# Question #0ee47

Jun 29, 2015

The net ionic equation is

$C {l}_{\left(a q\right)}^{-} + A {g}_{\left(a q\right)}^{+} \to A g C {l}_{\left(s\right)}$

#### Explanation:

Start by writing the balanced chemical equation for this double replacement reaction

$M g C {l}_{2 \left(a q\right)} + 2 A g N {O}_{3 \left(a q\right)} \to 2 A g C {l}_{\left(s\right)} + M g {\left(N {O}_{3}\right)}_{2 \left(a q\right)}$

The reaction between magnesium chloride and silver nitrate takes place in aqueous solution, which means that you have

$M g C {l}_{2 \left(a q\right)} \to M {g}_{\left(a q\right)}^{2 +} + 2 C {l}^{-}$

and

$A g N {O}_{3 \left(a q\right)} \to A {g}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-}$

The two products of this double replacement reaction are silver chloride, $A g C l$, which is an insoluble compound, and magnesium nitrate, $M g {\left(N {O}_{3}\right)}_{2}$, which is a soluble compound, i.e. it will exist as ions in solution.

Therefore, the complete ionic equation will be

$M {g}_{\left(a q\right)}^{2 +} + 2 C {l}^{-} + 2 A {g}_{\left(a q\right)}^{+} + 2 N {O}_{3 \left(a q\right)}^{-} \to 2 A g C {l}_{\left(s\right)} \downarrow + M {g}_{\left(a q\right)}^{2 +} + 2 N {O}_{3 \left(a q\right)}$

To get the ney ionic equation, cancel the spectator ions, i.e.ions that are present on both sides of the equation.

$\cancel{M {g}_{\left(a q\right)}^{2 +}} + 2 C {l}^{-} + 2 A {g}_{\left(a q\right)}^{+} + \cancel{2 N {O}_{3 \left(a q\right)}^{-}} \to 2 A g C {l}_{\left(s\right)} \downarrow + \cancel{M {g}_{\left(a q\right)}^{2 +}} + \cancel{2 N {O}_{3 \left(a q\right)}}$

$\overline{2 C {l}^{-} + 2 A {g}_{\left(a q\right)}^{+} \to 2 A g C {l}_{\left(s\right)} \downarrow}$

Simplify the equation by 2 to get

$C {l}^{-} + A {g}_{\left(a q\right)}^{+} \to A g C {l}_{\left(s\right)} \downarrow$