 6$. On the right - hand side, there is 1 $\text{Na}$ atom in $\text{NaC}_2\text{H}_3\text{O}_2$. So we put a 6 in front of $\text{NaC}_2\text{H}_3\text{O}_2$.
The equation becomes: $\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2 + 2\text{Na}_3\text{PO}_4
ightarrow6\text{NaC}_2\text{H}_3\text{O}_2+\text{Ba}_3(\text{PO}_4)_2(s)$

Step3: Balance barium ions

On the right - hand side, there are 3 $\text{Ba}$ atoms in $\text{Ba}_3(\text{PO}_4)_2$. On the left - hand side, there is 1 $\text{Ba}$ atom in $\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2$. So we put a 3 in front of $\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2$.
The balanced equation is: $3\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2+2\text{Na}_3\text{PO}_4
ightarrow6\text{NaC}_2\text{H}_3\text{O}_2+\text{Ba}_3(\text{PO}_4)_2(s)$

Answer:

The balanced equation is $3\text{Ba}(\text{C}_2\text{H}_3\text{O}_2)_2 + 2\text{Na}_3\text{PO}_4=6\text{NaC}_2\text{H}_3\text{O}_2+\text{Ba}_3(\text{PO}_4)_2(s)$