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Report (any type)/Brief report
22/09/2021
##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)|)))##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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