Question

Question #ae9c0

Answer 1

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

Explanation:

You're dealing with a double replacement reaction in which two soluble ionic compounds 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 compounds 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)|)))`