# How do you write the formula (overall) equation, the complete ionic equation, and the net ionic equation for the reaction between aqueous solutions of calcium chloride and silver nitrate?

Nov 30, 2016

Overall Equation:
$C a C {l}_{2} \left(a q\right)$ + $2 A g N {O}_{3} \left(a q\right)$$\rightarrow$$2 A g C l \left(s\right)$ + $C a {\left(N {O}_{3}\right)}_{2} \left(a q\right)$

Complete Ionic Equation:
$C {a}^{2 +} \left(a q\right) + 2 C {l}^{-} \left(a q\right) + 2 A {g}^{+} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right) \rightarrow 2 A g C l \left(s\right) + C {a}^{2 +} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right)$

Net Ionic Equation:
$2 C {l}^{-} \left(a q\right) + 2 A {g}^{+} \left(a q\right) \rightarrow 2 A g C l \left(s\right)$

#### Explanation:

How to find the Overall Equation:

The given compounds are reactants: Calcium Chloride $\left(C a C {l}_{2}\right)$ and Silver Nitrate $\left(A g N {O}_{3}^{-}\right)$.

To find the products, you simply combine the two 'inner' and two 'outer' elements together. In this case, the two 'inner' elements are Silver and Chloride, while the two 'outer' elements are Calcium and Nitrate. Keep in mind that during this process, subscripts are ignored during the conversion.

Since Silver has a charge of +, and Chlorine's charge is -, you do not need to make any subscript changes here. However, since Calcium's charge is 2+, while Nitrate's charge is -, you have to balance the net charge to zero, which means adding a 2 subscript for nitrate.

Then you balance.

To find the states, use a solubility table to determine which products are soluble: Anything with nitrate is soluble (thus making $C a {\left(N {O}_{3}\right)}_{2}$ aqueous), but Cl is insoluble with Ag (thus making $A g C l$ a solid).

How to find the Complete Ionic Equation:

To find the CIE, you have to separate each aqueous compound into its constituent ions. If a compound is solid or liquid, leave it as it is.

As you can see in the answer, charges are expressed in this, and subscripts are written as coefficients to the left of the equation.

How to find the Net Ionic Equation:

This is simplest process out of the three. All you have to do is cross out common ions that are on both sides to the arrow.

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

As you can see, we are left with $2 C {l}^{-} \left(a q\right) + 2 A {g}^{+} \left(a q\right) \rightarrow 2 A g C l \left(s\right)$

Side Note:

Questions like these usually ask for the Spectator Ions, which is simply the crossed out ions we saw when we were finding the Net Ionic Equation, which is $C {a}^{2 +} \left(a q\right) \mathmr{and} 2 N {O}_{3}^{-} \left(a q\right)$ in this case.