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Zn + FeCl2 ------> ZnCl2 + Fe (for Iron(II) ) or 3Zn + 2FeCl3 ------> 3ZnCl2 + 2Fe (for Iron(III) )
K2CO3 + BaCl2 = 2 KCI + BaCO3(s)
When silver nitrate reacts with hydrochloride a white precipitate of Silver Chloride is formed.
The mixture of silver chloride with hydrochloric acid produces the complex ion [AgCl2] with a charge of -1. This is what will precipitate from the reaction.
Forms a yellow precipitate
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Zn + FeCl2 ------> ZnCl2 + Fe (for Iron(II) ) or 3Zn + 2FeCl3 ------> 3ZnCl2 + 2Fe (for Iron(III) )
K2CO3 + BaCl2 = 2 KCI + BaCO3(s)
A white precipitate of Silver Chloride is formed
When silver nitrate reacts with hydrochloride a white precipitate of Silver Chloride is formed.
The mixture of silver chloride with hydrochloric acid produces the complex ion [AgCl2] with a charge of -1. This is what will precipitate from the reaction.
The reaction is: NaCl + AgNO3 = NaNO3 + AgCl(s) Silver chloride is a white precipitate.
Forms a yellow precipitate
yes
A white precipitate of mercury(I) chloride is formed when a small amount of tin chloride SnCl2 is put into a solution of mercury(II) chloride (HgCl2); adding more SnCl2 turns this precipitate black as metallic mercury is formed.
This equation can be written as 3 BaCl2 (aq) + Al2(SO4)3 (aq) => 2 AlCl3 (aq) + 3 BaSO4 (s).
Barium reacts with halogens, (fluorine, chlorine, bromine, and iodine), and oxygen. It also reacts with oxidizing agents, such as potassium chlorate, and acids such as sulfuric acid and nitric acid.