# Write the chemical equation and net ionic equation?

## I struggle a lot when it comes to completing chemical equations. Could someone please explain how to write the chemical equation and net ionic equation for $L {i}_{3} P {O}_{4} + M g S {O}_{4} \rightarrow$?

Mar 24, 2016

here's what I got.

#### Explanation:

You're dealing with two soluble ionic compounds, which implies that they dissociate completely in aqueous solution to form cations and anions.

This means that you can write the two reactants as

${\text{Li"_3"PO"_text(4(aq]) -> 3"Li"_text((aq])^(+) + "PO}}_{\textrm{4 \left(a q\right]}}^{3 -}$

${\text{MgSO"_text(4(aq]) -> "Mg"_text((aq])^(2+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

Now, when you mix an aqueous solution of lithium phosphate and an aqueous solution of magnesium phosphate, a double replacement reaction takes place.

The magnesium cations, ${\text{Mg}}^{2 +}$, and the phosphate anions, ${\text{PO}}_{4}^{3 -}$, will form magnesium phosphate, "Mg"_3("PO"_4)_2, an insoluble solid that precipitates out of solution.

The other product of the reaction will be aqueous lithium sulfate, ${\text{Li"_2"SO}}_{4}$.

The balanced chemical equation for this reaction looks like this

$\textcolor{b l u e}{2} {\text{Li"_3"PO"_text(4(aq]) + color(red)(3)"MgSO"_text(4(aq]) -> "Mg"_3("PO"_4)_text(2(s]) darr + 3"Li"_2"SO}}_{\textrm{4 \left(a q\right]}}$

You'll get a better idea of what is going on here by trying to balance the unbalanced complete ionic equation,which looks like this

$3 {\text{Li"_text((aq])^(+) + "PO"_text(4(aq])^(3-) + "Mg"_text((aq])^(2+) + "SO"_text(4(aq])^(2-) -> "Mg"_3("PO"_4)_text(2(s]) darr + 2"Li"_text((aq])^(2+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

Focus on balancing the ions that are a part of the precipitate first. Notice that one mole of magnesium phosphate contains three moles of magnesium cations and two moles of phosphate anions.

Look at the ions present on the reactants' side. You need to multiply the magnesium cations by $\textcolor{red}{3}$ and the phosphate anions by $\textcolor{b l u e}{2}$.

But remember, these ions are delivered to the solution by an ionic compound, which means that when you multiply an ion, you're actually multiplying an ionic compound.

You will have

$\textcolor{b l u e}{2} \times \overbrace{\left(3 \text{Li"_text((aq])^(+) + "PO"_text(4(aq])^(3-)))^(color(blue)("lithium phosphate")) + color(red)(3) xx overbrace(("Mg"_text((aq])^(2+) + "SO"_text(4(aq])^(2-)))^(color(red)("magnesium sulfate}\right)} \to$ ${\text{Mg"_3("PO"_4)_text(2(s]) darr + 2"Li"_text((aq])^(2+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

Now focus on balancing the lithium cations and the sulfate anions. Notice that you start with $3$ lithium cations on the reactants' side, but that you multiply them by $\textcolor{b l u e}{2}$ to get a total of $6$.

This means that you're going to have to multiply the lithium cations present on the product's side by $3$ to balance them out.

Doing this will also balance out the sulfate anions, since you'll end up with $3$ sulfate anions on both sides of the equation.

$\textcolor{b l u e}{2} \times \overbrace{\left(3 \text{Li"_text((aq])^(+) + "PO"_text(4(aq])^(3-)))^(color(blue)("lithium phosphate")) + color(red)(3) xx overbrace(("Mg"_text((aq])^(2+) + "SO"_text(4(aq])^(2-)))^(color(red)("magnesium sulfate}\right)} \to$ "Mg"_3("PO"_4)_text(2(s]) darr + 3 xx (2"Li"_text((aq])^(2+) + "SO"_text(4(aq])^(2-))

This is equivalent to

$6 {\text{Li"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) + 3"Mg"_text((aq])^(2+) + 3"SO"_text(4(aq])^(2-) -> "Mg"_3("PO"_4)_text(2(s]) darr + 6"Li"_text((aq])^(2+) + 3"SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

Now, to get the net ionic equation, remove spectator ions, i.e. the ions that are present on both sides of the equation

$\textcolor{red}{\cancel{\textcolor{b l a c k}{6 {\text{Li"_text((aq])^(+)))) + 2"PO"_text(4(aq])^(3-) + 3"Mg"_text((aq])^(2+) + color(red)(cancel(color(black)(3"SO"_text(4(aq])^(2-)))) -> "Mg"_3("PO"_4)_text(2(s]) darr + color(red)(cancel(color(black)(6"Li"_text((aq])^(2+)))) + color(red)(cancel(color(black)(3"SO}}_{\textrm{4 \left(a q\right]}}^{2 -}}}}$

You will thus have

color(green)(|bar(ul(color(white)(a/a)color(black)(3"Mg"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) -> "Mg"_3("PO"_4)_text(2(s]) darr)color(white)(a/a)|)))