Sodium Phosphate + Zinc Nitrate

You lot're dealing with a double replacement reaction in which two soluble ionic compounds react to class 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#, volition 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(four(aq])^(3-)#

The overall balanced chemic equation for this reaction looks like this

#color(red)(3)"Zn"("NO"_3)_text(2(aq]) + color(blueish)(2)"Li"_3"PO"_text(four(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 - practise not forget to use the corresponding stoichiometric coefficients!

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

This will exist equivalent to

#iii"Zn"_text((aq])^(2+) + 6"NO"_text(three(aq])^(-) + half-dozen"Li"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) -> "Zn"_3("PO"_4)_text(2(due south]) darr + half dozen"Li"_text((aq])^(+) + half-dozen"NO"_text(three(aq])^(-)#

To go the net ionic equation, eliminate spectator ions, which are ions that tin be institute on both sides of the equation

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

This will get y'all

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

Sodium Phosphate + Zinc Nitrate,

Source: https://socratic.org/questions/56e6e37c11ef6b71170ae9c0

Posted by: laraobeft1996.blogspot.com

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