Answer :
Sure, let's work through the process for determining the balanced, net ionic equation for the reaction between aqueous sodium phosphate (Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]) and aqueous iron (III) chloride (FeCl[tex]\(_3\)[/tex]).
1. Balanced Molecular Equation:
First, let's write the balanced molecular equation for the reaction:
[tex]\[ \text{Na}_3\text{PO}_4\text{(aq)} + \text{FeCl}_3\text{(aq)} \rightarrow 3 \text{NaCl (aq)} + \text{FePO}_4\text{ (s)} \][/tex]
Here, sodium phosphate reacts with iron (III) chloride to produce sodium chloride and solid iron (III) phosphate.
2. Complete Ionic Equation:
Next, we break down all the soluble ionic compounds (those in the aqueous state) into their respective ions:
[tex]\[ 3\text{Na}^+ \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} + \text{Fe}^{3+} \text{(aq)} + 3\text{Cl}^- \text{(aq)} \rightarrow 3\text{Na}^+ \text{(aq)} + 3\text{Cl}^- \text{(aq)} + \text{FePO}_4\text{ (s)} \][/tex]
3. Net Ionic Equation:
In the complete ionic equation, some ions appear on both sides of the equation. These are called spectator ions and do not participate in the reaction. To find the net ionic equation, we remove the spectator ions (Na[tex]\(^+\)[/tex] and Cl[tex]\(^-\)[/tex]) from both sides:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]
This balanced net ionic equation shows only the species that participate in the actual chemical change.
So, the balanced, net ionic equation for the reaction is:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]
This matches with the second given option:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]
1. Balanced Molecular Equation:
First, let's write the balanced molecular equation for the reaction:
[tex]\[ \text{Na}_3\text{PO}_4\text{(aq)} + \text{FeCl}_3\text{(aq)} \rightarrow 3 \text{NaCl (aq)} + \text{FePO}_4\text{ (s)} \][/tex]
Here, sodium phosphate reacts with iron (III) chloride to produce sodium chloride and solid iron (III) phosphate.
2. Complete Ionic Equation:
Next, we break down all the soluble ionic compounds (those in the aqueous state) into their respective ions:
[tex]\[ 3\text{Na}^+ \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} + \text{Fe}^{3+} \text{(aq)} + 3\text{Cl}^- \text{(aq)} \rightarrow 3\text{Na}^+ \text{(aq)} + 3\text{Cl}^- \text{(aq)} + \text{FePO}_4\text{ (s)} \][/tex]
3. Net Ionic Equation:
In the complete ionic equation, some ions appear on both sides of the equation. These are called spectator ions and do not participate in the reaction. To find the net ionic equation, we remove the spectator ions (Na[tex]\(^+\)[/tex] and Cl[tex]\(^-\)[/tex]) from both sides:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]
This balanced net ionic equation shows only the species that participate in the actual chemical change.
So, the balanced, net ionic equation for the reaction is:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]
This matches with the second given option:
[tex]\[ \text{Fe}^{3+} \text{(aq)} + \text{PO}_4^{3-} \text{(aq)} \rightarrow \text{FePO}_4\text{ (s)} \][/tex]