balance the following chemical equation (if necessary): na_{3}po_{4}(aq)+nicl_{2}(aq)\to ni_{3}(po_{4})_{2}(s)+nacl(aq)

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# Explanation:
## Step1: Count atoms on both sides
On the left - hand side (LHS):
- Sodium ($Na$): $3$ atoms (from $Na_3PO_4$)
- Phosphorus ($P$): $1$ atom (from $Na_3PO_4$)
- Oxygen ($O$): $4$ atoms (from $Na_3PO_4$)
- Nickel ($Ni$): $1$ atom (from $NiCl_2$)
- Chlorine ($Cl$): $2$ atoms (from $NiCl_2$)

On the right - hand side (RHS):
- Nickel ($Ni$): $1$ atom (from $Ni_3(PO_4)_2$)
- Phosphorus ($P$): $2$ atoms (from $Ni_3(PO_4)_2$)
- Oxygen ($O$): $8$ atoms (from $Ni_3(PO_4)_2$)
- Sodium ($Na$): $1$ atom (from $NaCl$)
- Chlorine ($Cl$): $1$ atom (from $NaCl$)

## Step2: Balance $Ni$ atoms
To balance $Ni$ atoms, put a coefficient of $3$ in front of $NiCl_2$ on the LHS.
The equation becomes $Na_3PO_4+3NiCl_2
ightarrow Ni_3(PO_4)_2 + NaCl$

## Step3: Balance $PO_4$ groups
To balance $PO_4$ groups, put a coefficient of $2$ in front of $Na_3PO_4$ on the LHS.
The equation becomes $2Na_3PO_4 + 3NiCl_2
ightarrow Ni_3(PO_4)_2+NaCl$

## Step4: Balance $Na$ and $Cl$ atoms
Now, for $Na$ atoms: on the LHS, we have $2\times3 = 6$ $Na$ atoms (from $2Na_3PO_4$). On the RHS, to balance $Na$ atoms, put a coefficient of $6$ in front of $NaCl$.
The balanced equation is $2Na_3PO_4+3NiCl_2 = Ni_3(PO_4)_2+6NaCl$

# Answer:
$2Na_3PO_4+3NiCl_2 = Ni_3(PO_4)_2+6NaCl$

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QUESTION IMAGE

Question

Explanation:

Step1: Count atoms on both sides

Step2: Balance \(Ni\) atoms

Step3: Balance \(PO_4\) groups

Step4: Balance \(Na\) and \(Cl\) atoms

Answer: