What is the net ionic equation that occurs when zinc nitrate and lithium phosphate are mixed together in water?

##3"Zn"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) -> "Zn"_3("PO"_4)_text(2(s]) darr##

You're dealing with a in which two soluble react to form an insoluble solid that precipitates out of solution.

Zinc nitrate, ##"Zn"("NO"_3)_2##, will dissociate completely in aqueous solution to form zinc cations, ##"Zn"^(2+)##, and nitrate anions, ##"NO"_3^(-)##

##"Zn"("NO"_3)_text(2(aq]) -> "Zn"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)##

Lithium phosphate, ##"Li"_3"PO"_4##, will dissociate completely in aqueous solution to form lithium cations, ##"Li"^(+)##, and phosphate anions, ##"PO"_4^(3-)##

##"Li"_3"PO"_text(4(aq]) -> 3"Li"_text((aq])^(+) + "PO"_text(4(aq])^(3-)##

The overall balanced chemical equation for this reaction looks like this

##color(red)(3)"Zn"("NO"_3)_text(2(aq]) + color(blue)(2)"Li"_3"PO"_text(4(aq]) -> "Zn"_3("PO"_4)_text(2(s]) darr + 6"LiNO"_text(3(aq])##

The reaction produces zinc phosphate, ##"Zn"_3"PO"_4##, a white insoluble solid that precipitates out of solution.

Notice that the reaction also produces aqueous lithium nitrate, ##"LiNO"_3##, another soluble ionic compound that exists as ions in solution.

To get the complete ionic equation, split the known soluble ionic into ions - do not forget to use the corresponding stoichiometric coefficients!

##color(red)(3) xx overbrace(["Zn"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)])^(color(purple)("zinc nitrate")) + color(blue)(2) xx overbrace([3"Li"_text((aq])^(+) + "PO"_text(4(aq])^(3-)])^(color(brown)("lithium phosphate")) -> "Zn"_3("PO"_4)_text(2(s])## ##color(white)(a/a)darr## ##+ 6 xx overbrace(["Li"_text((aq])^(+) + "NO"_text(3(aq])^(-)])^color(black)("lithium nitrate")##

This will be equivalent to

##3"Zn"_text((aq])^(2+) + 6"NO"_text(3(aq])^(-) + 6"Li"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) -> "Zn"_3("PO"_4)_text(2(s]) darr + 6"Li"_text((aq])^(+) + 6"NO"_text(3(aq])^(-)##

To get the net ionic equation, eliminate spectator ions, which are ions that can be found on both sides of the equation

##3"Zn"_text((aq])^(2+) + color(red)(cancel(color(black)(6"NO"_text(3(aq])^(-)))) + color(red)(cancel(color(black)(6"Li"_text((aq])^(+)))) + 2"PO"_text(4(aq])^(3-) -> "Zn"_3("PO"_4)_text(2(s]) darr + color(red)(cancel(color(black)(6"Li"_text((aq])^(+)))) + color(red)(cancel(color(black)(6"NO"_text(3(aq])^(-))))##

This will get you

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