What occurs when #"(i) silver nitrate reacts with zinc chloride"#; #"(ii) sodium carbonate reacts with barium chloride"#; and #"(iii) sodium carbonate reacts with barium chloride"#?

Question

Mateo Aguilar · Accepted Answer

Well we gots (i):

#Ag^+ + Cl^(-) rarr AgCl(s)darr# And this represents the net ionic equation for.... #2AgNO_3(aq) + ZnCl_2(aq) rarr 2AgCl(s)darr + ZnNO_3(aq)# Silver halides are as soluble as bricks.... And (ii) ...... #Ba^(2+) + CO_3^(2-) rarr BaCO_3(s)darr# ...the which is the net ionic equation for.... #Na_2CO_3(aq) + BaCl_2(aq) rarr 2NaCl(aq) + BaCO_3(s)darr# ...barium carbonate is also very insoluble.... As always, the net ionic equation is the ionic equation that represents macroscopic observable chemical change. How did I know that silver chloride and barium carbonate are insoluble? How else but by experiment. You simply have to learn the solubilities, according to the scheme your teacher supplies.

Wyatt Adkins · Answer

Warning! Long Answer. Here's what I get.

You have, of course, memorized the solubility rules. You know that A. #"AgNO"_3"(aq)" + "ZnCl"_2"(aq)"# 1) Skeleton equation #"AgNO"_3 + "ZnCl"_2 → "AgCl" + "Zn"("NO"_3)_2# 2) With states #"AgNO"_3"(aq)" + "ZnCl"_2"(aq)" → "AgCl(s)" + "Zn"("NO"_3)_2"(aq)"# 3) Balanced equation #"2AgNO"_3"(aq)" + "ZnCl"_2"(aq)" → "2AgCl(s)" + "Zn"("NO"_3)_2"(aq)"# 4) Ionic equation #"2Ag"^"+""(aq)" + "2NO"_3^"-""(aq)" + "Zn"^"2+""(aq)" + "2Cl"^"-""(aq)" → "2AgCl(s)" + "Zn"^"2+""(aq)" + "2NO"_3^"-""(aq)"# 5) Cancel spectator ions #"2Ag"^"+""(aq)" + color(red)(cancel(color(black)("2NO"_3^"-""(aq)"))) + color(red)(cancel(color(black)("Zn"^"2+""(aq)"))) + "2Cl"^"-""(aq)" → "2AgCl(s)" + color(red)(cancel(color(black)("Zn"^"2+""(aq)"))) + color(red)(cancel(color(black)("2NO"_3^"-""(aq)")))# 5) Net ionic equation #color(red)(cancel(color(black)(2)))"Ag"^"+""(aq)" + color(red)(cancel(color(black)(2)))"Cl"^"-""(aq)" → color(red)(cancel(color(black)(2)))"AgCl(s)"# #color(white)(mmmm)color(blue)("Ag"^"+""(aq)" + "Cl"^"-""(aq)" → "AgCl(s)""# B. #"Na"_2"CO"_3"(aq)" + "BaCl"_2"(aq)"# 1) Skeleton equation #"Na"_2"CO"_3 + "BaCl"_2 → "NaCl" + "BaCO"_3# 2) With states #"Na"_2"CO"_3"(aq)" + "BaCl"_2"(aq)" → "NaCl(aq)" + "BaCO"_3"(s)"# 3) Balanced equation #"Na"_2"CO"_3"(aq)" + "BaCl"_2"(aq)" → "2NaCl(aq)" + "BaCO"_3"(s)"# 4) Ionic equation #"2Na"^"+""(aq)" + "CO"_3^"2-""(aq)" + "Ba"^"2+""(aq)" + "2Cl"^"-""(aq)" → "2Na"^"+""(aq)" + "2Cl"^"-""(aq)" + "BaCO"_3"(s)"# 5) Cancel spectator ions #color(red)(cancel(color(black)("2Na"^"+""(aq)"))) + "CO"_3^"2-""(aq)" + "Ba"^"2+""(aq)" + color(red)(cancel(color(black)("2Cl"^"-""(aq)"))) → color(red)(cancel(color(black)("2Na"^"+""(aq)"))) + color(red)(cancel(color(black)("2Cl"^"-""(aq)"))) + "BaCO"_3"(s)"# 5) Net ionic equation #color(blue)("CO"_3^"2-""(aq)" + "Ba"^"2+""(aq)" → "BaCO"_3"(s)")#

Peyton Adams · Answer

undefined (i) When silver nitrate reacts with zinc chloride, silver chloride precipitate and zinc nitrate are formed.

(ii) When sodium carbonate reacts with barium chloride, barium carbonate precipitate and sodium chloride are formed.

(iii) It seems there's a repetition in the question. The correct third reaction could be "sodium sulfate reacts with barium chloride". In this case, barium sulfate precipitate and sodium chloride are formed.